Orbits as M decreases? [Archive]

Ken S. Tucker

Apr7-04, 09:27 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>Orbits as M decreases?\n\nAs the central mass Mass decreases, (our Sun expelling\nmatter for example), will the eccentricity of the planets\norbits remain constant as Mass reduces by dM/dt ?\n\n(start with e=0, how does that evolve?)\n\nThanks\nKen S. Tucker\n\nPS: So far as know, I don\'t know how the dM/dt factor\nis reliable in the usual orbital equations, and couldn\'t find\na reference.\nkst\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>Orbits as M decreases?

As the central mass Mass decreases, (our Sun expelling
matter for example), will the eccentricity of the planets
orbits remain constant as Mass reduces by dM/dt ?

(start with e=0, how does that evolve?)

Thanks
Ken S. Tucker

PS: So far as know, I don't know how the dM/dt factor
is reliable in the usual orbital equations, and couldn't find
a reference.
kst

ebunn@lfa221051.richmond.edu

Apr8-04, 05:50 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\nIn article <2202379a.0404060934.2f471b70@posting.google.com>, \nKen S. Tucker <dynamics@vianet.on.ca> wrote:\n\n>As the central mass Mass decreases, (our Sun expelling\n>matter for example), will the eccentricity of the planets\n>orbits remain constant as Mass reduces by dM/dt ?\n\nFirst, let me point out that in our actual solar system, this is a\nvery slow process. In fact, the changes in, say, Earth\'s orbit due to\nsolar mass loss are much smaller than the changes due to the\nperturbations from other planets.\n\nStill, as a purely theoretical exercise, we can figure out the effects\non orbits due to solar mass loss, ignoring other perturbations.\nTo make the problem precise, I\'ll assume that\n\n1. Newtonian gravity can be applied.\n\n2. The mass distribution of the Sun is spherically symmetric,\nincluding the solar wind, which will be taken to be blowing off\nequally in all directions.\n\n3. The rate of mass loss is very slow. To be specific, the rate of\nmass loss, (1/M) (dM/dt), is much, much less than the frequency of\na planet\'s orbit. In other words, during one year the Sun loses a\nnegligible fraction of its mass.\n\n[Note: M here means the mass contained within the Earth\'s orbit.\nAssuming spherical symmetry, mass outside of the orbit doesn\'t\ncontribute. So dM/dt means the mass per unit time crossing outward\nacross the surface of a sphere at Earth\'s radius. In principle,\nunless the Earth\'s orbit is circular, that radius is time-varying,\nbut as long as the rate of mass loss is very slow, changes in M due\nto cyclic changes in r aren\'t important.]\n\nGiven assumption 3, there\'s a standard way to attack this sort of\nproblem, which is to find "adiabatic invariants." Goldstein\'s\n"Classical Mechanics" discusses this in detail, for instance. My\nmemory of this stuff is pretty distant, but I\'m pretty sure I\'ve got\nthe details right. If not, I\'m sure someone will correct me.\n\nFor a completely integrable system (which the Kepler problem is), it\nturns out that you can find adiabatic invariants by describing the\nsystem in terms of "action-angle variables" (again, see Goldstein for\ndetails). The action variables are adiabatic invariants, which means\nthat they remain constant, in the adiabatic approximation (assumption\n3 above).\n\nFor the Kepler problem, it turns out that the eccentricity can be\nexpressed purely in terms of the action variables, so the eccentricity is an\nadiabatic invariant.\n\nThe bottom line, therefore, is that the orbit maintains its eccentricity\nas the Sun loses mass.\n\nOne of the action variables can be expressed purely in terms of Ma,\nthe product of the Sun\'s mass M and the semimajor axis a. So Ma is\nalso an adiabatic invariant, which means that the semimajor axis\nchanges inversely with the mass.\n\nThe adiabatic assumption is important here. If it fails, then the\neccentricity is not invariant. One easy way to see this is to\nconsider the opposite assumption, namely the sudden approximation:\nSay the Sun\'s mass instantaneously changed by some amount.\n\n[That is, suppose the Sun emitted a great big spherical shell of\nmatter that blew off into space, and that shell crossed the Earth\'s\norbit at some instant, so that M, the mass within Earth\'s orbit,\nsuddenly dropped.]\n\nSay the Earth\'s orbit was initially circular. Immediately after the\nmass loss, the Sun\'s speed would be the same, but the speed required\nfor a circular orbit would have declined. So the orbit would now be\nelliptical, with perihelion at the location where the Earth was at the\nmoment of mass loss.\n\nAs usual, cases in between the two extremes, where neither the\nadiabatic approximation nor the sudden approximation is good enough,\nwould be much harder to deal with. You\'d probably just have to\nintegrate the equations of motion numerically. (Continuing to assume\nspherical symmetry, you\'d still have angular momentum conservation,\nso you\'d only need to keep track of one degree of freedom, say r as\na function of time.) For our solar system, though, the adiabatic\napproximation is extremely good: according to\n\nhttp://home.comcast.net/~pdnoerd/SMassLoss.html\n\nthe Sun loses about one ten-trillionth of its mass per year. In\nother words, the adiabatic approximation is the approximation that\n\n10^(-13) << 1.\n\nBy the way, this web page, which I found after I\'d written the stuff\nabout adiabatic invariants above, also says that the eccentricity\nremains constant and the orbital semimajor axes are inversely\nproportional to the Sun\'s mass, so I think I must have gotten that\nstuff roughly right.\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\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 <2202379a.0404060934.2f471b70@posting.google.com>,
Ken S. Tucker <dynamics@vianet.on.ca> wrote:

