View Full Version : Rocket engine efficiency
Brad Eckert
Jun17-04, 04:14 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>Hi all,\n\nAFAIK, the acceleration of a rocket is constant. This is opposed to\nthe acceleration of a car, which decreases as the transmission\nswitches to higher gears.\n\nIf acceleration is constant, velocity is proportional to time and the\nkinetic energy of the rocket goes as the square of the flight time. At\nsome point, the per-second increase in the kinetic energy of the\nrocket would exceed the energy provided by burning the fuel.\n\nSince that would violate the laws of thermodynamics, this reasoning\nhas to break down somewhere. It could be that:\n\n1. The acceleration isn\'t constant. Or...\n2. The rocket can\'t carry enough fuel to reach this condition.\n\nCan someone explain this paradox?\n\n--\nBrad\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 all,
AFAIK, the acceleration of a rocket is constant. This is opposed to
the acceleration of a car, which decreases as the transmission
switches to higher gears.
If acceleration is constant, velocity is proportional to time and the
kinetic energy of the rocket goes as the square of the flight time. At
some point, the per-second increase in the kinetic energy of the
rocket would exceed the energy provided by burning the fuel.
Since that would violate the laws of thermodynamics, this reasoning
has to break down somewhere. It could be that:
1. The acceleration isn't constant. Or...
2. The rocket can't carry enough fuel to reach this condition.
Can someone explain this paradox?
--
Brad
Uncle Al
Jun17-04, 06:16 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>Brad Eckert wrote:\n>\n> Hi all,\n>\n> AFAIK, the acceleration of a rocket is constant. This is opposed to\n> the acceleration of a car, which decreases as the transmission\n> switches to higher gears.\n>\n> If acceleration is constant, velocity is proportional to time and the\n> kinetic energy of the rocket goes as the square of the flight time. At\n> some point, the per-second increase in the kinetic energy of the\n> rocket would exceed the energy provided by burning the fuel.\n>\n> Since that would violate the laws of thermodynamics, this reasoning\n> has to break down somewhere. It could be that:\n>\n> 1. The acceleration isn\'t constant. Or...\n> 2. The rocket can\'t carry enough fuel to reach this condition.\n>\n> Can someone explain this paradox?\n\nA rocket tossing stuff out its rear doesn\'t have constant mass vs.\ntime.\n\nGoogle\n"rocket equation" 1910 hits\n\nhttp://www.neofuel.com/optimum/\n\n--\nUncle Al\nhttp://www.mazepath.com/uncleal/qz.pdf\nhttp://www.mazepath.com/uncleal/eotvos.htm\n(The parity Eotvos experiment is queued)\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>Brad Eckert wrote:
>
> Hi all,
>
> AFAIK, the acceleration of a rocket is constant. This is opposed to
> the acceleration of a car, which decreases as the transmission
> switches to higher gears.
>
> If acceleration is constant, velocity is proportional to time and the
> kinetic energy of the rocket goes as the square of the flight time. At
> some point, the per-second increase in the kinetic energy of the
> rocket would exceed the energy provided by burning the fuel.
>
> Since that would violate the laws of thermodynamics, this reasoning
> has to break down somewhere. It could be that:
>
> 1. The acceleration isn't constant. Or...
> 2. The rocket can't carry enough fuel to reach this condition.
>
> Can someone explain this paradox?
A rocket tossing stuff out its rear doesn't have constant mass vs.
time.
