View Full Version : [SOLVED] gravitational force -airplane motion and earths rotation
surendra
Aug20-04, 04:47 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nHi,\nI was having a hot debate with my friend in which we were arguing about\nairplane motion.\nWhat my friend was saying was that when the airplanes they fly, because\nof the earth\'s rotation from west to east, it takes them(planes) shorter\ntime to go from east to west compared to from west to east.\nIn other words, a plane flying from London to New York will take less\ntime compared to a plane flying from New York to London. I was\ncontending that they will take equal time.\n\nAre there any experts around who can answer who is correct and with reason?\n\n\nI was arguing with them that when the earth rotates the plane also\nrotates with the earth and they have almost similar speed of rotation.\nI was arguing that the revolution of moon around the earth(or earth\naround the sun, etc., ) is like rotating with a hammer-throw (an olympic\nsport in which there is a heavy ball tied at the end of a chain, which\nthe atheletes has to throw the furthest). When we are still the\nhammer-throw is close to us and still. But when we start rotating the\nhammer throw revolves around us, and it starts going farther away from\nour body. It is the speed of our rotation which decides how far it is\naway from our body. So, in case of heavenly bodies(earth and moon) it is\nthe gravitational force. I was further contending that if the earth\nslowly stops rotating, then the moon will collapse into the earth. For,\nexample if we slow down our speed of rotation the hammer-throw starts\ncoming closer and closer to the body.\nMy explanation for the moon not doing revolution with the same speed as\nearth\'s rotation (or even planets not taking the same time to revolve\naround the sun) was that as we go away from earth (or sun) the\ngravitational forces decreases and so the angular velocity required by\nmoon to counteract also decreases and hence it takes longer time to do\none complete revolution as compared to earth\'s rotation.\n\nSo, my net explanation was that indeed plane will not rotate with the\nsame speed as earth because of decrease in gravitationale attraction but\nnot because it can stay still and not revolve around the earth.\n\nI will like to hear the opinion of advice of experts on this.\nAny help in settling the argument will be appreciated.\n\nThank you,\nSurendra\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi,
I was having a hot debate with my friend in which we were arguing about
airplane motion.
What my friend was saying was that when the airplanes they fly, because
of the earth's rotation from west to east, it takes them(planes) shorter
time to go from east to west compared to from west to east.
In other words, a plane flying from London to New York will take less
time compared to a plane flying from New York to London. I was
contending that they will take equal time.
Are there any experts around who can answer who is correct and with reason?
I was arguing with them that when the earth rotates the plane also
rotates with the earth and they have almost similar speed of rotation.
I was arguing that the revolution of moon around the earth(or earth
around the sun, etc., ) is like rotating with a hammer-throw (an olympic
sport in which there is a heavy ball tied at the end of a chain, which
the atheletes has to throw the furthest). When we are still the
hammer-throw is close to us and still. But when we start rotating the
hammer throw revolves around us, and it starts going farther away from
our body. It is the speed of our rotation which decides how far it is
away from our body. So, in case of heavenly bodies(earth and moon) it is
the gravitational force. I was further contending that if the earth
slowly stops rotating, then the moon will collapse into the earth. For,
example if we slow down our speed of rotation the hammer-throw starts
coming closer and closer to the body.
My explanation for the moon not doing revolution with the same speed as
earth's rotation (or even planets not taking the same time to revolve
around the sun) was that as we go away from earth (or sun) the
gravitational forces decreases and so the angular velocity required by
moon to counteract also decreases and hence it takes longer time to do
one complete revolution as compared to earth's rotation.
So, my net explanation was that indeed plane will not rotate with the
same speed as earth because of decrease in gravitationale attraction but
not because it can stay still and not revolve around the earth.
I will like to hear the opinion of advice of experts on this.
Any help in settling the argument will be appreciated.
Thank you,
Surendra
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nOn 20 Aug 2004 05:47:01 -0400, surendra <efuzzyone@netscape.net>\nwrote:\n\n>I was having a hot debate with my friend in which we were arguing about\n>airplane motion.\n>What my friend was saying was that when the airplanes they fly, because\n>of the earth\'s rotation from west to east, it takes them(planes) shorter\n>time to go from east to west compared to from west to east.\n>In other words, a plane flying from London to New York will take less\n>time compared to a plane flying from New York to London. I was\n>contending that they will take equal time.\n>\n>Are there any experts around who can answer who is correct and with reason?\n\nI wouldn\'t call myself an expert, but the atmosphere rotates along\nwith the earth (if it didn\'t it would be windy... a lot). The plane\'s\nspeed is relative to the atmosphere (and thus relative to the earth),\nso both ways take equally long (provided there is no wind).\n\nYou can check your friend\'s idea by taking a look at flight times. For\ninstance, take a look at the flight route between New York and Los\nAngeles. The flight to New York is shorter (which is exactly the\nopposite of what you would expect if your friend were right). The\nreason for the difference in flight times are probably wind currents\n(I\'m guessing).\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 20 Aug 2004 05:47:01 -0400, surendra <efuzzyone@netscape.net>
wrote:
>I was having a hot debate with my friend in which we were arguing about
>airplane motion.
>What my friend was saying was that when the airplanes they fly, because
>of the earth's rotation from west to east, it takes them(planes) shorter
>time to go from east to west compared to from west to east.
>In other words, a plane flying from London to New York will take less
>time compared to a plane flying from New York to London. I was
>contending that they will take equal time.
>
>Are there any experts around who can answer who is correct and with reason?
I wouldn't call myself an expert, but the atmosphere rotates along
with the earth (if it didn't it would be windy... a lot). The plane's
speed is relative to the atmosphere (and thus relative to the earth),
so both ways take equally long (provided there is no wind).
You can check your friend's idea by taking a look at flight times. For
instance, take a look at the flight route between New York and Los
Angeles. The flight to New York is shorter (which is exactly the
opposite of what you would expect if your friend were right). The
reason for the difference in flight times are probably wind currents
(I'm guessing).
Androcles
Aug20-04, 08:30 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n"surendra" <efuzzyone@netscape.net> wrote in message\nnews:cg3248\\$5cd\\$1@news.asu.edu...\n|\ n| Hi,\n| I was having a hot debate with my friend in which we were arguing about\n| airplane motion.\n| What my friend was saying was that when the airplanes they fly, because\n| of the earth\'s rotation from west to east, it takes them(planes) shorter\n| time to go from east to west compared to from west to east.\n| In other words, a plane flying from London to New York will take less\n| time compared to a plane flying from New York to London. I was\n| contending that they will take equal time.\n|\n| Are there any experts around who can answer who is correct and with\nreason?\n|\n\n\n\nhttp://www.virgin-atlantic.com/bookflightsandmore/bookflights/searchFlight.do\n\nLondon (LHR)\n20 AUG 14:00 New York (JFK)\n20 AUG 16:40 =\n2:40 + 5 hours (EST - GMT =5:00) = 7:40 outbound\n\n\nNew York (EWR)\n21 AUG 08:20 London (LHR)\n21 AUG 20:00\n= 11:40 - 5 (GMT - EST = -5:00) = 6:40 inbound.\n\nThe reason for the 1:00 hour difference is the high wind blowing the plane\nalong (or opposing it), and nothing else. If we ignore the jet stream, the\ntimes will be the same. A real flight will be subject to various delays, of\ncourse, and you may get in early or late.\n\n\n\n\n| I was arguing with them that when the earth rotates the plane also\n| rotates with the earth and they have almost similar speed of rotation.\n\nWell, all motion is relative. New York is about 40 degrees North of the\nequator and London 52 degrees latitiude, So we\'ll take 45 as the mean.\nThe equator, relative to the Sun, is about 25,000 miles in diameter and\ncompletes onw revolution in one day, so it os moving at 25,000/24 miles per\nhour, or a bit over 1000 mph. The pole has no distance to go, and at 45\ndegrees we use the cosine, so cos(45) = 0.707, multiply that by 1000, so New\nYork and London are moving at about 700 mph relative to the Sun.\nUsing round figures, they are about 2800 miles apart, the plane takes 7 hour\ns, so it moves at 400 mph ground speed (the air is thinner at 30,000 ft, so\nthe planes \'speedometer\' (actually called the \'air speed indicator\') only\nrecords 250 mph air speed. From East to West, then, the plane moves at 700 +\n400 = 1100 mph relative to the sun, travels 2800 miles at 400 mph = 7 hours.\nComing back, West to East, it travels 700-400 = 300 mph relative to the sun,\nbut still takes 2800 miles at 400 mph (relative to the ground) = 7 hours to\nreturn. So you book a day flight during the outbound trip, you leave at\n2:00 in the afternoon and arrive in New York at 4:40 the same afternoon,\nlocal time, and coming back you only travel at 300 mph, relative to the sun,\ntaking off at 8:20 am and arriving at 10:00 in the evening. Jet lag is a\nkiller, you gain a day going and lose a day coming back.\n\n\n\n\n| I was arguing that the revolution of moon around the earth(or earth\n| around the sun, etc., ) is like rotating with a hammer-throw (an olympic\n| sport in which there is a heavy ball tied at the end of a chain, which\n| the atheletes has to throw the furthest). When we are still the\n| hammer-throw is close to us and still. But when we start rotating the\n| hammer throw revolves around us, and it starts going farther away from\n| our body. It is the speed of our rotation which decides how far it is\n| away from our body. So, in case of heavenly bodies(earth and moon) it is\n| the gravitational force. I was further contending that if the earth\n| slowly stops rotating, then the moon will collapse into the earth.\n\nNo, sorry, you are way off with that one. There is no chain tying the moon\nthe earth. The moon takes 28 days to orbit the earth, and the Earth rotates\non its own axis in one day. The nearest you have to your idea is a\ncommunications satellite that people point dish antennae at to get TV\npictures and send phone calls through, and they are at 22,300 miles high.\nThey take 24 hours to go around the earth once, and because the earth turns\nonce every 24 hours, the dishes always point to the satellite. If the earth\nslowed down, we would have to move the satellites further away, but that\nwould be deliberate on our part. If the earth stopped dead, with the Sun\noverhead at one place, it would not affect the satellites or the moon.\nAndrocles.\n\nFor,\n| example if we slow down our speed of rotation the hammer-throw starts\n| coming closer and closer to the body.\n| My explanation for the moon not doing revolution with the same speed as\n| earth\'s rotation (or even planets not taking the same time to revolve\n| around the sun) was that as we go away from earth (or sun) the\n| gravitational forces decreases and so the angular velocity required by\n| moon to counteract also decreases and hence it takes longer time to do\n| one complete revolution as compared to earth\'s rotation.\n|\n| So, my net explanation was that indeed plane will not rotate with the\n| same speed as earth because of decrease in gravitationale attraction but\n| not because it can stay still and not revolve around the earth.\n|\n| I will like to hear the opinion of advice of experts on this.\n| Any help in settling the argument will be appreciated.\n|\n| Thank you,\n| Surendra\n|\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"surendra" <efuzzyone@netscape.net> wrote in message
news:cg3248$5cd$1@news.asu.edu...
|
| Hi,
| I was having a hot debate with my friend in which we were arguing about
| airplane motion.
| What my friend was saying was that when the airplanes they fly, because
| of the earth's rotation from west to east, it takes them(planes) shorter
| time to go from east to west compared to from west to east.
| In other words, a plane flying from London to New York will take less
| time compared to a plane flying from New York to London. I was
| contending that they will take equal time.
|
| Are there any experts around who can answer who is correct and with
reason?
|
http://www.virgin-atlantic.com/bookflightsandmore/bookflights/searchFlight.do
London (LHR)
20 AUG 14:00 New York (JFK)
20 AUG 16:40 =
2:40 + 5 hours (EST - GMT =5:00) = 7:40 outbound
New York (EWR)
21 AUG 08:20 London (LHR)
21 AUG 20:00
= 11:40 - 5 (GMT - EST = -5:00) = 6:40 inbound.
The reason for the 1:00 hour difference is the high wind blowing the plane
along (or opposing it), and nothing else. If we ignore the jet stream, the
times will be the same. A real flight will be subject to various delays, of
course, and you may get in early or late.
| I was arguing with them that when the earth rotates the plane also
| rotates with the earth and they have almost similar speed of rotation.
Well, all motion is relative. New York is about 40 degrees North of the
equator and London 52 degrees latitiude, So we'll take 45 as the mean.
The equator, relative to the Sun, is about 25,000 miles in diameter and
completes onw revolution in one day, so it os moving at 25,000/24 miles per
hour, or a bit over 1000 mph. The pole has no distance to go, and at 45
degrees we use the cosine, so cos(45) = .707, multiply that by 1000, so New
York and London are moving at about 700 mph relative to the Sun.
Using round figures, they are about 2800 miles apart, the plane takes 7 hour
s, so it moves at 400 mph ground speed (the air is thinner at 30,000 ft, so
the planes 'speedometer' (actually called the 'air speed indicator') only
records 250 mph air speed. From East to West, then, the plane moves at 700 +
400 = 1100 mph relative to the sun, travels 2800 miles at 400 mph = 7 hours.
Coming back, West to East, it travels 700-400 = 300 mph relative to the sun,
but still takes 2800 miles at 400 mph (relative to the ground) = 7 hours to
return. So you book a day flight during the outbound trip, you leave at
2:00 in the afternoon and arrive in New York at 4:40 the same afternoon,
local time, and coming back you only travel at 300 mph, relative to the sun,
taking off at 8:20 am and arriving at 10:00 in the evening. Jet lag is a
killer, you gain a day going and lose a day coming back.
| I was arguing that the revolution of moon around the earth(or earth
| around the sun, etc., ) is like rotating with a hammer-throw (an olympic
| sport in which there is a heavy ball tied at the end of a chain, which
| the atheletes has to throw the furthest). When we are still the
| hammer-throw is close to us and still. But when we start rotating the
| hammer throw revolves around us, and it starts going farther away from
| our body. It is the speed of our rotation which decides how far it is
| away from our body. So, in case of heavenly bodies(earth and moon) it is
| the gravitational force. I was further contending that if the earth
| slowly stops rotating, then the moon will collapse into the earth.