>As the central mass Mass decreases, (our Sun expelling
>matter for example), will the eccentricity of the planets
>orbits remain constant as Mass reduces by dM/dt ?

First, let me point out that in our actual solar system, this is a
very slow process. In fact, the changes in, say, Earth's orbit due to
solar mass loss are much smaller than the changes due to the
perturbations from other planets.

Still, as a purely theoretical exercise, we can figure out the effects
on orbits due to solar mass loss, ignoring other perturbations.
To make the problem precise, I'll assume that

1. Newtonian gravity can be applied.

2. The mass distribution of the Sun is spherically symmetric,
including the solar wind, which will be taken to be blowing off
equally in all directions.

3. The rate of mass loss is very slow. To be specific, the rate of
mass loss, (1/M) (dM/dt), is much, much less than the frequency of
a planet's orbit. In other words, during one year the Sun loses a
negligible fraction of its mass.

[Note: M here means the mass contained within the Earth's orbit.
Assuming spherical symmetry, mass outside of the orbit doesn't
contribute. So dM/dt means the mass per unit time crossing outward
across the surface of a sphere at Earth's radius. In principle,
unless the Earth's orbit is circular, that radius is time-varying,
but as long as the rate of mass loss is very slow, changes in M due
to cyclic changes in r aren't important.]

Given assumption 3, there's a standard way to attack this sort of
problem, which is to find "adiabatic invariants." Goldstein's
"Classical Mechanics" discusses this in detail, for instance. My
memory of this stuff is pretty distant, but I'm pretty sure I've got
the details right. If not, I'm sure someone will correct me.

For a completely integrable system (which the Kepler problem is), it
turns out that you can find adiabatic invariants by describing the
system in terms of "action-angle variables" (again, see Goldstein for
details). The action variables are adiabatic invariants, which means
that they remain constant, in the adiabatic approximation (assumption
3 above).

For the Kepler problem, it turns out that the eccentricity can be
expressed purely in terms of the action variables, so the eccentricity is an
adiabatic invariant.

The bottom line, therefore, is that the orbit maintains its eccentricity
as the Sun loses mass.

One of the action variables can be expressed purely in terms of Ma,
the product of the Sun's mass M and the semimajor axis a. So Ma is
also an adiabatic invariant, which means that the semimajor axis
changes inversely with the mass.

The adiabatic assumption is important here. If it fails, then the
eccentricity is not invariant. One easy way to see this is to
consider the opposite assumption, namely the sudden approximation:
Say the Sun's mass instantaneously changed by some amount.

[That is, suppose the Sun emitted a great big spherical shell of
matter that blew off into space, and that shell crossed the Earth's
orbit at some instant, so that M, the mass within Earth's orbit,
suddenly dropped.]

Say the Earth's orbit was initially circular. Immediately after the
mass loss, the Sun's speed would be the same, but the speed required
for a circular orbit would have declined. So the orbit would now be
elliptical, with perihelion at the location where the Earth was at the
moment of mass loss.