Google
"rocket equation" 1910 hits
http://www.neofuel.com/optimum/
--
Uncle Al
http://www.mazepath.com/uncleal/qz.pdf
http://www.mazepath.com/uncleal/eotvos.htm
(The parity Eotvos experiment is queued)
Jerzy Karczmarczuk
Jun18-04, 04:40 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>\nUncle Al wrote:\n> Brad Eckert wrote:\n\n>>AFAIK, the acceleration of a rocket is constant....\n>>If acceleration is constant, velocity is proportional to time and the\n>>kinetic energy of the rocket goes as the square of the flight time. At\n>>some point, the per-second increase in the kinetic energy of the\n>>rocket would exceed the energy provided by burning the fuel.\n>>\n>>Since that would violate the laws of thermodynamics, this reasoning\n>>has to break down somewhere. It could be that:\n>>\n>>1. The acceleration isn\'t constant. Or...\n>>2. The rocket can\'t carry enough fuel to reach this condition.\n>>\n>>Can someone explain this paradox?\n>\n> A rocket tossing stuff out its rear doesn\'t have constant mass vs.\n> time.\n\nThere is another factor.\nAlso contributing is the fact that matter thrown backwards by the rocket\n*decelerates*, may have the kinetic energy *less* than when transported\nby the rocket as combustible, provided that the rocket has enough speed,\nrelatively to the speed of the burned fuel. This won\'t work obviously\nfor photonic rockets, but the relation between the energy of fuel transpor-\nted and fuel discarded is anyway not trivial.\nThis question looks as, well, perhaps not as homework, but something\nof the kind...\n\nJerzy Karczmarczuk\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>Uncle Al wrote:
> Brad Eckert wrote:
>>AFAIK, the acceleration of a rocket is constant....
>>If acceleration is constant, velocity is proportional to time and the
>>kinetic energy of the rocket goes as the square of the flight time. At
>>some point, the per-second increase in the kinetic energy of the
>>rocket would exceed the energy provided by burning the fuel.
>>
>>Since that would violate the laws of thermodynamics, this reasoning
>>has to break down somewhere. It could be that:
>>
>>1. The acceleration isn't constant. Or...
>>2. The rocket can't carry enough fuel to reach this condition.
>>
>>Can someone explain this paradox?
>
> A rocket tossing stuff out its rear doesn't have constant mass vs.
> time.
There is another factor.
Also contributing is the fact that matter thrown backwards by the rocket
*decelerates*, may have the kinetic energy *less* than when transported
by the rocket as combustible, provided that the rocket has enough speed,
relatively to the speed of the burned fuel. This won't work obviously
for photonic rockets, but the relation between the energy of fuel transpor-
ted and fuel discarded is anyway not trivial.
This question looks as, well, perhaps not as homework, but something
of the kind...
Jerzy Karczmarczuk
Russell Blackadar
Jun18-04, 01:24 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>Brad Eckert wrote:\n> Hi all,\n>\n> AFAIK, the acceleration of a rocket is constant.\n\nIt\'s constant if you can neglect the change in mass due to\nthe burning of the fuel. That may or may not be a good\nassumption, depending on the situation, but even if it\nwere not, the effect of decreased rocket mass would be to\n*increase*, not lessen, the rocket\'s acceleration. So, your\nalternative (1) is certainly not the answer to your paradox.\n\nThis is opposed to\n> the acceleration of a car, which decreases as the transmission\n> switches to higher gears.\n>\n> If acceleration is constant, velocity is proportional to time and the\n> kinetic energy of the rocket goes as the square of the flight time. At\n> some point, the per-second increase in the kinetic energy of the\n> rocket would exceed the energy provided by burning the fuel.\n\nAlso, at all times before that (except for one instant) it\nwill be less, and that would be just as much of a violation\nof energy conservation. *Equality* is what\'s required. And\nso, your alternative (2) is not the answer either.\n\nCorrection -- it would be a violation *if* you have taken\nall of the energy in the system into account. Have you?\n(Hint, hint.)\n\n>\n> Since that would violate the laws of thermodynamics, this reasoning\n> has to break down somewhere. It could be that:\n>\n> 1. The acceleration isn\'t constant. Or...\n> 2. The rocket can\'t carry enough fuel to reach this condition.\n\n3. You forgot to include the energy of something important.\n\n>\n> Can someone explain this paradox?\n\nNote that for the accelerating car, there is a reaction\nforce that pushes the earth backwards; but since the delta\nv of the earth is very small, and v^2 much tinier, the KE\nchange of the earth is negligible even though M of the earth\nis large. In the rocket case the stuff that gets pushed back\n(propellant) is much less massive, so its v is very large and\nits KE can\'t be neglected. In other words, the physics of the\ncar scenario is really the same as that of the rocket, it just\nhas radically different numbers.\n\nTotal KE is the KE of rocket + KE of the ejected propellant.\nAt first (in the launchpad frame) the latter is by far the\ngreater, but as the rocket gains speed, the speed (relative\nto the launchpad) at which the propellant is thrown backward\nbecomes less and less. If you calculate it out (an instructive\nexercise) you find that the *total* KE increases linearly with\ntime, exactly as it should.\n\nBtw you reach the same conclusion no matter what inertial frame\nyou do the calculation in. You will get different values for\nthe KE, of course, depending on frame.\n\n>\n> --\n> Brad\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>Brad Eckert wrote:
> Hi all,
>
> AFAIK, the acceleration of a rocket is constant.