No, sorry, you are way off with that one. There is no chain tying the moon
the earth. The moon takes 28 days to orbit the earth, and the Earth rotates
on its own axis in one day. The nearest you have to your idea is a
communications satellite that people point dish antennae at to get TV
pictures and send phone calls through, and they are at 22,300 miles high.
They take 24 hours to go around the earth once, and because the earth turns
once every 24 hours, the dishes always point to the satellite. If the earth
slowed down, we would have to move the satellites further away, but that
would be deliberate on our part. If the earth stopped dead, with the Sun
overhead at one place, it would not affect the satellites or the moon.
Androcles.
For,
| example if we slow down our speed of rotation the hammer-throw starts
| coming closer and closer to the body.
| My explanation for the moon not doing revolution with the same speed as
| earth's rotation (or even planets not taking the same time to revolve
| around the sun) was that as we go away from earth (or sun) the
| gravitational forces decreases and so the angular velocity required by
| moon to counteract also decreases and hence it takes longer time to do
| one complete revolution as compared to earth's rotation.
|
| So, my net explanation was that indeed plane will not rotate with the
| same speed as earth because of decrease in gravitationale attraction but
| not because it can stay still and not revolve around the earth.
|
| I will like to hear the opinion of advice of experts on this.
| Any help in settling the argument will be appreciated.
|
| Thank you,
| Surendra
|
Uncle Al
Aug20-04, 10:45 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nsurendra wrote:\n>\n> Hi,\n> I was having a hot debate with my friend in which we were arguing about\n> airplane motion.\n> What my friend was saying was that when the airplanes they fly, because\n> of the earth\'s rotation from west to east, it takes them(planes) shorter\n> time to go from east to west compared to from west to east.\n> In other words, a plane flying from London to New York will take less\n> time compared to a plane flying from New York to London. I was\n> contending that they will take equal time.\n>\n> Are there any experts around who can answer who is correct and with reason?\n\n[snip]\n\n1)\n<http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html>\nHafele-Keating Experiment\n\n2) Motion of the atmosphere vs. the ground and vs. the fixed\nstars. (Ignore the jet streamfor fundamental analysis.)\n\n3) Coriolus acceleration. A plane gains or loses measurable\nlift from the vertical component. It shows up in fuel\nconsumption statistics.\n\n4) <http://www.mathpages.com/rr/s2-07/2-07.htm>\nSagnac effect.\n\n5) <http://scienceworld.wolfram.com/\nphysics/Lense-ThirringEffect.html>\nGravitomagnetic effects like Lense-Thirring frame dragging.\n\nOrbiting E-W vs. W-E around a spinning planet in vacuum is not\nthe same trip. Add atmosphere and whatnot and the differences\nmount. The question is, "how big is the total difference?" It\nis very small.\n\n--\nUncle Al\nhttp://www.mazepath.com/uncleal/\n(Toxic URL! Unsafe for children and most mammals)\nhttp://www.mazepath.com/uncleal/qz.pdf\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>surendra wrote:
>
> Hi,
> I was having a hot debate with my friend in which we were arguing about
> airplane motion.
> What my friend was saying was that when the airplanes they fly, because
> of the earth's rotation from west to east, it takes them(planes) shorter
> time to go from east to west compared to from west to east.
> In other words, a plane flying from London to New York will take less
> time compared to a plane flying from New York to London. I was
> contending that they will take equal time.
>
> Are there any experts around who can answer who is correct and with reason?
[snip]
1)
<http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html>
Hafele-Keating Experiment
2) Motion of the atmosphere vs. the ground and vs. the fixed
stars. (Ignore the jet streamfor fundamental analysis.)
3) Coriolus acceleration. A plane gains or loses measurable
lift from the vertical component. It shows up in fuel
consumption statistics.
4) <http://www.mathpages.com/rr/s2-07/2-07.htm>
Sagnac effect.
5) <http://scienceworld.wolfram.com/
physics/Lense-ThirringEffect.html>
Gravitomagnetic effects like Lense-Thirring frame dragging.
Orbiting E-W vs. W-E around a spinning planet in vacuum is not
the same trip. Add atmosphere and whatnot and the differences
mount. The question is, "how big is the total difference?" It
is very small.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
Nick Maclaren
Aug20-04, 10:45 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nIn article <OTmVc.3584\\$wd7.43453605@news-text.cableinet.net>,\n"Androcles" <androc1es@nospamblueyonder.co.uk> writes:\n|>\n|> The reason for the 1:00 hour difference is the high wind blowing the plane\n|> along (or opposing it), and nothing else. If we ignore the jet stream, the\n|> times will be the same. A real flight will be subject to various delays, of\n|> course, and you may get in early or late.\n\nA slight nitpick: the precise value of the difference is chosen\nto fit landing schedules, and the target speed of the aircraft\nis set accordingly. You cannot therefore use the magnitude of\nthe difference to calculate the expected speed of the jet stream.\nThe jet stream is, however, the only major reason for the difference.\n\n\nRegards,\nNick Maclaren.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <OTmVc.3584$wd7.43453605@news-text.cableinet.net>,
"Androcles" <androc1es@nospamblueyonder.co.uk> writes:
|>
|> The reason for the 1:00 hour difference is the high wind blowing the plane
|> along (or opposing it), and nothing else. If we ignore the jet stream, the
|> times will be the same. A real flight will be subject to various delays, of
|> course, and you may get in early or late.
A slight nitpick: the precise value of the difference is chosen
to fit landing schedules, and the target speed of the aircraft
is set accordingly. You cannot therefore use the magnitude of
the difference to calculate the expected speed of the jet stream.
The jet stream is, however, the only major reason for the difference.
Regards,
Nick Maclaren.
Randy Poe
Aug20-04, 02:18 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nsurendra <efuzzyone@netscape.net> wrote in message news:<cg3248\\$5cd\\$1@news.asu.edu>...\n> Hi,\n> I was having a hot debate with my friend in which we were arguing about\n> airplane motion.\n> What my friend was saying was that when the airplanes they fly, because\n> of the earth\'s rotation from west to east, it takes them(planes) shorter\n> time to go from east to west compared to from west to east.\n> In other words, a plane flying from London to New York will take less\n> time compared to a plane flying from New York to London. I was\n> contending that they will take equal time.\n>\n> Are there any experts around who can answer who is correct and with reason?\n\nLet\'s look at two other cases first. Let\'s assume you are\nwalking from East to West, and from West to East. Do you agree\nthat these two trips should take the same amount of time?\nThe reason they do is that you are walking relative to\nthe ground, which is rotating with the earth. Your ground\nvelocity is the same in both directions.\n\nNow consider something in space moving at a speed of 1000 km/hr\npast the earth. This motion is relative to the center of the\nearth, and is independent of the rotation. So as the earth\nrotates under it, it would indeed take longer to fly from\nNew York to London rather than London to New York.\n\nNow, what about an airplane? Well it is not flying in space\nor on the ground, but through the air. If the engines are\ncapable of 800 km/hr, then that speed is relative to the\nair. The circumference of the earth is 25000 miles (40000\nkm) at the equator, and it goes around in 24 hours, so\nthe ground at the equator is moving along at 1000 mph\n(1600 km/hr). If the air weren\'t moving at approximately\nthe same speed as the ground, you\'d feel a 1000 mph wind\nall day. Obviously you don\'t, and the winds you do feel are\nonly a small fraction of those speeds.\n\nBecause of the prevailing winds, it actually takes an\naircraft longer to go from London to New York than from\nNew York to London. You can do an online search for\nflight times, and you\'ll see it\'s about an hour longer\nfor a 747 to cross west to east than east to west.\n\n> I was arguing with them that when the earth rotates the plane also\n> rotates with the earth and they have almost similar speed of rotation.\n\nThat\'s correct. The atmosphere is mostly carried around at\nthe same speed as the ground. There are slight differences,\nand this is what we feel as wind.\n\n> I was further contending that if the earth\n> slowly stops rotating, then the moon will collapse into the earth. For,\n> example if we slow down our speed of rotation the hammer-throw starts\n> coming closer and closer to the body.\n\nThese two situations are not comparable. The earth\'s rotation\ndoes not have an effect on the moon\'s orbit, except a subtle\none called "tidal locking". Some day, millions of years from\nnow, the earth will rotate with the moon so that the moon\nwill appear to be geosynchronous. This has already happened\nto the moon, so it always presents the same face to us.\n\nBut that won\'t affect the gravitational force. Unlike the\nhammer thrower, it is not the rotation of the earth that\nkeeps the moon moving. There\'s no chain. The moon experiences\nno friction and needs no pull from a chain to keep it\nmoving forward. The earth provides a pull that keeps it\nmoving in an orbit.\n\n> My explanation for the moon not doing revolution with the same speed as\n> earth\'s rotation (or even planets not taking the same time to revolve\n> around the sun) was that as we go away from earth (or sun) the\n> gravitational forces decreases and so the angular velocity required by\n> moon to counteract also decreases and hence it takes longer time to do\n> one complete revolution as compared to earth\'s rotation.\n\nThis analysis is correct. Low earth satellites orbit in\nabout 90 minutes. Geosynchronous satellites, tens of\nthousands of km higher, orbit in 24 hours. The moon, even\nfarther out, orbits in about 28 days. The period depends\non the distance.\n\n> So, my net explanation was that indeed plane will not rotate with the\n> same speed as earth because of decrease in gravitationale attraction but\n> not because it can stay still and not revolve around the earth.\n\nThe moving air is what makes it revolve around the earth, at\nclose to the same speed as the ground. Air has a lot of\nfriction. Even satellites experience some air friction for\nthe first few hundred km of altitude, enough to drag them\ndown.\n\n- Randy\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>surendra <efuzzyone@netscape.net> wrote in message news:<cg3248$5cd$1@news.asu.edu>...
> Hi,
> I was having a hot debate with my friend in which we were arguing about
> airplane motion.
> What my friend was saying was that when the airplanes they fly, because
> of the earth's rotation from west to east, it takes them(planes) shorter
> time to go from east to west compared to from west to east.
> In other words, a plane flying from London to New York will take less
> time compared to a plane flying from New York to London. I was
> contending that they will take equal time.
>
> Are there any experts around who can answer who is correct and with reason?
Let's look at two other cases first. Let's assume you are
walking from East to West, and from West to East. Do you agree
that these two trips should take the same amount of time?
The reason they do is that you are walking relative to
the ground, which is rotating with the earth. Your ground
velocity is the same in both directions.
Now consider something in space moving at a speed of 1000 km/hr
past the earth. This motion is relative to the center of the
earth, and is independent of the rotation. So as the earth
rotates under it, it would indeed take longer to fly from
New York to London rather than London to New York.
Now, what about an airplane? Well it is not flying in space
or on the ground, but through the air. If the engines are
capable of 800 km/hr, then that speed is relative to the
air. The circumference of the earth is 25000 miles (40000
km) at the equator, and it goes around in 24 hours, so
the ground at the equator is moving along at 1000 mph
(1600 km/hr). If the air weren't moving at approximately
the same speed as the ground, you'd feel a 1000 mph wind
all day. Obviously you don't, and the winds you do feel are
only a small fraction of those speeds.
Because of the prevailing winds, it actually takes an
aircraft longer to go from London to New York than from
New York to London. You can do an online search for
flight times, and you'll see it's about an hour longer
for a 747 to cross west to east than east to west.
> I was arguing with them that when the earth rotates the plane also
> rotates with the earth and they have almost similar speed of rotation.
That's correct. The atmosphere is mostly carried around at
the same speed as the ground. There are slight differences,
and this is what we feel as wind.
> I was further contending that if the earth
> slowly stops rotating, then the moon will collapse into the earth. For,
> example if we slow down our speed of rotation the hammer-throw starts
> coming closer and closer to the body.
These two situations are not comparable. The earth's rotation
does not have an effect on the moon's orbit, except a subtle
one called "tidal locking". Some day, millions of years from
now, the earth will rotate with the moon so that the moon
will appear to be geosynchronous. This has already happened
to the moon, so it always presents the same face to us.
But that won't affect the gravitational force. Unlike the
hammer thrower, it is not the rotation of the earth that
keeps the moon moving. There's no chain. The moon experiences
no friction and needs no pull from a chain to keep it
moving forward. The earth provides a pull that keeps it
moving in an orbit.
> My explanation for the moon not doing revolution with the same speed as
> earth's rotation (or even planets not taking the same time to revolve
> around the sun) was that as we go away from earth (or sun) the
> gravitational forces decreases and so the angular velocity required by
> moon to counteract also decreases and hence it takes longer time to do
> one complete revolution as compared to earth's rotation.
This analysis is correct. Low earth satellites orbit in
about 90 minutes. Geosynchronous satellites, tens of
thousands of km higher, orbit in 24 hours. The moon, even
farther out, orbits in about 28 days. The period depends
on the distance.
> So, my net explanation was that indeed plane will not rotate with the
> same speed as earth because of decrease in gravitationale attraction but
> not because it can stay still and not revolve around the earth.
The moving air is what makes it revolve around the earth, at
close to the same speed as the ground. Air has a lot of
friction. Even satellites experience some air friction for
the first few hundred km of altitude, enough to drag them
down.
- Randy
surendra
Aug23-04, 04:06 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nRandy Poe wrote:\n> surendra <efuzzyone@netscape.net> wrote in message news:<cg3248\\$5cd\\$1@news.asu.edu>...\n>\n>>Hi,\ n>>I was having a hot debate with my friend in which we were arguing about\n>>airplane motion.\n>>What my friend was saying was that when the airplanes they fly, because\n>>of the earth\'s rotation from west to east, it takes them(planes) shorter\n>>time to go from east to west compared to from west to east.\n>>In other words, a plane flying from London to New York will take less\n>>time compared to a plane flying from New York to London. I was\n>>contending that they will take equal time.\n>>\n>>Are there any experts around who can answer who is correct and with reason?\n>\n>\n> Let\'s look at two other cases first. Let\'s assume you are\n\n<snip>\n\n\nHi,\nThanks for all the wonderful explanation. But I still have few questions.\n\nWhy do most heavenly bodies rotate?\n\nAnd why does the atmosphere rotate with the earth? Why doesn\'t it just\nstay still and erode the surface of earth? Does it moves because of the\nfriction with the surface of the earth?\n\nAlso why do planets revolve around the sun, why don\'t they shoot away\nor drop onto the surface due to gravitation?\n\nIs it just a coincedence that they are in circular motion? Or is it\ntheir intial kinetic energy which doesn\'t let them drop onto the surface\nand the gravitation which doesn\'t lets them escape?\n\n- Surendra\n\n\n[Moderator\'s note: Formatting improved by moderator. Please see\nhttp://www-stud.uni-essen.de/~sb0264/HowToPost.html -usc]\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Randy Poe wrote:
> surendra <efuzzyone@netscape.net> wrote in message news:<cg3248$5cd$1@news.asu.edu>...