As usual, cases in between the two extremes, where neither the
adiabatic approximation nor the sudden approximation is good enough,
would be much harder to deal with. You'd probably just have to
integrate the equations of motion numerically. (Continuing to assume
spherical symmetry, you'd still have angular momentum conservation,
so you'd only need to keep track of one degree of freedom, say r as
a function of time.) For our solar system, though, the adiabatic
approximation is extremely good: according to

http://home.comcast.net/~pdnoerd/SMassLoss.html

the Sun loses about one ten-trillionth of its mass per year. In
other words, the adiabatic approximation is the approximation that

10^(-13) << 1[/itex].

By the way, this web page, which I found after I'd written the stuff
about adiabatic invariants above, also says that the eccentricity
remains constant and the orbital semimajor axes are inversely
proportional to the Sun's mass, so I think I must have gotten that
stuff roughly right.

[itex]-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

Ken S. Tucker

Apr9-04, 02:22 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>ebunn@lfa221051.richmond.edu wrote in message news:<c51dkj\\$eit\\$1@lfa222122.richmond.edu>...\ n>In article <2202379a.0404060934.2f471b70@posting.google.com>, \n>Ken S. Tucker <dynamics@vianet.on.ca> wrote:\n>\n>>As the central mass Mass decreases, (our Sun expelling\n>>matter for example), will the eccentricity of the planets\n>>orbits remain constant as Mass reduces by dM/dt ?\n>\n>First, let me point out that in our actual solar system, this is a\n>very slow process. In fact, the changes in, say, Earth\'s orbit due to\n>solar mass loss are much smaller than the changes due to the\n>perturbations from other planets.\n>\n>Still, as a purely theoretical exercise, we can figure out the effects\n>on orbits due to solar mass loss, ignoring other perturbations.\n>To make the problem precise, I\'ll assume that\n>\n>1. Newtonian gravity can be applied.\n>\n>2. The mass distribution of the Sun is spherically symmetric,\n> including the solar wind, which will be taken to be blowing off\n> equally in all directions.\n>\n>3. The rate of mass loss is very slow. To be specific, the rate of\n> mass loss, (1/M) (dM/dt), is much, much less than the frequency of\n> a planet\'s orbit. In other words, during one year the Sun loses a\n> negligible fraction of its mass.\n>\n>[Note: M here means the mass contained within the Earth\'s orbit.\n> Assuming spherical symmetry, mass outside of the orbit doesn\'t\n> contribute. So dM/dt means the mass per unit time crossing outward\n> across the surface of a sphere at Earth\'s radius. In principle,\n> unless the Earth\'s orbit is circular, that radius is time-varying,\n> but as long as the rate of mass loss is very slow, changes in M due\n> to cyclic changes in r aren\'t important.]\n>\n>Given assumption 3, there\'s a standard way to attack this sort of\n>problem, which is to find "adiabatic invariants." Goldstein\'s\n>"Classical Mechanics" discusses this in detail, for instance. My\n>memory of this stuff is pretty distant, but I\'m pretty sure I\'ve got\n>the details right. If not, I\'m sure someone will correct me.\n>\n>For a completely integrable system (which the Kepler problem is), it\n>turns out that you can find adiabatic invariants by describing the\n>system in terms of "action-angle variables" (again, see Goldstein for\n>details). The action variables are adiabatic invariants, which means\n>that they remain constant, in the adiabatic approximation (assumption\n>3 above).\n>\n>For the Kepler problem, it turns out that the eccentricity can be\n>expressed purely in terms of the action variables, so the eccentricity is an\n>adiabatic invariant.\n>\n>The bottom line, therefore, is that the orbit maintains its eccentricity\n>as the Sun loses mass.\n\n>One of the action variables can be expressed purely in terms of Ma,\n>the product of the Sun\'s mass M and the semimajor axis a. So Ma is\n>also an adiabatic invariant, which means that the semimajor axis\n>changes inversely with the mass.\n>\n>The adiabatic assumption is important here. If it fails, then the\n>eccentricity is not invariant.\n***\n>One easy way to see this is to\n>consider the opposite assumption, namely the sudden approximation:\n>Say the Sun\'s mass instantaneously changed by some amount.\n>\n>[That is, suppose the Sun emitted a great big spherical shell of\n> matter that blew off into space, and that shell crossed the Earth\'s\n> orbit at some instant, so that M, the mass within Earth\'s orbit,\n> suddenly dropped.]\n>\n>Say the Earth\'s orbit was initially circular. Immediately after the\n>mass loss, the Sun\'s speed would be the same, but the speed required\n>for a circular orbit would have declined. So the orbit would now be\n>elliptical, with perihelion at the location where the Earth was at the\n>moment of mass loss.