It's constant if you can neglect the change in mass due to
the burning of the fuel. That may or may not be a good
assumption, depending on the situation, but even if it
were not, the effect of decreased rocket mass would be to
*increase*, not lessen, the rocket's acceleration. So, your
alternative (1) is certainly not the answer to your paradox.
This is opposed to
> the acceleration of a car, which decreases as the transmission
> switches to higher gears.
>
> If acceleration is constant, velocity is proportional to time and the
> kinetic energy of the rocket goes as the square of the flight time. At
> some point, the per-second increase in the kinetic energy of the
> rocket would exceed the energy provided by burning the fuel.
Also, at all times before that (except for one instant) it
will be less, and that would be just as much of a violation
of energy conservation. *Equality* is what's required. And
so, your alternative (2) is not the answer either.
Correction -- it would be a violation *if* you have taken
all of the energy in the system into account. Have you?
(Hint, hint.)
>
> Since that would violate the laws of thermodynamics, this reasoning
> has to break down somewhere. It could be that:
>
> 1. The acceleration isn't constant. Or...
> 2. The rocket can't carry enough fuel to reach this condition.
3. You forgot to include the energy of something important.
>
> Can someone explain this paradox?
Note that for the accelerating car, there is a reaction
force that pushes the earth backwards; but since the \delta
v of the earth is very small, and v^2 much tinier, the KE
change of the earth is negligible even though M of the earth
is large. In the rocket case the stuff that gets pushed back
(propellant) is much less massive, so its v is very large and
its KE can't be neglected. In other words, the physics of the
car scenario is really the same as that of the rocket, it just
has radically different numbers.
Total KE is the KE of rocket + KE of the ejected propellant.
At first (in the launchpad frame) the latter is by far the
greater, but as the rocket gains speed, the speed (relative
to the launchpad) at which the propellant is thrown backward
becomes less and less. If you calculate it out (an instructive
exercise) you find that the *total* KE increases linearly with
time, exactly as it should.
Btw you reach the same conclusion no matter what inertial frame
you do the calculation in. You will get different values for
the KE, of course, depending on frame.
>
> --
> Brad
>
G. R. L. Cowan
Jun18-04, 01:27 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>Jerzy Karczmarczuk included:\n>\n> Uncle Al wrote:\n> > Brad Eckert wrote:\n>\n> >>AFAIK, the acceleration of a rocket is constant....\n> >>If acceleration is constant, velocity is proportional to time and the\n> >>kinetic energy of the rocket goes as the square of the flight time. At\n> >>some point, the per-second increase in the kinetic energy of the\n> >>rocket would exceed the energy provided by burning the fuel.\n> >>\n> >>Since that would violate the laws of thermodynamics, this reasoning\n> >>has to break down somewhere. It could be that:\n> >>\n> >>1. The acceleration isn\'t constant. Or...\n> >>2. The rocket can\'t carry enough fuel to reach this condition.\n> >>\n> >>Can someone explain this paradox?\n> >\n> > A rocket tossing stuff out its rear doesn\'t have constant mass vs.\n> > time.\n>\n> There is another factor.\n> Also contributing is the fact that matter thrown backwards by the rocket\n> *decelerates*, may have the kinetic energy *less* than when transported\n> by the rocket as combustible ...\n> This won\'t work obviously for photonic rockets ...\n\nIt will, though. The aft-flying photons can\'t slow down,\nbut they can get less energetic.\n\nSupposing the rocket is accelerating\ndirectly away from a black protosolar nebula,\nthe rate of heating where the photon beam hits the nebula\nwill diminish (even though it remains angularly large\nenough to catch the whole beam), due to redshift,\nas the rocket accelerates.\nThe diminution will be made up by an equal increase\nin the rocket\'s rate of change of kinetic energy.\n\n\n--- Graham Cowan\nhttp://www.eagle.ca/~gcowan/Paper_for_11th_CHC.doc --\nHow individual mobility gains nuclear cachet.\nLink if you want it to happen\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>Jerzy Karczmarczuk included:
>
> Uncle Al wrote:
> > Brad Eckert wrote:
>
> >>AFAIK, the acceleration of a rocket is constant....