>
>>Hi,
>>I was having a hot debate with my friend in which we were arguing about
>>airplane motion.
>>What my friend was saying was that when the airplanes they fly, because
>>of the earth's rotation from west to east, it takes them(planes) shorter
>>time to go from east to west compared to from west to east.
>>In other words, a plane flying from London to New York will take less
>>time compared to a plane flying from New York to London. I was
>>contending that they will take equal time.
>>
>>Are there any experts around who can answer who is correct and with reason?
>
>
> Let's look at two other cases first. Let's assume you are
<snip>
Hi,
Thanks for all the wonderful explanation. But I still have few questions.
Why do most heavenly bodies rotate?
And why does the atmosphere rotate with the earth? Why doesn't it just
stay still and erode the surface of earth? Does it moves because of the
friction with the surface of the earth?
Also why do planets revolve around the sun, why don't they shoot away
or drop onto the surface due to gravitation?
Is it just a coincedence that they are in circular motion? Or is it
their intial kinetic energy which doesn't let them drop onto the surface
and the gravitation which doesn't lets them escape?
- Surendra
[Moderator's note: Formatting improved by moderator. Please see
http://www-stud.uni-essen.de/~sb0264/HowToPost.html -usc]
Martin Stone
Aug24-04, 04:52 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"surendra" <efuzzyone@netscape.net> wrote in message\nnews:cg3248\\$5cd\\$1@news.asu.edu...\n>\ n> Hi,\n> I was having a hot debate with my friend in which we were arguing about\n> airplane motion.\n> What my friend was saying was that when the airplanes they fly, because\n> of the earth\'s rotation from west to east, it takes them(planes) shorter\n> time to go from east to west compared to from west to east.\n> In other words, a plane flying from London to New York will take less\n> time compared to a plane flying from New York to London. I was\n> contending that they will take equal time.\n\nLet\'s pretend this is an exam and make some assumptions. I\'ll assume that\nthere is no weather (particularly wind) and that the air spins at exactly\nthe same rate as the earth with no lag and that we can ignore relativistic\neffects as negligable. In that case you\'re right.\n\nIn the real world there is weather, and I would have thought crossing\nweather systems has far more effect than the atmosphere at high altitudes\nlagging behind the earth\'s surface (which is also technically weather I\nguess) - I don\'t know at what altitude this becomes significant (if ever)\nbut I\'ve heard that "round the world ballooners" use it. That may be\n"tosh". Weather may mean your friend IS right, but his reasons were not.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"surendra" <efuzzyone@netscape.net> wrote in message
news:cg3248$5cd$1@news.asu.edu...
>
> Hi,
> I was having a hot debate with my friend in which we were arguing about
> airplane motion.
> What my friend was saying was that when the airplanes they fly, because
> of the earth's rotation from west to east, it takes them(planes) shorter
> time to go from east to west compared to from west to east.
> In other words, a plane flying from London to New York will take less
> time compared to a plane flying from New York to London. I was
> contending that they will take equal time.
Let's pretend this is an exam and make some assumptions. I'll assume that
there is no weather (particularly wind) and that the air spins at exactly
the same rate as the earth with no lag and that we can ignore relativistic
effects as negligable. In that case you're right.
In the real world there is weather, and I would have thought crossing
weather systems has far more effect than the atmosphere at high altitudes
lagging behind the earth's surface (which is also technically weather I
guess) - I don't know at what altitude this becomes significant (if ever)
but I've heard that "round the world ballooners" use it. That may be
"tosh". Weather may mean your friend IS right, but his reasons were not.
George Kinley
Aug24-04, 04:54 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>No Way wrote:\n\n>\n> On 20 Aug 2004 05:47:01 -0400, surendra <efuzzyone@netscape.net>\n> wrote:\n>\n> > I was having a hot debate with my friend in which we were arguing\n> > about airplane motion.\n> > What my friend was saying was that when the airplanes they fly,\n> > because of the earth\'s rotation from west to east, it takes\n> > them(planes) shorter time to go from east to west compared to from\n> > west to east. In other words, a plane flying from London to New\n> > York will take less time compared to a plane flying from New York\n> > to London. I was contending that they will take equal time.\n> >\n> > Are there any experts around who can answer who is correct and with\n> > reason?\n>\n> I wouldn\'t call myself an expert, but the atmosphere rotates along\n> with the earth (if it didn\'t it would be windy... a lot). The plane\'s\n> speed is relative to the atmosphere (and thus relative to the earth),\n> so both ways take equally long (provided there is no wind).\n>\n> You can check your friend\'s idea by taking a look at flight times. For\n> instance, take a look at the flight route between New York and Los\n> Angeles. The flight to New York is shorter (which is exactly the\n> opposite of what you would expect if your friend were right). The\n> reason for the difference in flight times are probably wind currents\n> (I\'m guessing).\nLast month I flew from London to Los Angeles, flight form LON->LA was\nlonger then reverse, the reason of Difference flight time is not\nbecause of air current , but flight route it takes,\n\nI am not sure of Relative speed of aircraft and earth but if ther is no\natmosphere then flight time from east to west should be less the west\nto east , if flight route is exactally same\n\n\n\n--\n-Gk\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>No Way wrote:
>
> On 20 Aug 2004 05:47:01 -0400, surendra <efuzzyone@netscape.net>
> wrote:
>
> > I was having a hot debate with my friend in which we were arguing
> > about airplane motion.
> > What my friend was saying was that when the airplanes they fly,
> > because of the earth's rotation from west to east, it takes
> > them(planes) shorter time to go from east to west compared to from
> > west to east. In other words, a plane flying from London to New
> > York will take less time compared to a plane flying from New York
> > to London. I was contending that they will take equal time.
> >
> > Are there any experts around who can answer who is correct and with
> > reason?
>
> I wouldn't call myself an expert, but the atmosphere rotates along
> with the earth (if it didn't it would be windy... a lot). The plane's
> speed is relative to the atmosphere (and thus relative to the earth),
> so both ways take equally long (provided there is no wind).
>
> You can check your friend's idea by taking a look at flight times. For
> instance, take a look at the flight route between New York and Los
> Angeles. The flight to New York is shorter (which is exactly the
> opposite of what you would expect if your friend were right). The
> reason for the difference in flight times are probably wind currents
> (I'm guessing).
Last month I flew from London to Los Angeles, flight form LON->LA was
longer then reverse, the reason of Difference flight time is not
because of air current , but flight route it takes,
I am not sure of Relative speed of aircraft and earth but if ther is no
atmosphere then flight time from east to west should be less the west
to east , if flight route is exactally same
--
-Gk
Sam Wormley
Aug24-04, 04:54 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>surendra wrote:\n>\n> What my friend was saying was that when the airplanes they fly, because\n> of the earth\'s rotation from west to east, it takes them(planes) shorter\n> time to go from east to west compared to from west to east.\n\nIt\'s the winds.\nhttp://virga.sfsu.edu/crws/jetstream.html\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>surendra wrote:
>
> What my friend was saying was that when the airplanes they fly, because
> of the earth's rotation from west to east, it takes them(planes) shorter
> time to go from east to west compared to from west to east.
It's the winds.
http://virga.sfsu.edu/crws/jetstream.html
Moataz H. Emam
Aug24-04, 04:54 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>surendra wrote:\n> I was having a hot debate with my friend in which we were arguing about\n> airplane motion.\n\nHmm! I am tempted to say that it takes the same time each way, your\nargument is correct. However, I travel to Egypt from the US once or\ntwice a year regularly, and it does take an hour longer going West to\nEast. Is this a wind effect kind of thing? If so, doesn\'t the wind EVER\nchange direction?\n\n--\nMoataz H. Emam\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>surendra wrote:
> I was having a hot debate with my friend in which we were arguing about
> airplane motion.
Hmm! I am tempted to say that it takes the same time each way, your
argument is correct. However, I travel to Egypt from the US once or
twice a year regularly, and it does take an hour longer going West to
East. Is this a wind effect kind of thing? If so, doesn't the wind EVER
change direction?
--
Moataz H. Emam
Androcles
Aug24-04, 04:54 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"Nick Maclaren" <nmm1@cus.cam.ac.uk> wrote in message\nnews:cg4v2i\\$hlm\\$1@pegasus.csx.cam.ac. uk...\n|\n| In article <OTmVc.3584\\$wd7.43453605@news-text.cableinet.net>,\n| "Androcles" <androc1es@nospamblueyonder.co.uk> writes:\n| |>\n| |> The reason for the 1:00 hour difference is the high wind blowing the\nplane\n| |> along (or opposing it), and nothing else. If we ignore the jet stream,\nthe\n| |> times will be the same. A real flight will be subject to various\ndelays, of\n| |> course, and you may get in early or late.\n|\n| A slight nitpick: the precise value of the difference is chosen\n| to fit landing schedules, and the target speed of the aircraft\n| is set accordingly.\n\nI\'ve flown the Atlantic many times, and rarely have I arrived exactly on\ntime.\nMy worst trip was from Detmold in Germany to Pittsburgh Pennsyslvania taking\na total of 24.5 hours. 8 hours of that were two scheduled layovers,\nFrankfurt and Boston. It isn\'t fun when the plane gets in after it should\nhave taken off, then when you do board it, it taxies to the runway, turns\naround and comes back for an instrument replacement.\nHowever, that is just a personal war story from twenty years ago.\n\nSince we are nitpicking, the target air speed of a commercial airliner is\nactually set to conserve fuel. If there is a wind assist and the plane gets\nin early, the passengers never complain and neither does the airline. Only\nwhen the control tower cannot land the plane for safety/traffic\nconsiderations will it be placed on the "stack", circling in an oval and\ndescending a further 500 ft as the lowest plane takes up its final approach.\nThis is when you are likely to experience the vortex of the plane ahead, the\nturbulence bouncing the plane around, which is not all that comfortable at\n3,000 feet with the left wing down 30 degrees.\n\n| You cannot therefore use the magnitude of\n| the difference to calculate the expected speed of the jet stream.\n\nSo where did anyone suggest you could?\nOh, that was Nick Maclaren, wasn\'t it?\nIt\'s your own nit you are picking :-)\n\n| The jet stream is, however, the only major reason for the difference.\n|\n| Regards,\n| Nick Maclaren.\nJust so.\nAndrocles\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Nick Maclaren" <nmm1@cus.cam.ac.uk> wrote in message
news:cg4v2i$hlm$1@pegasus.csx.cam.ac.uk...
|
| In article <OTmVc.3584$wd7.43453605@news-text.cableinet.net>,
| "Androcles" <androc1es@nospamblueyonder.co.uk> writes:
| |>| |> The reason for the 1:00 hour difference is the high wind blowing the
plane
| |> along (or opposing it), and nothing else. If we ignore the jet stream,
the
| |> times will be the same. A real flight will be subject to various
delays, of
| |> course, and you may get in early or late.
|
| A slight nitpick: the precise value of the difference is chosen
| to fit landing schedules, and the target speed of the aircraft
| is set accordingly.
I've flown the Atlantic many times, and rarely have I arrived exactly on
time.
My worst trip was from Detmold in Germany to Pittsburgh Pennsyslvania taking
a total of 24.5 hours. 8 hours of that were two scheduled layovers,
Frankfurt and Boston. It isn't fun when the plane gets in after it should
have taken off, then when you do board it, it taxies to the runway, turns
around and comes back for an instrument replacement.
However, that is just a personal war story from twenty years ago.
Since we are nitpicking, the target air speed of a commercial airliner is
actually set to conserve fuel. If there is a wind assist and the plane gets
in early, the passengers never complain and neither does the airline. Only
when the control tower cannot land the plane for safety/traffic
considerations will it be placed on the "stack", circling in an oval and
descending a further 500 ft as the lowest plane takes up its final approach.
This is when you are likely to experience the vortex of the plane ahead, the
turbulence bouncing the plane around, which is not all that comfortable at
3,000 feet with the left wing down 30 degrees.
| You cannot therefore use the magnitude of
| the difference to calculate the expected speed of the jet stream.
So where did anyone suggest you could?
Oh, that was Nick Maclaren, wasn't it?
It's your own nit you are picking :-)
| The jet stream is, however, the only major reason for the difference.
|
| Regards,
| Nick Maclaren.
Just so.