\n>\n>As usual, cases in between the two extremes, where neither the\n>adiabatic approximation nor the sudden approximation is good enough,\n>would be much harder to deal with. You\'d probably just have to\n>integrate the equations of motion numerically. (Continuing to assume\n>spherical symmetry, you\'d still have angular momentum conservation,\n>so you\'d only need to keep track of one degree of freedom, say r as\n>a function of time.) For our solar system, though, the adiabatic\n>approximation is extremely good: according to\n>\n>http://home.comcast.net/~pdnoerd/SMassLoss.html\n###\n>the Sun loses about one ten-trillionth of its mass per year. In\n>other words, the adiabatic approximation is the approximation that\n>\n>10^(-13) << 1.\n>\n>By the way, this web page, which I found after I\'d written the stuff\n>about adiabatic invariants above, also says that the eccentricity\n>remains constant and the orbital semimajor axes are inversely\n>proportional to the Sun\'s mass, so I think I must have gotten that\n>stuff roughly right.\n>-Ted\n\nThank you Ted, I did not expect so much energy from my\nsimple question. You are evidentially serious about this\nproblem. I studied the site ### and respect the implied\nscaling therein, and so to your remark *** I\'ll try a basic\nscaling factor.\n\nBeginning with a circular orbit described conventionally by\n\nE=T+V\n\nwhere T=kinetic energy and V = potential energy,\nwe can find 0 = T+V-E\n\nI can multiply that by X to get,\n\n0 = X*T + X*V - X*E\n\nUsually in orbital equations we expand the kinetic\nenergy to be\n\nT=k/r^2\n\nwhere k conserves angular momentum,\n\nand the potential is V = M/r (G=1).\n\nGetting back to X, lets use X to decrease M\nby a rate m=X*M, where the coefficient X describes\nthe rate at which M loses mass. In the orbital\nequations this becomes,\n\n0 = Xk/r^2 + XM/r -XE\n\nAs a reminder, the central mass is m = XM, where\nX is less than one, and is not a constant, it is declining as\ncental Mass M reduces to m.\n\nAfter some time we have a difference of mass M=>m and\nradius r=>R giving an orbit,\n\n0 = Xk/R^2 + m/R - XE =0\n\nNow we cannot divide out the coefficient X, because\nthe matter from M is lost, so we have a new orbital\nequation based on "m".\n\nAt this point I\'m flustered, I think I haven\'t a necessary\nconservation constraint to close.\nThanks again,\nKen S. Tucker\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>ebunn@lfa221051.richmond.edu wrote in message news:<c51dkj$eit$1@lfa222122.richmond.edu>...
>In article <2202379a.0404060934.2f471b70@posting.google.com>,
>Ken S. Tucker <dynamics@vianet.on.ca> wrote:
>
>>As the central mass Mass decreases, (our Sun expelling
>>matter for example), will the eccentricity of the planets
>>orbits remain constant as Mass reduces by dM/dt ?
>
>First, let me point out that in our actual solar system, this is a
>very slow process. In fact, the changes in, say, Earth's orbit due to
>solar mass loss are much smaller than the changes due to the
>perturbations from other planets.
>
>Still, as a purely theoretical exercise, we can figure out the effects
>on orbits due to solar mass loss, ignoring other perturbations.
>To make the problem precise, I'll assume that
>
>1. Newtonian gravity can be applied.
>
>2. The mass distribution of the Sun is spherically symmetric,
> including the solar wind, which will be taken to be blowing off
> equally in all directions.
>
>3. The rate of mass loss is very slow. To be specific, the rate of
> mass loss, (1/M) (dM/dt), is much, much less than the frequency of
> a planet's orbit. In other words, during one year the Sun loses a
> negligible fraction of its mass.
>
>[Note: M here means the mass contained within the Earth's orbit.
> Assuming spherical symmetry, mass outside of the orbit doesn't
> contribute. So dM/dt means the mass per unit time crossing outward
> across the surface of a sphere at Earth's radius. In principle,
> unless the Earth's orbit is circular, that radius is time-varying,
> but as long as the rate of mass loss is very slow, changes in M due
> to cyclic changes in r aren't important.]
>
>Given assumption 3, there's a standard way to attack this sort of
>problem, which is to find "adiabatic invariants." Goldstein's
>"Classical Mechanics" discusses this in detail, for instance. My
>memory of this stuff is pretty distant, but I'm pretty sure I've got
>the details right. If not, I'm sure someone will correct me.
>
>For a completely integrable system (which the Kepler problem is), it
>turns out that you can find adiabatic invariants by describing the
>system in terms of "action-angle variables" (again, see Goldstein for
>details). The action variables are adiabatic invariants, which means
>that they remain constant, in the adiabatic approximation (assumption
>3 above).
>
>For the Kepler problem, it turns out that the eccentricity can be
>expressed purely in terms of the action variables, so the eccentricity is an
>adiabatic invariant.
>
>The bottom line, therefore, is that the orbit maintains its eccentricity
>as the Sun loses mass.