> >>If acceleration is constant, velocity is proportional to time and the
> >>kinetic energy of the rocket goes as the square of the flight time. At
> >>some point, the per-second increase in the kinetic energy of the
> >>rocket would exceed the energy provided by burning the fuel.
> >>
> >>Since that would violate the laws of thermodynamics, this reasoning
> >>has to break down somewhere. It could be that:
> >>
> >>1. The acceleration isn't constant. Or...
> >>2. The rocket can't carry enough fuel to reach this condition.
> >>
> >>Can someone explain this paradox?
> >
> > A rocket tossing stuff out its rear doesn't have constant mass vs.
> > time.
>
> There is another factor.
> Also contributing is the fact that matter thrown backwards by the rocket
> *decelerates*, may have the kinetic energy *less* than when transported
> by the rocket as combustible ...
> This won't work obviously for photonic rockets ...
It will, though. The aft-flying photons can't slow down,
but they can get less energetic.
Supposing the rocket is accelerating
directly away from a black protosolar nebula,
the rate of heating where the photon beam hits the nebula
will diminish (even though it remains angularly large
enough to catch the whole beam), due to redshift,
as the rocket accelerates.
The diminution will be made up by an equal increase
in the rocket's rate of change of kinetic energy.
--- Graham Cowan
http://www.eagle.ca/~gcowan/Paper_for_11th_CHC.doc --
How individual mobility gains nuclear cachet.
Link if you want it to happen
Gerard Westendorp
Jun22-04, 03:13 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>Russell Blackadar wrote:\n\n[..]\n\n\n>\n> Total KE is the KE of rocket + KE of the ejected propellant.\n> At first (in the launchpad frame) the latter is by far the\n> greater, but as the rocket gains speed, the speed (relative\n> to the launchpad) at which the propellant is thrown backward\n> becomes less and less. If you calculate it out (an instructive\n> exercise) you find that the *total* KE increases linearly with\n> time, exactly as it should.\n>\n\n\nInteresting.\n\nI calculated the rocket velocity (v) if fuel mass is ejected at\nrate (m), with velocity (-w) *relative to the rocket* and the\ninitial mass of [rocket + fuel] is (M0):\n\nv = w ln( 1 - mt/M0)\n\n\nThe mass (M) decreases linearly:\n\nM = M0-mt\n\nI you plot graphs of v, m and 1/2 Mv^2, you see that v goes\nto infinity as M goes to zero (A rocket consisting of pure\nfuel).\n\nBut 1/2 Mv^2 reaches a maximum somewhere at 0.87 M0/m.\n\n(I usually make errors, though)\nGerard\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>Russell Blackadar wrote:
[..]
>
> Total KE is the KE of rocket + KE of the ejected propellant.
> At first (in the launchpad frame) the latter is by far the
> greater, but as the rocket gains speed, the speed (relative
> to the launchpad) at which the propellant is thrown backward
> becomes less and less. If you calculate it out (an instructive
> exercise) you find that the *total* KE increases linearly with
> time, exactly as it should.
>
Interesting.
I calculated the rocket velocity (v) if fuel mass is ejected at
rate (m), with velocity (-w) *relative to the rocket* and the
initial mass of [rocket + fuel] is (M0):
v = w ln( 1 - mt/M0)
The mass (M) decreases linearly:
M = M0-mt
I you plot graphs of v, m and 1/2 Mv^2, you see that v goes
to infinity as M goes to zero (A rocket consisting of pure
fuel).