Androcles
John T Lowry
Aug24-04, 04:54 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"surendra" <efuzzyone@netscape.net> wrote in message\nnews:cg3248\\$5cd\\$1@news.asu.edu...\n>\ n> Hi,\n> I was having a hot debate with my friend in which we were arguing\nabout\n> airplane motion.\n> What my friend was saying was that when the airplanes they fly,\nbecause\n> of the earth\'s rotation from west to east, it takes them(planes)\nshorter\n> time to go from east to west compared to from west to east.\n> In other words, a plane flying from London to New York will take less\n> time compared to a plane flying from New York to London. I was\n> contending that they will take equal time.\n>\n> Are there any experts around who can answer who is correct and with\nreason?\n>\n>\n> I was arguing with them that when the earth rotates the plane also\n> rotates with the earth and they have almost similar speed of rotation.\n> I was arguing that the revolution of moon around the earth(or earth\n> around the sun, etc., ) is like rotating with a hammer-throw (an\nolympic\n> sport in which there is a heavy ball tied at the end of a chain, which\n> the atheletes has to throw the furthest). When we are still the\n> hammer-throw is close to us and still. But when we start rotating the\n> hammer throw revolves around us, and it starts going farther away from\n> our body. It is the speed of our rotation which decides how far it is\n> away from our body. So, in case of heavenly bodies(earth and moon) it\nis\n> the gravitational force. I was further contending that if the earth\n> slowly stops rotating, then the moon will collapse into the earth.\nFor,\n> example if we slow down our speed of rotation the hammer-throw starts\n> coming closer and closer to the body.\n> My explanation for the moon not doing revolution with the same speed\nas\n> earth\'s rotation (or even planets not taking the same time to revolve\n> around the sun) was that as we go away from earth (or sun) the\n> gravitational forces decreases and so the angular velocity required by\n> moon to counteract also decreases and hence it takes longer time to do\n> one complete revolution as compared to earth\'s rotation.\n>\n> So, my net explanation was that indeed plane will not rotate with the\n> same speed as earth because of decrease in gravitationale attraction\nbut\n> not because it can stay still and not revolve around the earth.\n>\n> I will like to hear the opinion of advice of experts on this.\n> Any help in settling the argument will be appreciated.\n>\n> Thank you,\n> Surendra\n>\n\n\nHere is a snip from an article I wrote on this subject. The Physics\nTeacher rejected it because it was too complicated, suggesting I send it\nto the American Journal of Physics. AMJ rejected it because it was too\nsimple! Oh well....\n\nThe airplanes (one headed straight east, the other straight west, were\ntaken over the equator, no wind, at altitude 2000 m, speed 200 m/s (389\nKTAS).\n***\nBecause centrifugal terms for the eastbound and westbound airplanes are\nidentical (since that term depends only on the rotation of the\nnon-inertial frame and the common location of the two objects), it might\nappear there is no dynamical difference between the airplanes. But\n("just to be on the safe side") compute the Coriolis terms. On the\neastbound airplane E, the Coriolis acceleration is 2SV outwards (to\nairplane E, upwards). On the westbound airplane W, 2SV inwards\n(downwards to airplane W). Unconstrained airplane W would be forced down\nby this Coriolis force and so to counteract W\'s higher effective\nacceleration of gravity - higher than E\'s by 4SV = 0.0583 m sec!2 -\nairplane W requires more lift. Higher lift is accomplished by W\'s taking\na slightly higher angle of attack, the angle with which its wings meet\nthe relative wind. But more lift means more induced drag (drag due to\nlift), hence - everything else being equal, as we have specified - a\nslightly higher fuel consumption rate for the westbound airplane. Since\nthe acceleration of gravity near the equator is 9.780 m sec!2, there is\na lifting acceleration difference of 0.0060 g. For a 10,000 lbf airplane\n(the scales having read somewhat light because of the centrifugal\npseudoforce), perhaps at this air speed a World War II fighter, that is\nan effective weight reduction of 60 lbf just for having turned from\nwestbound to eastbound. In practice, however, negligible. We apologize\nthat we don\'t have space here for a meaningful discussion of induced\ndrag.\n\n***\n\nSo Uncle Al is right (as usual). An Airbus captain friend of mine, Capt.\nMike, though totally untutored (his school in Rouen was destroyed during\nthe D-Day invasion and he never made it back to school) in any of this,\npicked up the effect by careful analysis of fuel records (corrected for\nwind, etc., of course).\n\n\n\nJohn T. Lowry, PhD\n\nFlight Physics\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"surendra" <efuzzyone@netscape.net> wrote in message
news:cg3248$5cd$1@news.asu.edu...
>
> Hi,
> I was having a hot debate with my friend in which we were arguing
about
> airplane motion.
> What my friend was saying was that when the airplanes they fly,
because
> of the earth's rotation from west to east, it takes them(planes)
shorter
> time to go from east to west compared to from west to east.
> In other words, a plane flying from London to New York will take less
> time compared to a plane flying from New York to London. I was
> contending that they will take equal time.
>
> Are there any experts around who can answer who is correct and with
reason?
>
>
> I was arguing with them that when the earth rotates the plane also
> rotates with the earth and they have almost similar speed of rotation.
> I was arguing that the revolution of moon around the earth(or earth
> around the sun, etc., ) is like rotating with a hammer-throw (an
olympic
> sport in which there is a heavy ball tied at the end of a chain, which
> the atheletes has to throw the furthest). When we are still the
> hammer-throw is close to us and still. But when we start rotating the
> hammer throw revolves around us, and it starts going farther away from
> our body. It is the speed of our rotation which decides how far it is
> away from our body. So, in case of heavenly bodies(earth and moon) it
is
> the gravitational force. I was further contending that if the earth
> slowly stops rotating, then the moon will collapse into the earth.
For,
> example if we slow down our speed of rotation the hammer-throw starts
> coming closer and closer to the body.
> My explanation for the moon not doing revolution with the same speed
as
> earth's rotation (or even planets not taking the same time to revolve
> around the sun) was that as we go away from earth (or sun) the
> gravitational forces decreases and so the angular velocity required by
> moon to counteract also decreases and hence it takes longer time to do
> one complete revolution as compared to earth's rotation.
>
> So, my net explanation was that indeed plane will not rotate with the
> same speed as earth because of decrease in gravitationale attraction
but
> not because it can stay still and not revolve around the earth.
>
> I will like to hear the opinion of advice of experts on this.
> Any help in settling the argument will be appreciated.
>
> Thank you,
> Surendra
>
Here is a snip from an article I wrote on this subject. The Physics
Teacher rejected it because it was too complicated, suggesting I send it
to the American Journal of Physics. AMJ rejected it because it was too
simple! Oh well....
The airplanes (one headed straight east, the other straight west, were
taken over the equator, no wind, at altitude 2000 m, speed 200 m/s (389
KTAS).
***
Because centrifugal terms for the eastbound and westbound airplanes are
identical (since that term depends only on the rotation of the
non-inertial frame and the common location of the two objects), it might
appear there is no dynamical difference between the airplanes. But
("just to be on the safe side") compute the Coriolis terms. On the
eastbound airplane E, the Coriolis acceleration is 2SV outwards (to
airplane E, upwards). On the westbound airplane W, 2SV inwards
(downwards to airplane W). Unconstrained airplane W would be forced down
by this Coriolis force and so to counteract W's higher effective
acceleration of gravity - higher than E's by 4SV = .0583 m sec!2 -
airplane W requires more lift. Higher lift is accomplished by W's taking
a slightly higher angle of attack, the angle with which its wings meet
the relative wind. But more lift means more induced drag (drag due to
lift), hence - everything else being equal, as we have specified - a
slightly higher fuel consumption rate for the westbound airplane. Since
the acceleration of gravity near the equator is 9.780 m sec!2, there is
a lifting acceleration difference of .0060 g. For a 10,000 lbf airplane
(the scales having read somewhat light because of the centrifugal
pseudoforce), perhaps at this air speed a World War II fighter, that is
an effective weight reduction of 60 lbf just for having turned from
westbound to eastbound. In practice, however, negligible. We apologize
that we don't have space here for a meaningful discussion of induced
drag.
***
So Uncle Al is right (as usual). An Airbus captain friend of mine, Capt.
Mike, though totally untutored (his school in Rouen was destroyed during
the D-Day invasion and he never made it back to school) in any of this,
picked up the effect by careful analysis of fuel records (corrected for
wind, etc., of course).
John T. Lowry, PhD
Flight Physics
surendra
Aug24-04, 04:56 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hi,\nThanks for all the wonderful explanation. But I still have few questions.\n\n\n\nWhy does most heavenly bodies rotates?\nAnd why does the atmosphere rotates with earth? Why doesn\'t it just\nstays still and erode the surface of earth. Does it moves because of the\nfriction with the surface of earth? Does the altitiude makes any\ndifferene to the speed relative to the surface of earth?\n\n\nAlso why does planets revolve around the sun, why don\'t they shoot away\nor drop onto the surface due to gravitation?\nIs it just a coincedence that they are in circular motion. Or is it\ntheir intial kinetic energy which doesn\'t let them drop onto the surface\nand the gravitation which doesn\'t lets them escape?\n\nThanks in advance.\n- Surendra\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi,
Thanks for all the wonderful explanation. But I still have few questions.
Why does most heavenly bodies rotates?
And why does the atmosphere rotates with earth? Why doesn't it just
stays still and erode the surface of earth. Does it moves because of the
friction with the surface of earth? Does the altitiude makes any
differene to the speed relative to the surface of earth?
Also why does planets revolve around the sun, why don't they shoot away
or drop onto the surface due to gravitation?
Is it just a coincedence that they are in circular motion. Or is it
their intial kinetic energy which doesn't let them drop onto the surface
and the gravitation which doesn't lets them escape?
Thanks in advance.
- Surendra
Edward Green
Aug24-04, 04:56 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0408200950.5e8baf79@posting.google. com>...\n> surendra <efuzzyone@netscape.net> wrote in message news:<cg3248\\$5cd\\$1@news.asu.edu>...\n> > Hi,\n> > I was having a hot debate with my friend in which we were arguing about\n> > airplane motion.\n> > What my friend was saying was that when the airplanes they fly, because\n> > of the earth\'s rotation from west to east, it takes them(planes) shorter\n> > time to go from east to west compared to from west to east.\n> > In other words, a plane flying from London to New York will take less\n> > time compared to a plane flying from New York to London. I was\n> > contending that they will take equal time.\n> >\n> > Are there any experts around who can answer who is correct and with reason?\n>\n> Let\'s look at two other cases first. Let\'s assume you are\n> walking from East to West, and from West to East. Do you agree\n> that these two trips should take the same amount of time?\n> The reason they do is that you are walking relative to\n> the ground, which is rotating with the earth. Your ground\n> velocity is the same in both directions.\n>\n> Now consider something in space moving at a speed of 1000 km/hr\n> past the earth. This motion is relative to the center of the\n> earth, and is independent of the rotation. So as the earth\n> rotates under it, it would indeed take longer to fly from\n> New York to London rather than London to New York.\n>\n> Now, what about an airplane? Well it is not flying in space\n> or on the ground, but through the air. If the engines are\n> capable of 800 km/hr, then that speed is relative to the\n> air. The circumference of the earth is 25000 miles (40000\n> km) at the equator, and it goes around in 24 hours, so\n> the ground at the equator is moving along at 1000 mph\n> (1600 km/hr). If the air weren\'t moving at approximately\n> the same speed as the ground, you\'d feel a 1000 mph wind\n> all day. Obviously you don\'t, and the winds you do feel are\n> only a small fraction of those speeds.\n>\n> Because of the prevailing winds, it actually takes an\n> aircraft longer to go from London to New York than from\n> New York to London. You can do an online search for\n> flight times, and you\'ll see it\'s about an hour longer\n> for a 747 to cross west to east than east to west.\n\nTry as I might, I can\'t make "west to east" correspond to flying from\nLondon to New York. I\'m not sure if this is an irregularity in my\nunderstanding of compass directions (for a long time I had trouble\nwith the idea that the "west wind" was out of the west, rather than\nheading west), or a typo ...\n\nBut anyway, pedantically, we might mention one effect tending to agree\nwith the OP\'s friend: say an object ascended 1000km straight up into\nspace over London, with no other acceleration, hung out, and\neventually descended. If it hung out long enough, it would find\nitself descending in Canada somewhere, since the tangential velocity\nit had on the Earth\'s surface would not be the tangential velocity\ncorresponding to the same 2pi/24hr rotation rate at 1000km altitude;\nso the Earth would turn under it, thus somewhat confirming to the idea\nof the OP\'s friend. Of course, we could also describe the same\ntrajectory relative to the rotating Earth by the "Coriolos force".\n\nI\'m not claiming this is significant for aircraft.\n\n<...>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0408200950.5e8baf79@posting.google.com>...
> surendra <efuzzyone@netscape.net> wrote in message news:<cg3248$5cd$1@news.asu.edu>...
> > Hi,
> > I was having a hot debate with my friend in which we were arguing about
> > airplane motion.
> > What my friend was saying was that when the airplanes they fly, because
> > of the earth's rotation from west to east, it takes them(planes) shorter
> > time to go from east to west compared to from west to east.
> > In other words, a plane flying from London to New York will take less
> > time compared to a plane flying from New York to London. I was
> > contending that they will take equal time.
> >
> > Are there any experts around who can answer who is correct and with reason?
>
> Let's look at two other cases first. Let's assume you are
> walking from East to West, and from West to East. Do you agree
> that these two trips should take the same amount of time?
> The reason they do is that you are walking relative to
> the ground, which is rotating with the earth. Your ground
> velocity is the same in both directions.
>
> Now consider something in space moving at a speed of 1000 km/hr
> past the earth. This motion is relative to the center of the
> earth, and is independent of the rotation. So as the earth
> rotates under it, it would indeed take longer to fly from
> New York to London rather than London to New York.
>
> Now, what about an airplane? Well it is not flying in space
> or on the ground, but through the air. If the engines are
> capable of 800 km/hr, then that speed is relative to the
> air. The circumference of the earth is 25000 miles (40000
> km) at the equator, and it goes around in 24 hours, so
> the ground at the equator is moving along at 1000 mph
> (1600 km/hr). If the air weren't moving at approximately
> the same speed as the ground, you'd feel a 1000 mph wind
> all day. Obviously you don't, and the winds you do feel are
> only a small fraction of those speeds.
>
> Because of the prevailing winds, it actually takes an
> aircraft longer to go from London to New York than from
> New York to London. You can do an online search for
> flight times, and you'll see it's about an hour longer
> for a 747 to cross west to east than east to west.
Try as I might, I can't make "west to east" correspond to flying from
London to New York. I'm not sure if this is an irregularity in my
understanding of compass directions (for a long time I had trouble
with the idea that the "west wind" was out of the west, rather than
heading west), or a typo ...
But anyway, pedantically, we might mention one effect tending to agree
with the OP's friend: say an object ascended 1000km straight up into
space over London, with no other acceleration, hung out, and
eventually descended. If it hung out long enough, it would find
itself descending in Canada somewhere, since the tangential velocity
it had on the Earth's surface would not be the tangential velocity
corresponding to the same 2pi/24hr rotation rate at 1000km altitude;
so the Earth would turn under it, thus somewhat confirming to the idea
of the OP's friend. Of course, we could also describe the same
trajectory relative to the rotating Earth by the "Coriolos force".