>One of the action variables can be expressed purely in terms of Ma,
>the product of the Sun's mass M and the semimajor axis a. So Ma is
>also an adiabatic invariant, which means that the semimajor axis
>changes inversely with the mass.
>
>The adiabatic assumption is important here. If it fails, then the
>eccentricity is not invariant.
***
>One easy way to see this is to
>consider the opposite assumption, namely the sudden approximation:
>Say the Sun's mass instantaneously changed by some amount.
>
>[That is, suppose the Sun emitted a great big spherical shell of
> matter that blew off into space, and that shell crossed the Earth's
> orbit at some instant, so that M, the mass within Earth's orbit,
> suddenly dropped.]
>
>Say the Earth's orbit was initially circular. Immediately after the
>mass loss, the Sun's speed would be the same, but the speed required
>for a circular orbit would have declined. So the orbit would now be
>elliptical, with perihelion at the location where the Earth was at the
>moment of mass loss.
>
>As usual, cases in between the two extremes, where neither the
>adiabatic approximation nor the sudden approximation is good enough,
>would be much harder to deal with. You'd probably just have to
>integrate the equations of motion numerically. (Continuing to assume
>spherical symmetry, you'd still have angular momentum conservation,
>so you'd only need to keep track of one degree of freedom, say r as
>a function of time.) For our solar system, though, the adiabatic
>approximation is extremely good: according to
>
>http://home.comcast.net/~pdnoerd/SMassLoss.html
###
>the Sun loses about one ten-trillionth of its mass per year. In
>other words, the adiabatic approximation is the approximation that
>
>10^(-13) << 1.
>
>By the way, this web page, which I found after I'd written the stuff
>about adiabatic invariants above, also says that the eccentricity
>remains constant and the orbital semimajor axes are inversely
>proportional to the Sun's mass, so I think I must have gotten that
>stuff roughly right.
>-Ted

Thank you Ted, I did not expect so much energy from my
simple question. You are evidentially serious about this
problem. I studied the site ### and respect the implied
scaling therein, and so to your remark *** I'll try a basic
scaling factor.

Beginning with a circular orbit described conventionally by

E=T+V

where T=kinetic energy and V = potential energy,
we can find = T+V-E

I can multiply that by X to get,

= X*T + X*V - X*E

Usually in orbital equations we expand the kinetic
energy to be

T=k/r^2

where k conserves angular momentum,

and the potential is V = M/r (G=1).

Getting back to X, lets use X to decrease M
by a rate m=X*M, where the coefficient X describes
the rate at which M loses mass. In the orbital
equations this becomes,

= Xk/r^2 + XM/r -XE

As a reminder, the central mass is m = XM, where
X is less than one, and is not a constant, it is declining as
cental Mass M reduces to m.

After some time we have a difference of mass M=>m and
radius r=>R giving an orbit,

= Xk/R^2 + m/R - XE =0

Now we cannot divide out the coefficient X, because
the matter from M is lost, so we have a new orbital
equation based on "m".