But 1/2 Mv^2 reaches a maximum somewhere at .87 M0/m.
(I usually make errors, though)
Gerard
A fundemental principle of how rockets work is that the 'rocket' looses mass as it goes up, since it's dropping mass out the back in order to go forward. A surface-to-orbit rocket like the Space Shuttle's launch mass is more than 70% fuel. So, while the rocket is accelerating, it's dropping mass like mad.
Russell Blackadar
Jun25-04, 03:44 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>Gerard Westendorp wrote:\n\n> Russell Blackadar wrote:\n\n[snip my remarks about total KE]\n\n>\n> Interesting.\n>\n> I calculated the rocket velocity (v) if fuel mass is ejected at\n> rate (m), with velocity (-w) *relative to the rocket* and the\n> initial mass of [rocket + fuel] is (M0):\n>\n> v = w ln( 1 - mt/M0)\n\nYes, that\'s one form of the "rocket equation" mentioned\nby Uncle Al.\n\n>\n>\n> The mass (M) decreases linearly:\n>\n> M = M0-mt\n>\n> I you plot graphs of v, m and 1/2 Mv^2, you see that v goes\n> to infinity as M goes to zero (A rocket consisting of pure\n> fuel).\n\nYes. By then the approximation of constant a is\n*definitely* no good. ;-)\n\n>\n> But 1/2 Mv^2 reaches a maximum somewhere at 0.87 M0/m.\n\nDidn\'t check your numbers, but indeed the derivative\n(see below) does pass through zero before M=0 so it\nseems you are right.\n\nOf course (as no doubt you know) in my earlier post\nI was talking about the total *system* KE which *does*\nincrease linearly with time. The time derivative of\nyour rocket KE via product and chain rules is\n\n0.5mw^2 [ -(ln(1-mt/M0))^2 + 2 ln(1-mt/M0) ]\n\nwhich is almost exactly cancelled by the propellant\ndKE/dt given by:\n\n0.5 m (v-w)^2\n\nIf you calculate the latter out (i.e. plug in your\nexpression for v) you find that the difference between\nthe two is the constant term 1/2 mw^2. Exactly as it\nshould be. Note, that is exactly the energy you would\nhave expended per unit time if you\'d fired the rocket\nengine in a laboratory test on a fixed mount.\n\n>\n> (I usually make errors, though)\n\nBut here you did not, I think.\n\n> Gerard\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>Gerard Westendorp wrote:
> Russell Blackadar wrote:
[snip my remarks about total KE]
>
> Interesting.
>
> I calculated the rocket velocity (v) if fuel mass is ejected at
> rate (m), with velocity (-w) *relative to the rocket* and the
> initial mass of [rocket + fuel] is (M0):
>
> v = w ln( 1 - mt/M0)
Yes, that's one form of the "rocket equation" mentioned
by Uncle Al.
>
>
> The mass (M) decreases linearly:
>
> M = M0-mt
>
> I you plot graphs of v, m and 1/2 Mv^2, you see that v goes
> to infinity as M goes to zero (A rocket consisting of pure
> fuel).
Yes. By then the approximation of constant a is
*definitely* no good. ;-)
>
> But 1/2 Mv^2 reaches a maximum somewhere at .87 M0/m.
Didn't check your numbers, but indeed the derivative
(see below) does pass through zero before M=0 so it
seems you are right.
Of course (as no doubt you know) in my earlier post
I was talking about the total *system* KE which *does*
increase linearly with time. The time derivative of
your rocket KE via product and chain rules is
.5mw^2 [ -(ln(1-mt/M0))^2 + 2 ln(1-mt/M0) ]
which is almost exactly cancelled by the propellant
dKE/dt given by:
.5 m (v-w)^2
If you calculate the latter out (i.e. plug in your
expression for v) you find that the difference between
the two is the constant term 1/2 mw^2. Exactly as it
should be. Note, that is exactly the energy you would
have expended per unit time if you'd fired the rocket
engine in a laboratory test on a fixed mount.
>
> (I usually make errors, though)
But here you did not, I think.