I'm not claiming this is significant for aircraft.
<...>
Wolfgang Beirl
Aug24-04, 04:56 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>surendra <efuzzyone@netscape.net> wrote in message news:<cg3248\\$5cd\\$1@news.asu.edu>...\n> Hi,\n> I was having a hot debate with my friend in which we were arguing about\n> airplane motion.\n> What my friend was saying was that when the airplanes they fly, because\n> of the earth\'s rotation from west to east, it takes them(planes) shorter\n> time to go from east to west compared to from west to east.\n> In other words, a plane flying from London to New York will take less\n> time compared to a plane flying from New York to London. I was\n> contending that they will take equal time.\n>\n> Are there any experts around who can answer who is correct and with reason?\n>\n>\n> I was arguing with them that when the earth rotates the plane also\n> rotates with the earth and they have almost similar speed of rotation.\n> I was arguing that the revolution of moon around the earth(or earth\n> around the sun, etc., ) is like rotating with a hammer-throw (an olympic\n> sport in which there is a heavy ball tied at the end of a chain, which\n> the atheletes has to throw the furthest). When we are still the\n> hammer-throw is close to us and still. But when we start rotating the\n> hammer throw revolves around us, and it starts going farther away from\n> our body. It is the speed of our rotation which decides how far it is\n> away from our body. So, in case of heavenly bodies(earth and moon) it is\n> the gravitational force. I was further contending that if the earth\n> slowly stops rotating, then the moon will collapse into the earth. For,\n> example if we slow down our speed of rotation the hammer-throw starts\n> coming closer and closer to the body.\n> My explanation for the moon not doing revolution with the same speed as\n> earth\'s rotation (or even planets not taking the same time to revolve\n> around the sun) was that as we go away from earth (or sun) the\n> gravitational forces decreases and so the angular velocity required by\n> moon to counteract also decreases and hence it takes longer time to do\n> one complete revolution as compared to earth\'s rotation.\n>\n> So, my net explanation was that indeed plane will not rotate with the\n> same speed as earth because of decrease in gravitationale attraction but\n> not because it can stay still and not revolve around the earth.\n>\n> I will like to hear the opinion of advice of experts on this.\n> Any help in settling the argument will be appreciated.\n>\n> Thank you,\n> Surendra\n\n\nAs people have pointed out on this thread already, the main difference\nof about an hour is caused by the jet stream. But the direction of the\njet stream is a direct result of the earth\'s rotation (Coreolis force).\nThus your friend is right in the end: There is a significant difference\nand its root cause is the rotation of the earth.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>surendra <efuzzyone@netscape.net> wrote in message news:<cg3248$5cd$1@news.asu.edu>...
> Hi,
> I was having a hot debate with my friend in which we were arguing about
> airplane motion.
> What my friend was saying was that when the airplanes they fly, because
> of the earth's rotation from west to east, it takes them(planes) shorter
> time to go from east to west compared to from west to east.
> In other words, a plane flying from London to New York will take less
> time compared to a plane flying from New York to London. I was
> contending that they will take equal time.
>
> Are there any experts around who can answer who is correct and with reason?
>
>
> I was arguing with them that when the earth rotates the plane also
> rotates with the earth and they have almost similar speed of rotation.
> I was arguing that the revolution of moon around the earth(or earth
> around the sun, etc., ) is like rotating with a hammer-throw (an olympic
> sport in which there is a heavy ball tied at the end of a chain, which
> the atheletes has to throw the furthest). When we are still the
> hammer-throw is close to us and still. But when we start rotating the
> hammer throw revolves around us, and it starts going farther away from
> our body. It is the speed of our rotation which decides how far it is
> away from our body. So, in case of heavenly bodies(earth and moon) it is
> the gravitational force. I was further contending that if the earth
> slowly stops rotating, then the moon will collapse into the earth. For,
> example if we slow down our speed of rotation the hammer-throw starts
> coming closer and closer to the body.
> My explanation for the moon not doing revolution with the same speed as
> earth's rotation (or even planets not taking the same time to revolve
> around the sun) was that as we go away from earth (or sun) the
> gravitational forces decreases and so the angular velocity required by
> moon to counteract also decreases and hence it takes longer time to do
> one complete revolution as compared to earth's rotation.
>
> So, my net explanation was that indeed plane will not rotate with the
> same speed as earth because of decrease in gravitationale attraction but
> not because it can stay still and not revolve around the earth.
>
> I will like to hear the opinion of advice of experts on this.
> Any help in settling the argument will be appreciated.
>
> Thank you,
> Surendra
As people have pointed out on this thread already, the main difference
of about an hour is caused by the jet stream. But the direction of the
jet stream is a direct result of the earth's rotation (Coreolis force).
Thus your friend is right in the end: There is a significant difference
and its root cause is the rotation of the earth.
robert j. kolker
Aug24-04, 11:30 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nMartin Stone wrote:\n\n> Let\'s pretend this is an exam and make some assumptions. I\'ll assume that\n> there is no weather (particularly wind) and that the air spins at exactly\n> the same rate as the earth with no lag and that we can ignore relativistic\n> effects as negligable. In that case you\'re right.\n\nThe motion of the air as it rises from the tropics is largely\ndertermined by the Coriolis acceleration.\n\nWhether moves from west to east in the norther hemisphere.\n\nSo a plane flying with the weather is going to make better time than a\nplane flying against the weather, all other things being equal.\n\nWe also launch most of our space probes as near to the equator as\npossible for the same reason.\n\nThe French, luck them, have Devil\'s Island (Guyana) which is very close\nto the equator from which they launch the ESA vessels.\n\nBob Kolker\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Martin Stone wrote:
> Let's pretend this is an exam and make some assumptions. I'll assume that
> there is no weather (particularly wind) and that the air spins at exactly
> the same rate as the earth with no lag and that we can ignore relativistic
> effects as negligable. In that case you're right.
The motion of the air as it rises from the tropics is largely
dertermined by the Coriolis acceleration.
Whether moves from west to east in the norther hemisphere.
So a plane flying with the weather is going to make better time than a
plane flying against the weather, all other things being equal.
We also launch most of our space probes as near to the equator as
possible for the same reason.
The French, luck them, have Devil's Island (Guyana) which is very close
to the equator from which they launch the ESA vessels.
Bob Kolker
N:dlzc D:aol T:com (dlzc)
Aug25-04, 02:42 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Dear Moataz H. Emam:\n\n"Moataz H. Emam" <memam@spam.physics.umass.edu> wrote in message\nnews:412626E0.2E573F29@spam.physics.umass .edu...\n> surendra wrote:\n> > I was having a hot debate with my friend in which we were arguing about\n> > airplane motion.\n>\n> Hmm! I am tempted to say that it takes the same time each way, your\n> argument is correct. However, I travel to Egypt from the US once or\n> twice a year regularly, and it does take an hour longer going West to\n> East. Is this a wind effect kind of thing? If so, doesn\'t the wind EVER\n> change direction?\n\nGoogle "jet stream". It wanders around, but always goes generally\nwest-to-east. Almost all airports have a runway that is west-to-east.\n\nDavid A. Smith\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Dear Moataz H. Emam:
"Moataz H. Emam" <memam@spam.physics.umass.edu> wrote in message
news:412626E0.2E573F29@spam.physics.umass.edu...
> surendra wrote:
> > I was having a hot debate with my friend in which we were arguing about
> > airplane motion.
>
> Hmm! I am tempted to say that it takes the same time each way, your
> argument is correct. However, I travel to Egypt from the US once or
> twice a year regularly, and it does take an hour longer going West to
> East. Is this a wind effect kind of thing? If so, doesn't the wind EVER
> change direction?
Google "jet stream". It wanders around, but always goes generally
west-to-east. Almost all airports have a runway that is west-to-east.
David A. Smith
surendra
Aug25-04, 02:44 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>I am not a physics expert(I have just taken high school physics) but I\nhave some serious objections to your argument.\n\nEdward Green wrote:\n> But anyway, pedantically, we might mention one effect tending to agree\n> with the OP\'s friend: say an object ascended 1000km straight up into\n> space over London, with no other acceleration, hung out, and\n> eventually descended. If it hung out long enough, it would find\n> itself descending in Canada somewhere, since the tangential velocity\n> it had on the Earth\'s surface would not be the tangential velocity\n> corresponding to the same 2pi/24hr rotation rate at 1000km altitude;\n> so the Earth would turn under it, thus somewhat confirming to the idea\n> of the OP\'s friend. Of course, we could also describe the same\n> trajectory relative to the rotating Earth by the "Coriolos force".\n>\n> I\'m not claiming this is significant for aircraft.\n>\n> <...>\n>\n\n\nWhen the object is on the surface of earth it has certain angular\nvelocity relative to the center of earth. As the object ascends,assuming\nthere is only a vertical force acting along a line from the center of\nearth to the CG of the object, the angular velocity relative to the\ncenter of earth should remain constant.\nThe tangential velocity will increase but so does the circumference of\nearth as the altitude increases. The object will have to travel a\ngreater distance relative to the center of earth compared to the land of\nLondon. And this increase in tangential velocity is what helps the\nobject to be in London and not recede to Asia.\nI hope, you are getting my point.\n\nSo, if the object was initially at London, it will remain in the sky\nabove London, and it will never reach Canada.\n\n- Surendra\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>I am not a physics expert(I have just taken high school physics) but I
have some serious objections to your argument.
Edward Green wrote:
> But anyway, pedantically, we might mention one effect tending to agree
> with the OP's friend: say an object ascended 1000km straight up into
> space over London, with no other acceleration, hung out, and
> eventually descended. If it hung out long enough, it would find
> itself descending in Canada somewhere, since the tangential velocity
> it had on the Earth's surface would not be the tangential velocity
> corresponding to the same 2pi/24hr rotation rate at 1000km altitude;
> so the Earth would turn under it, thus somewhat confirming to the idea
> of the OP's friend. Of course, we could also describe the same
> trajectory relative to the rotating Earth by the "Coriolos force".
>
> I'm not claiming this is significant for aircraft.
>
> <...>
>
When the object is on the surface of earth it has certain angular
velocity relative to the center of earth. As the object ascends,assuming
there is only a vertical force acting along a line from the center of
earth to the CG of the object, the angular velocity relative to the
center of earth should remain constant.
The tangential velocity will increase but so does the circumference of
earth as the altitude increases. The object will have to travel a
greater distance relative to the center of earth compared to the land of
London. And this increase in tangential velocity is what helps the
object to be in London and not recede to Asia.
I hope, you are getting my point.
So, if the object was initially at London, it will remain in the sky
above London, and it will never reach Canada.
- Surendra
Randy Poe
Aug25-04, 02:46 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>spamspamspam3@netzero.com (Edward Green) wrote in message news:<eca320d0.0408212025.4493b8e7@posting.google. com>...\n> poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0408200950.5e8baf79@posting.google. com>...\n> Try as I might, I can\'t make "west to east" correspond to flying from\n> London to New York.\n\nNitpicker.\n\n> I\'m not sure if this is an irregularity in my\n> understanding of compass directions (for a long time I had trouble\n> with the idea that the "west wind" was out of the west, rather than\n> heading west), or a typo ...\n\nTypo, of course. The point is that if there weren\'t that\npesky air to deal with, earth rotation would cause the\ntrip from New York to London to be longer than London to\nNew York, since the former involves chasing London as it\nrotates around, and the latter involves New York rushing\ntoward you.\n\nBut because of the winds, New York-London is *shorter* than\nLondon-New York.\n\n> But anyway, pedantically, we might mention one effect tending to agree\n> with the OP\'s friend: say an object ascended 1000km straight up into\n> space over London, with no other acceleration, hung out, and\n> eventually descended. If it hung out long enough, it would find\n> itself descending in Canada somewhere, since the tangential velocity\n> it had on the Earth\'s surface would not be the tangential velocity\n> corresponding to the same 2pi/24hr rotation rate at 1000km altitude;\n> so the Earth would turn under it, thus somewhat confirming to the idea\n> of the OP\'s friend. Of course, we could also describe the same\n> trajectory relative to the rotating Earth by the "Coriolos force".\n>\n> I\'m not claiming this is significant for aircraft.\n\nOnly in so far as the coriolis gives rise to air flow\npatterns.\n\n- Randy\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>spamspamspam3@netzero.com (Edward Green) wrote in message news:<eca320d0.0408212025.4493b8e7@posting.google.com>...
> poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0408200950.5e8baf79@posting.google.com>...
> Try as I might, I can't make "west to east" correspond to flying from
> London to New York.
Nitpicker.
> I'm not sure if this is an irregularity in my
> understanding of compass directions (for a long time I had trouble
> with the idea that the "west wind" was out of the west, rather than
> heading west), or a typo ...
Typo, of course. The point is that if there weren't that
pesky air to deal with, earth rotation would cause the
trip from New York to London to be longer than London to
New York, since the former involves chasing London as it
rotates around, and the latter involves New York rushing
toward you.
But because of the winds, New York-London is *shorter* than
London-New York.
> But anyway, pedantically, we might mention one effect tending to agree
> with the OP's friend: say an object ascended 1000km straight up into
> space over London, with no other acceleration, hung out, and
> eventually descended. If it hung out long enough, it would find
> itself descending in Canada somewhere, since the tangential velocity
> it had on the Earth's surface would not be the tangential velocity
> corresponding to the same 2pi/24hr rotation rate at 1000km altitude;
> so the Earth would turn under it, thus somewhat confirming to the idea
> of the OP's friend. Of course, we could also describe the same
> trajectory relative to the rotating Earth by the "Coriolos force".
>
> I'm not claiming this is significant for aircraft.
Only in so far as the coriolis gives rise to air flow
patterns.