> Gerard
>
Steven Gray
Jun27-04, 05:54 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>nospaambrad1@tinyboot.com (Brad Eckert) wrote in\nnews:7d4cc56.0406171231.2320a9f0@posting.googl e.com:\n\n> Hi all,\n>\n> AFAIK, the acceleration of a rocket is constant. This is opposed to\n> the acceleration of a car, which decreases as the transmission\n> switches to higher gears.\n\nThe acceleration of the rocket won\'t be constant because the mass of the\nrocket decreases as fuel is burned.\n>\n> If acceleration is constant, velocity is proportional to time and the\n> kinetic energy of the rocket goes as the square of the flight time. At\n> some point, the per-second increase in the kinetic energy of the\n> rocket would exceed the energy provided by burning the fuel.\n\nIn doing your energy calculations you need to stick to one frame of\nreference and you need to include the kinetic energy of the expended\nreaction mass.\n\nSuppose we use the frame of reference in which the rocket was at rest\nwhen the motor was first lit off and assume constant thrust (constant\nrate of fuel burning and constant exhaust velocity). The speed of the\nreaction mass is initially at a maximum and most of the energy is carried\nby the reaction mass. As the rocket accelerates in the opposite\ndirection, the speed of the newly burned reaction mass decreases\ndecreases in our FOR and more energy goes to the rocket, as you\'ve\nobserved. If you take everything into account, it all comes out even.\n\n--\nSteve Gray\nsgray2@cfl.rr.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>nospaambrad1@tinyboot.com (Brad Eckert) wrote in
news:7d4cc56.0406171231.2320a9f0@posting.google.co m:
> Hi all,
>
> AFAIK, the acceleration of a rocket is constant. This is opposed to
> the acceleration of a car, which decreases as the transmission
> switches to higher gears.
The acceleration of the rocket won't be constant because the mass of the
rocket decreases as fuel is burned.
>
> If acceleration is constant, velocity is proportional to time and the
> kinetic energy of the rocket goes as the square of the flight time. At
> some point, the per-second increase in the kinetic energy of the
> rocket would exceed the energy provided by burning the fuel.
In doing your energy calculations you need to stick to one frame of
reference and you need to include the kinetic energy of the expended
reaction mass.
Suppose we use the frame of reference in which the rocket was at rest
when the motor was first lit off and assume constant thrust (constant
rate of fuel burning and constant exhaust velocity). The speed of the
reaction mass is initially at a maximum and most of the energy is carried
by the reaction mass. As the rocket accelerates in the opposite
direction, the speed of the newly burned reaction mass decreases
decreases in our FOR and more energy goes to the rocket, as you've
observed. If you take everything into account, it all comes out even.
--
Steve Gray
sgray2@cfl.rr.com
<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>On Thu, 17 Jun 2004 23:16:21 +0000 (UTC), Uncle Al\n<UncleAl0@hate.spam.net> wrote:\n\n>Brad Eckert wrote:\n>>\n>> Hi all,\n>>\n>> AFAIK, the acceleration of a rocket is constant. This is opposed to\n>> the acceleration of a car, which decreases as the transmission\n>> switches to higher gears.\n>>\n>> If acceleration is constant, velocity is proportional to time and the\n>> kinetic energy of the rocket goes as the square of the flight time. At\n>> some point, the per-second increase in the kinetic energy of the\n>> rocket would exceed the energy provided by burning the fuel.\n>>\n>> Since that would violate the laws of thermodynamics, this reasoning\n>> has to break down somewhere. It could be that:\n>>\n>> 1. The acceleration isn\'t constant. Or...\n>> 2. The rocket can\'t carry enough fuel to reach this condition.\n>>\n>> Can someone explain this paradox?\n>\n>A rocket tossing stuff out its rear doesn\'t have constant mass vs.\n>time.\n>\n>Google\n>"rocket equation" 1910 hits\n\nI take for granted that considering the conservation of momentum\nbetween the rocket and it\'s exausts, the paradox will disappear...\n\nBut the mass ejected can be made small as pleased by increasing exaust\nvelocity, until speed of light is reached. Will the rocket equation\nhold even taking into account the mass-energy lost by the propulsion\nsystem [i.e. a pure photon-propelled rocket-there will be also\ndoppler-shift related questions in this case?]\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>On Thu, 17 Jun 2004 23:16:21 +0000 (UTC), Uncle Al
<UncleAl0@hate.spam.net> wrote:
>Brad Eckert wrote:
>>
>> Hi all,
>>
>> AFAIK, the acceleration of a rocket is constant. This is opposed to
>> the acceleration of a car, which decreases as the transmission
>> switches to higher gears.
>>
>> If acceleration is constant, velocity is proportional to time and the
>> kinetic energy of the rocket goes as the square of the flight time. At
>> some point, the per-second increase in the kinetic energy of the
>> rocket would exceed the energy provided by burning the fuel.
>>
>> Since that would violate the laws of thermodynamics, this reasoning
>> has to break down somewhere. It could be that:
>>
>> 1. The acceleration isn't constant. Or...
>> 2. The rocket can't carry enough fuel to reach this condition.
>>
>> Can someone explain this paradox?
>
>A rocket tossing stuff out its rear doesn't have constant mass vs.
>time.
>
>Google
>"rocket equation" 1910 hits
I take for granted that considering the conservation of momentum
between the rocket and it's exausts, the paradox will disappear...
But the mass ejected can be made small as pleased by increasing exaust
velocity, until speed of light is reached. Will the rocket equation
hold even taking into account the mass-energy lost by the propulsion
system [i.e. a pure photon-propelled rocket-there will be also
doppler-shift related questions in this case?]
Alfred Einstead
Jun27-04, 05:58 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>nospaambrad1@tinyboot.com (Brad Eckert) wrote:\n> If acceleration is constant, velocity is proportional to time and the\n> kinetic energy of the rocket goes as the square of the flight time. At\n> some point, the per-second increase in the kinetic energy of the\n> rocket would exceed the energy provided by burning the fuel.\n\nNone of those things matter. The dependence of the rocket\'s total\nchange in speed vs. the amount of fuel required is exponential, with\nan effective limit equal to something on the same order as the\nspeed of its exhaust.\n\nIt\'s not going to go much faster (relative to its initial state)\nthan its exhaust goes the other way (relative to the rocket).\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>nospaambrad1@tinyboot.com (Brad Eckert) wrote:
> If acceleration is constant, velocity is proportional to time and the
> kinetic energy of the rocket goes as the square of the flight time. At
> some point, the per-second increase in the kinetic energy of the
> rocket would exceed the energy provided by burning the fuel.
None of those things matter. The dependence of the rocket's total
change in speed vs. the amount of fuel required is exponential, with
an effective limit equal to something on the same order as the
speed of its exhaust.
It's not going to go much faster (relative to its initial state)
than its exhaust goes the other way (relative to the rocket).
Brad Eckert
Jun29-04, 04:38 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>Jerzy Karczmarczuk <karczma@info.unicaen.fr> wrote in message news:<cau3o1\\$71t\\$1@elingue.unicaen.fr>...\n> This question looks as, well, perhaps not as homework, but something\n> of the kind...\n\nYes, it would make a fine homework question. I\'ve haven\'t been to a\nphysics class in 15 years, just musing about rockets.\n\n-- Brad\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>Jerzy Karczmarczuk <karczma@info.unicaen.fr> wrote in message news:<cau3o1$71t$1@elingue.unicaen.fr>...
> This question looks as, well, perhaps not as homework, but something
> of the kind...
Yes, it would make a fine homework question. I've haven't been to a
physics class in 15 years, just musing about rockets.