- Randy
Randy Poe
Aug25-04, 02:46 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>surendra <efuzzyone@netscape.net> wrote in message news:<cg5q3g\\$e6v\\$1@news.asu.edu>...\n> Randy Poe wrote:\n> > surendra <efuzzyone@netscape.net> wrote in message news:<cg3248\\$5cd\\$1@news.asu.edu>...\n> >\n> >>Hi,\n> >>I was having a hot debate with my friend in which we were arguing about\n> >>airplane motion.\n> >>What my friend was saying was that when the airplanes they fly, because\n> >>of the earth\'s rotation from west to east, it takes them(planes) shorter\n> >>time to go from east to west compared to from west to east.\n> >>In other words, a plane flying from London to New York will take less\n> >>time compared to a plane flying from New York to London. I was\n> >>contending that they will take equal time.\n> >>\n> >>Are there any experts around who can answer who is correct and with reason?\n> >\n> >\n> > Let\'s look at two other cases first. Let\'s assume you are\n>\n> <snip>\n>\n>\n> Hi,\n> Thanks for all the wonderful explanation. But I still have few questions.\n>\n> Why do most heavenly bodies rotate?\n\nSpin leftover from angular momentum when they formed. Don\'t\nforget that the entire galaxy is in motion. Almost everything\nstarts out with a little bit of rotation.\n\n> And why does the atmosphere rotate with the earth? Why doesn\'t it just\n> stay still and erode the surface of earth? Does it moves because of the\n> friction with the surface of the earth?\n\nYes. The same reason that you rotate with the earth.\n\n>\n> Also why do planets revolve around the sun, why don\'t they shoot away\n> or drop onto the surface due to gravitation?\n\nBecause that\'s the effect of the inverse square law of\ngravitation: that paths are elliptical if closed, hyperbolic\nif open. You\'d only get falling straight in if the planet\nhappened to have no motion relative to the sun as it formed,\nin which case the gas would not have been out there long\nenough to condense anyway. If the planet formed from condensation\nof a gas cloud, that cloud itself was in orbit.\n\n> Is it just a coincedence that they are in circular motion?\n\nCircles as opposed to ellipses? No, that almost certainly has\nsomething to do with how they form.\n\nClosed orbits as opposed to crashing into the sun? No, not\ncoincidence. Almost any initial conditions you choose,\nso long as you don\'t have escape velocity, are going to\nresult in a closed elliptical orbit. It\'s just what you\nget when you start something moving in an inverse square\nlaw.\n\n\n> Or is it\n> their intial kinetic energy which doesn\'t let them drop onto the surface\n> and the gravitation which doesn\'t lets them escape?\n\nI guess that\'s one way of putting "this is what the inverse\nsquare law predicts".\n\n- Randy\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>surendra <efuzzyone@netscape.net> wrote in message news:<cg5q3g$e6v$1@news.asu.edu>...
> Randy Poe wrote:
> > surendra <efuzzyone@netscape.net> wrote in message news:<cg3248$5cd$1@news.asu.edu>...
> >
> >>Hi,
> >>I was having a hot debate with my friend in which we were arguing about
> >>airplane motion.
> >>What my friend was saying was that when the airplanes they fly, because
> >>of the earth's rotation from west to east, it takes them(planes) shorter
> >>time to go from east to west compared to from west to east.
> >>In other words, a plane flying from London to New York will take less
> >>time compared to a plane flying from New York to London. I was
> >>contending that they will take equal time.
> >>
> >>Are there any experts around who can answer who is correct and with reason?
> >
> >
> > Let's look at two other cases first. Let's assume you are
>
> <snip>
>
>
> Hi,
> Thanks for all the wonderful explanation. But I still have few questions.
>
> Why do most heavenly bodies rotate?
Spin leftover from angular momentum when they formed. Don't
forget that the entire galaxy is in motion. Almost everything
starts out with a little bit of rotation.
> And why does the atmosphere rotate with the earth? Why doesn't it just
> stay still and erode the surface of earth? Does it moves because of the
> friction with the surface of the earth?
Yes. The same reason that you rotate with the earth.
>
> Also why do planets revolve around the sun, why don't they shoot away
> or drop onto the surface due to gravitation?
Because that's the effect of the inverse square law of
gravitation: that paths are elliptical if closed, hyperbolic
if open. You'd only get falling straight in if the planet
happened to have no motion relative to the sun as it formed,
in which case the gas would not have been out there long
enough to condense anyway. If the planet formed from condensation
of a gas cloud, that cloud itself was in orbit.
> Is it just a coincedence that they are in circular motion?
Circles as opposed to ellipses? No, that almost certainly has
something to do with how they form.
Closed orbits as opposed to crashing into the sun? No, not
coincidence. Almost any initial conditions you choose,
so long as you don't have escape velocity, are going to
result in a closed elliptical orbit. It's just what you
get when you start something moving in an inverse square
law.
> Or is it
> their intial kinetic energy which doesn't let them drop onto the surface
> and the gravitation which doesn't lets them escape?
I guess that's one way of putting "this is what the inverse
square law predicts".
- Randy
Tom Roberts
Aug26-04, 04:07 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nsurendra wrote:\n> And why does the atmosphere rotates with earth?\n\nDrag from the surface is the primary driving force, augmented and\nmodified by the curious phenomenon we call "weather".\n\n\n> Does the altitiude makes any\n> differene to the speed relative to the surface of earth?\n\nYes. A modest difference (we rarely see 100 MPH winds within a few\nhundred feet of the surface; in the jet stream they\'re common). But this\nis small compared to the ~600-1000 MPH rotation of the surface. The\naltitude differential is due to the interaction of several effects,\nincluding weather, differential solar heating, pressure differentials\n(both vertical and horizontal), ionization, etc.\n\nThis modest difference for different altitudes is significant for\nlong-distance airplanes, and can affect flight times across the Atlantic\nby an hour or so (but there are other effects involved, because airline\nselect different routes eastbound and westbound, precisely to take\nadvantage of varying wind patterns).\n\n\n> Also why does planets revolve around the sun, why don\'t they shoot away\n> or drop onto the surface due to gravitation?\n\nBecause each planet must follow its local geodesic through spacetime.\nFor all planets, the geodesic encircles the sun in an excellent\napproximation to an ellipse in space. If this were not true we would not\ncall it a planet (c.f. comets, for which the ellipse is much more\neccentric; there are also non-periodic geodesics which simply pass\nthrough the solar system never to return, but we have no name for\nobjects on such trajectories [why this is so has nothing to do with\nphysics]).\n\n\nTom Roberts tjroberts@lucent.com\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>surendra wrote:
> And why does the atmosphere rotates with earth?
Drag from the surface is the primary driving force, augmented and
modified by the curious phenomenon we call "weather".
> Does the altitiude makes any
> differene to the speed relative to the surface of earth?
Yes. A modest difference (we rarely see 100 MPH winds within a few
hundred feet of the surface; in the jet stream they're common). But this
is small compared to the ~600-1000 MPH rotation of the surface. The
altitude differential is due to the interaction of several effects,
including weather, differential solar heating, pressure differentials
(both vertical and horizontal), ionization, etc.
This modest difference for different altitudes is significant for
long-distance airplanes, and can affect flight times across the Atlantic
by an hour or so (but there are other effects involved, because airline
select different routes eastbound and westbound, precisely to take
advantage of varying wind patterns).
> Also why does planets revolve around the sun, why don't they shoot away
> or drop onto the surface due to gravitation?
Because each planet must follow its local geodesic through spacetime.
For all planets, the geodesic encircles the sun in an excellent
approximation to an ellipse in space. If this were not true we would not
call it a planet (c.f. comets, for which the ellipse is much more
eccentric; there are also non-periodic geodesics which simply pass
through the solar system never to return, but we have no name for
objects on such trajectories [why this is so has nothing to do with
physics]).
Tom Roberts tjroberts@lucent.com
Androcles
Aug26-04, 04:29 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"surendra" <efuzzyone@netscape.net> wrote in message\nnews:cgg1kn\\$2t6\\$1@news.asu.edu...\n| I am not a physics expert(I have just taken high school physics) but I\n| have some serious objections to your argument.\n|\n| Edward Green wrote:\n| > But anyway, pedantically, we might mention one effect tending to agree\n| > with the OP\'s friend: say an object ascended 1000km straight up into\n| > space over London, with no other acceleration, hung out, and\n| > eventually descended. If it hung out long enough, it would find\n| > itself descending in Canada somewhere, since the tangential velocity\n| > it had on the Earth\'s surface would not be the tangential velocity\n| > corresponding to the same 2pi/24hr rotation rate at 1000km altitude;\n| > so the Earth would turn under it, thus somewhat confirming to the idea\n| > of the OP\'s friend. Of course, we could also describe the same\n| > trajectory relative to the rotating Earth by the "Coriolos force".\n| >\n| > I\'m not claiming this is significant for aircraft.\n| >\n| > <...>\n| >\n|\n|\n| When the object is on the surface of earth it has certain angular\n| velocity relative to the center of earth. As the object ascends,assuming\n| there is only a vertical force acting along a line from the center of\n| earth to the CG of the object, the angular velocity relative to the\n| center of earth should remain constant.\n| The tangential velocity will increase\n\nWHOA!!\nWhere does that idea come from?\nAngular *momentum* is conserved. That doesn\'t mean\ntangential velocity is.\nWhen an ice skater pulls her outspread arms into her body\nshe spins faster, and the reverse is true.\nYou are confusing conservation of momentum with some\nimagined conservation of velocity.\nAndrocles\n\n\n\n\n\nbut so does the circumference of\n| earth as the altitude increases. The object will have to travel a\n| greater distance relative to the center of earth compared to the land of\n| London. And this increase in tangential velocity is what helps the\n| object to be in London and not recede to Asia.\n| I hope, you are getting my point.\n|\n| So, if the object was initially at London, it will remain in the sky\n| above London, and it will never reach Canada.\n|\n| - Surendra\n|\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"surendra" <efuzzyone@netscape.net> wrote in message
news:cgg1kn$2t6$1@news.asu.edu...
| I am not a physics expert(I have just taken high school physics) but I
| have some serious objections to your argument.
|
| Edward Green wrote:
| > But anyway, pedantically, we might mention one effect tending to agree
| > with the OP's friend: say an object ascended 1000km straight up into
| > space over London, with no other acceleration, hung out, and
| > eventually descended. If it hung out long enough, it would find
| > itself descending in Canada somewhere, since the tangential velocity
| > it had on the Earth's surface would not be the tangential velocity
| > corresponding to the same 2pi/24hr rotation rate at 1000km altitude;
| > so the Earth would turn under it, thus somewhat confirming to the idea
| > of the OP's friend. Of course, we could also describe the same
| > trajectory relative to the rotating Earth by the "Coriolos force".
| >
| > I'm not claiming this is significant for aircraft.
| >
| > <...>
| >
|
|
| When the object is on the surface of earth it has certain angular
| velocity relative to the center of earth. As the object ascends,assuming
| there is only a vertical force acting along a line from the center of
| earth to the CG of the object, the angular velocity relative to the
| center of earth should remain constant.
| The tangential velocity will increase
WHOA!!
Where does that idea come from?
Angular *momentum* is conserved. That doesn't mean
tangential velocity is.
When an ice skater pulls her outspread arms into her body
she spins faster, and the reverse is true.
You are confusing conservation of momentum with some
imagined conservation of velocity.
Androcles
but so does the circumference of
| earth as the altitude increases. The object will have to travel a
| greater distance relative to the center of earth compared to the land of
| London. And this increase in tangential velocity is what helps the
| object to be in London and not recede to Asia.
| I hope, you are getting my point.
|
| So, if the object was initially at London, it will remain in the sky
| above London, and it will never reach Canada.
|
| - Surendra
|
Richard Saam
Aug26-04, 04:30 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>This is a multi-part message in MIME format.\n--------------040307030606090307070407\nContent-Type: text/plain; charset=us-ascii; format=flowed\nContent-Transfer-Encoding: 7bit\n\n\n\nJohn T Lowry wrote:\n\n>\n>Here is a snip from an article I wrote on this subject. The Physics\n>Teacher rejected it because it was too complicated, suggesting I send it\n>to the American Journal of Physics. AMJ rejected it because it was too\n>simple! Oh well....\n>\n>The airplanes (one headed straight east, the other straight west, were\n>taken over the equator, no wind, at altitude 2000 m, speed 200 m/s (389\n>KTAS).\n>***\n>Because centrifugal terms for the eastbound and westbound airplanes are\n>identical (since that term depends only on the rotation of the\n>non-inertial frame and the common location of the two objects), it might\n>appear there is no dynamical difference between the airplanes. But\n>("just to be on the safe side") compute the Coriolis terms. On the\n>eastbound airplane E, the Coriolis acceleration is 2SV outwards (to\n>airplane E, upwards). On the westbound airplane W, 2SV inwards\n>(downwards to airplane W). Unconstrained airplane W would be forced down\n>by this Coriolis force and so to counteract W\'s higher effective\n>acceleration of gravity - higher than E\'s by 4SV = 0.0583 m sec!2 -\n>airplane W requires more lift. Higher lift is accomplished by W\'s taking\n>a slightly higher angle of attack, the angle with which its wings meet\n>the relative wind. But more lift means more induced drag (drag due to\n>lift), hence - everything else being equal, as we have specified - a\n>slightly higher fuel consumption rate for the westbound airplane. Since\n>the acceleration of gravity near the equator is 9.780 m sec!2, there is\n>a lifting acceleration difference of 0.0060 g. For a 10,000 lbf airplane\n>(the scales having read somewhat light because of the centrifugal\n>pseudoforce), perhaps at this air speed a World War II fighter, that is\n>an effective weight reduction of 60 lbf just for having turned from\n>westbound to eastbound. In practice, however, negligible. We apologize\n>that we don\'t have space here for a meaningful discussion of induced\n>drag.\n>\n>***\n>\n>\n>\n>John T. Lowry, PhD\n>\n>Flight Physics\n>\n>\n>\nOn the subject of Coriolis Correction, the inside back cover of \'AIR\nALMANAC\' published by The United States Naval Observatory indicates a\ntable to correct aircraft celestial bubble sextant fixes for coriolis\neffect. The table heading states "To be applied by moving the position\nline a distance Z to starboard (right) of the track in northern\nlatitudes and to port (left) in southern latitudes". Corrections in\nnautical miles are in terms of Latitude and Ground Speed in Knots. The\ncorrection is zero at the equator for all Ground Speeds proceeding\ntowards maximum at the poles. At 45 degree latitude and 500 knots the\ncorrection is 9 nautical miles.\n\nAssuming an aircraft moving east at 45 degree north latitude, the 9\nnautical mile correction would be applied to the south.\n\nAssuming an aircraft moving west at 45 degree north latitude, the 9\nnautical mile correction would be applied to the north.\n\nLogic would indicate that\n(cos(45 + 9/60) - cos(45 - 9/60)) / cos(45)\nbe representative of difference in pseudo distance traveled by aircraft\ndue to coriolus effect.\n\nThe value is .00524 (.524%)\n\nSo for a long trip of say 4000 nautical miles there would be pseudo\ndistance difference of 21 nautical miles.\n\nThere should be no effect on opposing great circle routes.\n\nRichard\n\n>\n>\n\n\nThe airplanes (one headed straight east, the other straight west, were\ntaken over the equator, no wind, at altitude 2000 m, speed 200 m/s (389\nKTAS).\n***\nBecause centrifugal terms for the eastbound and westbound airplanes are\nidentical (since that term depends only on the rotation of the\nnon-inertial frame and the common location of the two objects), it might\nappear there is no dynamical difference between the airplanes. But\n("just to be on the safe side") compute the Coriolis terms. On the\neastbound airplane E, the Coriolis acceleration is 2SV outwards (to\nairplane E, upwards). On the westbound airplane W, 2SV inwards\n(downwards to airplane W). Unconstrained airplane W would be forced down\nby this Coriolis force and so to counteract W\'s higher effective\nacceleration of gravity - higher than E\'s by 4SV = 0.0583 m sec!2 -\nairplane W requires more lift. Higher lift is accomplished by W\'s taking\na slightly higher angle of attack, the angle with which its wings meet\nthe relative wind. But more lift means more induced drag (drag due to\nlift), hence - everything else being equal, as we have specified - a\nslightly higher fuel consumption rate for the westbound airplane. Since\nthe acceleration of gravity near the equator is 9.780 m sec!2, there is\na lifting acceleration difference of 0.0060 g. For a 10,000 lbf airplane\n(the scales having read somewhat light because of the centrifugal\npseudoforce), perhaps at this air speed a World War II fighter, that is\nan effective weight reduction of 60 lbf just for having turned from\nwestbound to eastbound. In practice, however, negligible. We apologize\nthat we don\'t have space here for a meaningful discussion of induced\ndrag.\n\n***\n\n\n\nJohn T. Lowry, PhD\n\nFlight Physics\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>This is a multi-part message in MIME format.