-- Brad
David Williams
Jun29-04, 05:43 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>-> None of those things matter. The dependence of the rocket\'s total\n-> change in speed vs. the amount of fuel required is exponential, with\n-> an effective limit equal to something on the same order as the\n-> speed of its exhaust.\n\n-> It\'s not going to go much faster (relative to its initial state)\n-> than its exhaust goes the other way (relative to the rocket).\n\nHardly!\n\nThis kind of thing can be modelled with a few lines of any computer\nlanguage. Here\'s a little thing in QBasic:\n\n------------------------------------------------------\n\nINPUT "Initial Mass of Rocket (including fuel)"; MR\nINPUT "Mass of Fuel"; MF\nINPUT "Exhaust Velocity"; EV\nVR = 0\' velocity of rocket\n\nFOR M = 0 TO MF STEP MF / 10000 \' mass MF/10000 is ejected in each step\nRM = MR - M\' instantaneous mass of rocket\nVR = VR + EV * (MF / 10000) / RM\nNEXT\n\nPRINT "Final velocity of rocket"; VR\n\n--------------------------------------------------------\n\nIn the loop, the rocket\'s instantaneous mass declines linearly, and its\nvelocity increases at each iteration as shown by the "VR = ..." line,\nwhich just conserves momentum as a small mass of exhaust is ejected.\n\nAny units can be used, so long as they are consistent.\n\nIf the rocket is initially almost all fuel, so MF is almost as great as\nMR, the final velocity of the rocket is *far* greater than the exhaust\nvelocity (i.e. the speed at which the exhaust leaves the rocket,\nrelative to the rocket itself). For example, if MF is 0.999 of MR, the\nfinal velocity is about seven times EV.\n\ndow\n\n(p.s. to Moderators: You may get two copies of this message, the first\none containing a stupid mistake, and this one, which has the mistake\nfixed. Sorry about that!)\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>-> None of those things matter. The dependence of the rocket's total
-> change in speed vs. the amount of fuel required is exponential, with
-> an effective limit equal to something on the same order as the
-> speed of its exhaust.
-> It's not going to go much faster (relative to its initial state)
-> than its exhaust goes the other way (relative to the rocket).
Hardly!
This kind of thing can be modelled with a few lines of any computer
language. Here's a little thing in QBasic:
------------------------------------------------------
INPUT "Initial Mass of Rocket (including fuel)"; MR
INPUT "Mass of Fuel"; MF
INPUT "Exhaust Velocity"; EV
VR = 0' velocity of rocket
FOR M = TO MF STEP MF / 10000 ' mass MF/10000 is ejected in each step
RM = MR - M' instantaneous mass of rocket
VR = VR + EV * (MF / 10000) / RM
NEXT
PRINT "Final velocity of rocket"; VR
--------------------------------------------------------
In the loop, the rocket's instantaneous mass declines linearly, and its
velocity increases at each iteration as shown by the "VR = ..." line,
which just conserves momentum as a small mass of exhaust is ejected.
Any units can be used, so long as they are consistent.
If the rocket is initially almost all fuel, so MF is almost as great as
MR, the final velocity of the rocket is *far* greater than the exhaust
velocity (i.e. the speed at which the exhaust leaves the rocket,
relative to the rocket itself). For example, if MF is .999 of MR, the
final velocity is about seven times EV.
dow
(p.s. to Moderators: You may get two copies of this message, the first
one containing a stupid mistake, and this one, which has the mistake
fixed. Sorry about that!)
superczar
Jun30-04, 05:39 PM
Well, first its given that the acceleration is constant. Also way number oneout of the paradox is given to be "the acceleration isn't constant". A contradiction. Therefore we don't take way number one out. Its not available.
Also way out #2 states the rocket can't carry enough fuel to reach the condition that the increase in KE of the rocket exceeds the amount available by burning fuel (all on a per second basis). Fuel amount carried or burned per second is not relevant as the KE of the rocket can at most increase as the energy of fuel burned. In practice most energy escapes uselessly in the exhaust and does not serve to increase the velocity of the rocket. Accordingly there is no time (even 0 sec) when the increase in KE of the rocket exceeds the energy of burning fuel, its just not physically possible. The tricky part is this statement: "At some point, the per-second increase in the kinetic energy of the rocket would exceed the energy provided by burning the fuel." That's assuming we continue to accelerate. We don't need to take the statement in quotes as true. We can always burn enough fuel to increase the KE as required, just assume we have it on board. Problem is if this is to run indefinitely the total acceleration will approach zero as the mass of the rocket (because of the unlimited amount of fuel necessary) approaches infinity. You wind up with an irresistable force/immovable object type situation (not a paradox).
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