--------------040307030606090307070407
Content-Type: text/plain; charset=us-ascii; format=flowed
Content-Transfer-Encoding: 7bit
John T Lowry wrote:
>
>Here is a snip from an article I wrote on this subject. The Physics
>Teacher rejected it because it was too complicated, suggesting I send it
>to the American Journal of Physics. AMJ rejected it because it was too
>simple! Oh well....
>
>The airplanes (one headed straight east, the other straight west, were
>taken over the equator, no wind, at altitude 2000 m, speed 200 m/s (389
>KTAS).
>***
>Because centrifugal terms for the eastbound and westbound airplanes are
>identical (since that term depends only on the rotation of the
>non-inertial frame and the common location of the two objects), it might
>appear there is no dynamical difference between the airplanes. But
>("just to be on the safe side") compute the Coriolis terms. On the
>eastbound airplane E, the Coriolis acceleration is 2SV outwards (to
>airplane E, upwards). On the westbound airplane W, 2SV inwards
>(downwards to airplane W). Unconstrained airplane W would be forced down
>by this Coriolis force and so to counteract W's higher effective
>acceleration of gravity - higher than E's by 4SV = .0583 m sec!2 -
>airplane W requires more lift. Higher lift is accomplished by W's taking
>a slightly higher angle of attack, the angle with which its wings meet
>the relative wind. But more lift means more induced drag (drag due to
>lift), hence - everything else being equal, as we have specified - a
>slightly higher fuel consumption rate for the westbound airplane. Since
>the acceleration of gravity near the equator is 9.780 m sec!2, there is
>a lifting acceleration difference of .0060 g. For a 10,000 lbf airplane
>(the scales having read somewhat light because of the centrifugal
>pseudoforce), perhaps at this air speed a World War II fighter, that is
>an effective weight reduction of 60 lbf just for having turned from
>westbound to eastbound. In practice, however, negligible. We apologize
>that we don't have space here for a meaningful discussion of induced
>drag.
>
>***
>
>
>
>John T. Lowry, PhD
>
>Flight Physics
>
>
>
On the subject of Coriolis Correction, the inside back cover of 'AIR
ALMANAC' published by The United States Naval Observatory indicates a
table to correct aircraft celestial bubble sextant fixes for coriolis
effect. The table heading states "To be applied by moving the position
line a distance Z to starboard (right) of the track in northern
latitudes and to port (left) in southern latitudes". Corrections in
nautical miles are in terms of Latitude and Ground Speed in Knots. The
correction is zero at the equator for all Ground Speeds proceeding
towards maximum at the poles. At 45 degree latitude and 500 knots the
correction is 9 nautical miles.
Assuming an aircraft moving east at 45 degree north latitude, the 9
nautical mile correction would be applied to the south.
Assuming an aircraft moving west at 45 degree north latitude, the 9
nautical mile correction would be applied to the north.
Logic would indicate that
(cos(45 + 9/60) - cos(45 - 9/60)) / cos(45)
be representative of difference in pseudo distance traveled by aircraft
due to coriolus effect.
The value is .00524 (.524%)
So for a long trip of say 4000 nautical miles there would be pseudo
distance difference of 21 nautical miles.
There should be no effect on opposing great circle routes.
Richard
>
>
The airplanes (one headed straight east, the other straight west, were
taken over the equator, no wind, at altitude 2000 m, speed 200 m/s (389
KTAS).
***
Because centrifugal terms for the eastbound and westbound airplanes are
identical (since that term depends only on the rotation of the
non-inertial frame and the common location of the two objects), it might
appear there is no dynamical difference between the airplanes. But
("just to be on the safe side") compute the Coriolis terms. On the
eastbound airplane E, the Coriolis acceleration is 2SV outwards (to
airplane E, upwards). On the westbound airplane W, 2SV inwards
(downwards to airplane W). Unconstrained airplane W would be forced down
by this Coriolis force and so to counteract W's higher effective
acceleration of gravity - higher than E's by 4SV = .0583 m sec!2 -
airplane W requires more lift. Higher lift is accomplished by W's taking
a slightly higher angle of attack, the angle with which its wings meet
the relative wind. But more lift means more induced drag (drag due to
lift), hence - everything else being equal, as we have specified - a
slightly higher fuel consumption rate for the westbound airplane. Since
the acceleration of gravity near the equator is 9.780 m sec!2, there is
a lifting acceleration difference of .0060 g. For a 10,000 lbf airplane
(the scales having read somewhat light because of the centrifugal
pseudoforce), perhaps at this air speed a World War II fighter, that is
an effective weight reduction of 60 lbf just for having turned from
westbound to eastbound. In practice, however, negligible. We apologize
that we don't have space here for a meaningful discussion of induced
drag.
***
John T. Lowry, PhD
Flight Physics
surendra
Aug27-04, 02:42 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>This is in contradiction with the earlier posts by other people(if I am\nunderstanding things correctly).\nCan I know your qualification in physics?\nWhat do you mean by modest?\nDo you intend to mean that the top layer of atmosphere rotates around\nthe earth because of different relative angular velocities?\nDoes the surface of earth also does this rotation?\n\n\n-Surendra\n\nTom Roberts wrote:\n\n> surendra wrote:\n>\n>>And why does the atmosphere rotates with earth?\n>\n>\n> Drag from the surface is the primary driving force, augmented and\n> modified by the curious phenomenon we call "weather".\n>\n>\n>\n>>Does the altitiude makes any\n>>differene to the speed relative to the surface of earth?\n>\n>\n> Yes. A modest difference (we rarely see 100 MPH winds within a few\n> hundred feet of the surface; in the jet stream they\'re common). But this\n> is small compared to the ~600-1000 MPH rotation of the surface. The\n> altitude differential is due to the interaction of several effects,\n> including weather, differential solar heating, pressure differentials\n> (both vertical and horizontal), ionization, etc.\n>\n> This modest difference for different altitudes is significant for\n> long-distance airplanes, and can affect flight times across the Atlantic\n> by an hour or so (but there are other effects involved, because airline\n> select different routes eastbound and westbound, precisely to take\n> advantage of varying wind patterns).\n>\n>\n>\n>>Also why does planets revolve around the sun, why don\'t they shoot away\n>>or drop onto the surface due to gravitation?\n>\n>\n> Because each planet must follow its local geodesic through spacetime.\n> For all planets, the geodesic encircles the sun in an excellent\n> approximation to an ellipse in space. If this were not true we would not\n> call it a planet (c.f. comets, for which the ellipse is much more\n> eccentric; there are also non-periodic geodesics which simply pass\n> through the solar system never to return, but we have no name for\n> objects on such trajectories [why this is so has nothing to do with\n> physics]).\n>\n>\n> Tom Roberts tjroberts@lucent.com\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>This is in contradiction with the earlier posts by other people(if I am
understanding things correctly).
Can I know your qualification in physics?
What do you mean by modest?
Do you intend to mean that the top layer of atmosphere rotates around
the earth because of different relative angular velocities?
Does the surface of earth also does this rotation?
-Surendra
Tom Roberts wrote:
> surendra wrote:
>
>>And why does the atmosphere rotates with earth?
>
>
> Drag from the surface is the primary driving force, augmented and
> modified by the curious phenomenon we call "weather".
>
>
>
>>Does the altitiude makes any
>>differene to the speed relative to the surface of earth?
>
>
> Yes. A modest difference (we rarely see 100 MPH winds within a few
> hundred feet of the surface; in the jet stream they're common). But this
> is small compared to the ~600-1000 MPH rotation of the surface. The
> altitude differential is due to the interaction of several effects,
> including weather, differential solar heating, pressure differentials
> (both vertical and horizontal), ionization, etc.
>
> This modest difference for different altitudes is significant for
> long-distance airplanes, and can affect flight times across the Atlantic
> by an hour or so (but there are other effects involved, because airline
> select different routes eastbound and westbound, precisely to take
> advantage of varying wind patterns).
>
>
>
>>Also why does planets revolve around the sun, why don't they shoot away
>>or drop onto the surface due to gravitation?
>
>
> Because each planet must follow its local geodesic through spacetime.
> For all planets, the geodesic encircles the sun in an excellent
> approximation to an ellipse in space. If this were not true we would not
> call it a planet (c.f. comets, for which the ellipse is much more
> eccentric; there are also non-periodic geodesics which simply pass
> through the solar system never to return, but we have no name for
> objects on such trajectories [why this is so has nothing to do with
> physics]).
>
>
> Tom Roberts tjroberts@lucent.com
surendra
Aug27-04, 02:42 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>> WHOA!!\n> Where does that idea come from?\n> Angular *momentum* is conserved. That doesn\'t mean\n> tangential velocity is.\n> When an ice skater pulls her outspread arms into her body\n> she spins faster, and the reverse is true.\n> You are confusing conservation of momentum with some\n> imagined conservation of velocity.\n> Androcles\n>\n\n\nIf the angular momentum is conserved and if the mass remains the same\nthen so does the angular velocity.\n\nAng. Momentum = mass * angular velocity\n\nWhen a spinning ice skater pulls her arm close to her body then does the\nangular velocity of her body changes?\n\n-Surendra\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>> WHOA!!
> Where does that idea come from?
> Angular *momentum* is conserved. That doesn't mean
> tangential velocity is.
> When an ice skater pulls her outspread arms into her body
> she spins faster, and the reverse is true.
> You are confusing conservation of momentum with some
> imagined conservation of velocity.
> Androcles
>
If the angular momentum is conserved and if the mass remains the same
then so does the angular velocity.
Ang. Momentum = mass * angular velocity
When a spinning ice skater pulls her arm close to her body then does the
angular velocity of her body changes?
-Surendra
Androcles
Aug28-04, 04:26 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"surendra" <efuzzyone@netscape.net> wrote in message\nnews:cgm6uh\\$83d\\$1@news.asu.edu...\n| > WHOA!!\n| > Where does that idea come from?\n| > Angular *momentum* is conserved. That doesn\'t mean\n| > tangential velocity is.\n| > When an ice skater pulls her outspread arms into her body\n| > she spins faster, and the reverse is true.\n| > You are confusing conservation of momentum with some\n| > imagined conservation of velocity.\n| > Androcles\n| >\n|\n|\n| If the angular momentum is conserved and if the mass remains the same\n| then so does the angular velocity.\n|\n| Ang. Momentum = mass * angular velocity\n|\n| When a spinning ice skater pulls her arm close to her body then does the\n| angular velocity of her body changes?\n\n| -Surendra\n\n\nTry this simple test, particularly Question 7.\nhttp://www.lboro.ac.uk/faculty/eng/engtlsc/Eng_Mech/tutorials/tut15_5.htm\nAndrocles.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"surendra" <efuzzyone@netscape.net> wrote in message
news:cgm6uh$83d$1@news.asu.edu...
| > WHOA!!
| > Where does that idea come from?
| > Angular *momentum* is conserved. That doesn't mean
| > tangential velocity is.
| > When an ice skater pulls her outspread arms into her body
| > she spins faster, and the reverse is true.
| > You are confusing conservation of momentum with some
| > imagined conservation of velocity.
| > Androcles
| >
|
|
| If the angular momentum is conserved and if the mass remains the same
| then so does the angular velocity.
|
| Ang. Momentum = mass * angular velocity
|
| When a spinning ice skater pulls her arm close to her body then does the
| angular velocity of her body changes?
| -Surendra
Try this simple test, particularly Question 7.
http://www.lboro.ac.uk/faculty/eng/engtlsc/Eng_Mech/tutorials/tut15_5.htm
Androcles.
surendra
Aug28-04, 04:30 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Androcles wrote:\n> "surendra" <efuzzyone@netscape.net> wrote in message\n> news:cgg1kn\\$2t6\\$1@news.asu.edu...\n> | I am not a physics expert(I have just taken high school physics) but I\n> | have some serious objections to your argument.\n> |\n> | Edward Green wrote:\n> | > But anyway, pedantically, we might mention one effect tending to agree\n> | > with the OP\'s friend: say an object ascended 1000km straight up into\n> | > space over London, with no other acceleration, hung out, and\n> | > eventually descended. If it hung out long enough, it would find\n> | > itself descending in Canada somewhere, since the tangential velocity\n> | > it had on the Earth\'s surface would not be the tangential velocity\n> | > corresponding to the same 2pi/24hr rotation rate at 1000km altitude;\n> | > so the Earth would turn under it, thus somewhat confirming to the idea\n> | > of the OP\'s friend. Of course, we could also describe the same\n> | > trajectory relative to the rotating Earth by the "Coriolos force".\n> | >\n> | > I\'m not claiming this is significant for aircraft.\n> | >\n> | > <...>\n> | >\n> |\n> |\n> | When the object is on the surface of earth it has certain angular\n> | velocity relative to the center of earth. As the object ascends,assuming\n> | there is only a vertical force acting along a line from the center of\n> | earth to the CG of the object, the angular velocity relative to the\n> | center of earth should remain constant.\n> | The tangential velocity will increase\n>\n> WHOA!!\n> Where does that idea come from?\n> Angular *momentum* is conserved. That doesn\'t mean\n> tangential velocity is.\n> When an ice skater pulls her outspread arms into her body\n> she spins faster, and the reverse is true.\n> You are confusing conservation of momentum with some\n> imagined conservation of velocity.\n> Androcles\n\n\nMoreover when the tangential velocity increases doesn\'t it has to cover\nmore distance as the circumference increases? So, doesn\'t the increase\nin tangential velocity and the increase in distance balance each other?\n\n-Surendra\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Androcles wrote:
> "surendra" <efuzzyone@netscape.net> wrote in message
> news:cgg1kn$2t6$1@news.asu.edu...
> | I am not a physics expert(I have just taken high school physics) but I
> | have some serious objections to your argument.
> |
> | Edward Green wrote:
> | > But anyway, pedantically, we might mention one effect tending to agree
> | > with the OP's friend: say an object ascended 1000km straight up into
> | > space over London, with no other acceleration, hung out, and
> | > eventually descended. If it hung out long enough, it would find
> | > itself descending in Canada somewhere, since the tangential velocity
> | > it had on the Earth's surface would not be the tangential velocity
> | > corresponding to the same 2pi/24hr rotation rate at 1000km altitude;
> | > so the Earth would turn under it, thus somewhat confirming to the idea
> | > of the OP's friend. Of course, we could also describe the same
> | > trajectory relative to the rotating Earth by the "Coriolos force".
> | >
> | > I'm not claiming this is significant for aircraft.
> | >
> | > <...>
> | >
> |
> |
> | When the object is on the surface of earth it has certain angular
> | velocity relative to the center of earth. As the object ascends,assuming
> | there is only a vertical force acting along a line from the center of
> | earth to the CG of the object, the angular velocity relative to the
> | center of earth should remain constant.
> | The tangential velocity will increase
>
> WHOA!!
> Where does that idea come from?
> Angular *momentum* is conserved. That doesn't mean
> tangential velocity is.
> When an ice skater pulls her outspread arms into her body
> she spins faster, and the reverse is true.
> You are confusing conservation of momentum with some
> imagined conservation of velocity.
> Androcles
Moreover when the tangential velocity increases doesn't it has to cover
more distance as the circumference increases? So, doesn't the increase
in tangential velocity and the increase in distance balance each other?
-Surendra
Tom Roberts
Aug28-04, 08:43 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nsurendra wrote:\n> [referring to my response]\n> This is in contradiction with the earlier posts by other people(if I am\n> understanding things correctly).\n\nHmmm. There are a lot of people around here who post nonsense\nmasquerading as physics. <shrug>\n\n\n> Can I know your qualification in physics?\n\nI earned in Ph.D. in physics in 1975, and am currently working in the\nfield of high energy physics (elementary particles).\n\n\n> What do you mean by modest?\n\nIn the context below, a few hundred MPH difference is "modest" compared\nto the 600-1000 MPH rotation of the earth wrt the Earth-Centered\nInertial (ECI) frame.\n\n\n> Do you intend to mean that the top layer of atmosphere rotates around\n> the earth because of different relative angular velocities?\n\nIf the upper atmosphere has a consistently-directed wind wrt the\nsurface, then it must have a different (average) angular velocity than\nthe surface. A look at a weather map shows that the upper atmosphere\ndoes indeed have a consistent velocity from west to east wrt the surface\n(at least where I live in the northern hemisphere).\n\nAs I said before, this is all modified by the curious phenomenon we call\n"weather".\n\n\n> Does the surface of earth also does this rotation?\n\nYes, the earth rotates. Where have you been living for the past 4\ncenturies???\n\n\nTom Roberts tjroberts@lucent.com\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>surendra wrote:
> [referring to my response]
> This is in contradiction with the earlier posts by other people(if I am
> understanding things correctly).
Hmmm. There are a lot of people around here who post nonsense
masquerading as physics. <shrug>
> Can I know your qualification in physics?
I earned in Ph.D. in physics in 1975, and am currently working in the
field of high energy physics (elementary particles).
> What do you mean by modest?
In the context below, a few hundred MPH difference is "modest" compared
to the 600-1000 MPH rotation of the earth wrt the Earth-Centered
Inertial (ECI) frame.
> Do you intend to mean that the top layer of atmosphere rotates around
> the earth because of different relative angular velocities?
If the upper atmosphere has a consistently-directed wind wrt the
surface, then it must have a different (average) angular velocity than
the surface. A look at a weather map shows that the upper atmosphere
does indeed have a consistent velocity from west to east wrt the surface
(at least where I live in the northern hemisphere).
As I said before, this is all modified by the curious phenomenon we call
"weather".
> Does the surface of earth also does this rotation?
Yes, the earth rotates. Where have you been living for the past 4
centuries???
Tom Roberts tjroberts@lucent.com
Franz Heymann
Aug28-04, 08:43 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n"surendra" <efuzzyone@netscape.net> wrote in message\nnews:cgm6uh\\$83d\\$1@news.asu.edu...\n\n[snip]\n\n> Ang. Momentum = mass * angular velocity\n\nNo.\n\nFranz\n\n[Moderator\'s note: For the definition see for instance\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/amom.html -usc]\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"surendra" <efuzzyone@netscape.net> wrote in message
news:cgm6uh$83d$1@news.asu.edu...
[snip]
> Ang. Momentum = mass * angular velocity
No.
Franz
[Moderator's note: For the definition see for instance
http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html -usc]
Edward Green
Aug29-04, 01:38 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0408241400.11130dcc@posting.google. com>...\n> spamspamspam3@netzero.com (Edward Green) wrote in message news:<eca320d0.0408212025.4493b8e7@posting.google. com>...\n> > poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0408200950.5e8baf79@posting.google. com>...\n> > Try as I might, I can\'t make "west to east" correspond to flying from\n> > London to New York.\n>\n> Nitpicker.\n\nWell, not 100%. Because of my residual childhood uncertainty about\nthe sense of certain expressions involving the compass, 95% nitpick,\n5% idiot check.\n\n> > I\'m not sure if this is an irregularity in my\n> > understanding of compass directions (for a long time I had trouble\n> > with the idea that the "west wind" was out of the west, rather than\n> > heading west), or a typo ...\n>\n> Typo, of course.\n\nPhew. ;-)\n\nBut now, I have a _genuine_ 99.5% pure nitpick. When you write:\n\n> The point is that if there weren\'t that\n> pesky air to deal with, earth rotation would cause the\n> trip from New York to London to be longer than London to\n> New York, since the former involves chasing London as it\n> rotates around, and the latter involves New York rushing\n> toward you.\n\nI would emphasize, for the fictitious system where you are sitting on\nthe surface of a vacuum covered Earth, that this is only true if there\nis an altitude variation, as per my learned comment aforesaid about\nthe Coriol<i>s force. The point is, if you start sitting on the\nsurface, you _already_ have the appropriate tangential speed for that\nheight, and if you used purely surface transportation neither London\nnor New York would be rushing towards you nor coursing from you like a\nhound, but behaving like a perfectly reasonable dweller on an infinite\nplane.\n\nThat is, if they were really on the same lattitude, which they aren\'t.\n_Damn_! Perfectly good first order nitpick ruined by second-order\nnitpick!\n\nIgnoring that annoyance we note that it is the inappropriateness of\nyour original tangential speed for your temporary altitude change --\nwhich can be expressed more learnedly in terms of the oolie-ish\nCoriolis force -- which accounts for all rushing and coursing in terms\nof your local quasi-inertial rest frame. Note we could reverse the\nmagical London NY transit using only altitude changes by dropping into\nan appropriately curved tunnel at NY.\n\nNow you have me curious about the lattitude effect. London is at a\nhigher lattitude than NY, hence, smaller tangential speed. So we\ncould say some similar things about latitude and north/south travel as\naltitude and up/down travel, which I leave to the reader.\n\n> But because of the winds, New York-London is *shorter* than\n> London-New York.\n>\n> > But anyway, pedantically, we might mention one effect tending to agree\n> > with the OP\'s friend: say an object ascended 1000km straight up into\n> > space over London, with no other acceleration, hung out, and\n> > eventually descended. If it hung out long enough, it would find\n> > itself descending in Canada somewhere, since the tangential velocity\n> > it had on the Earth\'s surface would not be the tangential velocity\n> > corresponding to the same 2pi/24hr rotation rate at 1000km altitude;\n> > so the Earth would turn under it, thus somewhat confirming to the idea\n> > of the OP\'s friend. Of course, we could also describe the same\n> > trajectory relative to the rotating Earth by the "Coriolos force".\n> >\n> > I\'m not claiming this is significant for aircraft.\n>\n> Only in so far as the coriolis gives rise to air flow\n> patterns.\n\nWell, not entirely: even if there we no interaction with the\natmosphere, we would still have to contend with the altitude and\nlatitude dependent versions of the Coriolis force, as above. In fact,\nnow I\'m not so ready to dismiss this as insignicant: the altitude\nchange may be trivial as a fraction of the Earth\'s radius, but the\nchange in the Earth\'s effective radius about its axis of rotation\nbetween London and NY is not -- though it may have negligable effect\ncompared to local motion of the atmophere. But academically speaking,\nif you integrated the reactive force supplied by your jets, and\nsubtracted the integral of the force of air resistance, you should\nfind a discrepancy in your calculated initial and final velocities\nover ground (zero on the tarmac).\n\nTurns out the "second order nitpick" is the significant effect.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0408241400.11130dcc@posting.google.com>...
> spamspamspam3@netzero.com (Edward Green) wrote in message news:<eca320d0.0408212025.4493b8e7@posting.google.com>...
> > poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0408200950.5e8baf79@posting.google.com>...
> > Try as I might, I can't make "west to east" correspond to flying from
> > London to New York.
>
> Nitpicker.
Well, not 100%. Because of my residual childhood uncertainty about
the sense of certain expressions involving the compass, 95% nitpick,
5% idiot check.
> > I'm not sure if this is an irregularity in my
> > understanding of compass directions (for a long time I had trouble
> > with the idea that the "west wind" was out of the west, rather than
> > heading west), or a typo ...
>
> Typo, of course.
Phew. ;-)
But now, I have a _genuine_ 99.5% pure nitpick. When you write:
> The point is that if there weren't that
> pesky air to deal with, earth rotation would cause the
> trip from New York to London to be longer than London to
> New York, since the former involves chasing London as it
> rotates around, and the latter involves New York rushing
> toward you.
I would emphasize, for the fictitious system where you are sitting on
the surface of a vacuum covered Earth, that this is only true if there
is an altitude variation, as per my learned comment aforesaid about
the Coriol<i>s force. The point is, if you start sitting on the
surface, you _already_ have the appropriate tangential speed for that
height, and if you used purely surface transportation neither London
nor New York would be rushing towards you nor coursing from you like a
hound, but behaving like a perfectly reasonable dweller on an infinite
plane.
That is, if they were really on the same lattitude, which they aren't.
_Damn_! Perfectly good first order nitpick ruined by second-order
nitpick!
Ignoring that annoyance we note that it is the inappropriateness of
your original tangential speed for your temporary altitude change --
which can be expressed more learnedly in terms of the oolie-ish
Coriolis force -- which accounts for all rushing and coursing in terms
of your local quasi-inertial rest frame. Note we could reverse the
magical London NY transit using only altitude changes by dropping into
an appropriately curved tunnel at NY.
Now you have me curious about the lattitude effect. London is at a
higher lattitude than NY, hence, smaller tangential speed. So we
could say some similar things about latitude and north/south travel as
altitude and up/down travel, which I leave to the reader.
> But because of the winds, New York-London is *shorter* than
> London-New York.
>
> > But anyway, pedantically, we might mention one effect tending to agree
> > with the OP's friend: say an object ascended 1000km straight up into
> > space over London, with no other acceleration, hung out, and
> > eventually descended. If it hung out long enough, it would find
> > itself descending in Canada somewhere, since the tangential velocity
> > it had on the Earth's surface would not be the tangential velocity
> > corresponding to the same 2pi/24hr rotation rate at 1000km altitude;
> > so the Earth would turn under it, thus somewhat confirming to the idea
> > of the OP's friend. Of course, we could also describe the same
> > trajectory relative to the rotating Earth by the "Coriolos force".
> >
> > I'm not claiming this is significant for aircraft.
>
> Only in so far as the coriolis gives rise to air flow
> patterns.
Well, not entirely: even if there we no interaction with the
atmosphere, we would still have to contend with the altitude and
latitude dependent versions of the Coriolis force, as above. In fact,
now I'm not so ready to dismiss this as insignicant: the altitude
change may be trivial as a fraction of the Earth's radius, but the
change in the Earth's effective radius about its axis of rotation
between London and NY is not -- though it may have negligable effect
compared to local motion of the atmophere. But academically speaking,
if you integrated the reactive force supplied by your jets, and
subtracted the integral of the force of air resistance, you should
find a discrepancy in your calculated initial and final velocities
over ground (zero on the tarmac).
Turns out the "second order nitpick" is the significant effect.
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