At this point I'm flustered, I think I haven't a necessary
conservation constraint to close.
Thanks again,
Ken S. Tucker

David Williams

Apr11-04, 11:43 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-> >As the central mass Mass decreases, (our Sun expelling\n-> >matter for example), will the eccentricity of the planets\n-> >orbits remain constant as Mass reduces by dM/dt ?\n\n-> First, let me point out that in our actual solar system, this is a\n-> very slow process. In fact, the changes in, say, Earth\'s orbit due to\n-> solar mass loss are much smaller than the changes due to the\n-> perturbations from other planets.\n\n-> Still, as a purely theoretical exercise, we can figure out the effects\n-> on orbits due to solar mass loss, ignoring other perturbations.\n-> To make the problem precise, I\'ll assume that\n\n-> 1. Newtonian gravity can be applied.\n\n-> 2. The mass distribution of the Sun is spherically symmetric,\n-> including the solar wind, which will be taken to be blowing off\n-> equally in all directions.\n\n-> 3. The rate of mass loss is very slow. To be specific, the rate of\n-> mass loss, (1/M) (dM/dt), is much, much less than the frequency of\n-> a planet\'s orbit. In other words, during one year the Sun loses a\n-> negligible fraction of its mass.\n\nWe need a fourth condition, that the mass-loss is continuous, rather\nthan periodic, otherwise different things happen. E.g.:\n\n-> The adiabatic assumption is important here. If it fails, then the\n-> eccentricity is not invariant. One easy way to see this is to\n-> consider the opposite assumption, namely the sudden approximation:\n-> Say the Sun\'s mass instantaneously changed by some amount.\n\n-> [That is, suppose the Sun emitted a great big spherical shell of\n-> matter that blew off into space, and that shell crossed the Earth\'s\n-> orbit at some instant, so that M, the mass within Earth\'s orbit,\n-> suddenly dropped.]\n\n-> Say the Earth\'s orbit was initially circular. Immediately after the\n-> mass loss, the Sun\'s speed would be the same, but the speed required\n-> for a circular orbit would have declined. So the orbit would now be\n-> elliptical, with perihelion at the location where the Earth was at the\n-> moment of mass loss.\n\nLet\'s suppose the initial loss of mass is very small, so the orbit\nbecomes just slightly elliptical. But suppose that, when the planet is\nnext at perihelion, another small loss of the star\'s mass occurs. The\nresult will be to raise the aphelion, leaving the perihelion unmoved,\nso the eccentricity of the orbit will increase. A long sequence of\nthese events, with every mass-loss occurring as the planet passes\nthrough perihelion, will lead to a substantial increase in the\neccentricity of the orbit, even though the loss of mass in each orbital\nrevolution is very small.\n\nIt is not utterly impossible for this to be a realistic scenario. For\nexample, if the planet is very massive and very close to the star, its\ntidal effect on the star could lead to fluctuations in the "solar wind"\nwhich would be synchronized with the planet\'s revolution around its\nelliptical orbit. This could lead to the orbit becoming more eccentric.\n\ndow\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>-> >As the central mass Mass decreases, (our Sun expelling
-> >matter for example), will the eccentricity of the planets
-> >orbits remain constant as Mass reduces by dM/dt ?

-> First, let me point out that in our actual solar system, this is a
-> very slow process. In fact, the changes in, say, Earth's orbit due to
-> solar mass loss are much smaller than the changes due to the
-> perturbations from other planets.

-> Still, as a purely theoretical exercise, we can figure out the effects
-> on orbits due to solar mass loss, ignoring other perturbations.
-> To make the problem precise, I'll assume that

-> 1. Newtonian gravity can be applied.

-> 2. The mass distribution of the Sun is spherically symmetric,
-> including the solar wind, which will be taken to be blowing off
-> equally in all directions.

-> 3. The rate of mass loss is very slow. To be specific, the rate of
-> mass loss, (1/M) (dM/dt), is much, much less than the frequency of
-> a planet's orbit. In other words, during one year the Sun loses a
-> negligible fraction of its mass.

We need a fourth condition, that the mass-loss is continuous, rather
than periodic, otherwise different things happen. E.g.:

-> The adiabatic assumption is important here. If it fails, then the
-> eccentricity is not invariant. One easy way to see this is to
-> consider the opposite assumption, namely the sudden approximation:
-> Say the Sun's mass instantaneously changed by some amount.

-> [That is, suppose the Sun emitted a great big spherical shell of
-> matter that blew off into space, and that shell crossed the Earth's
-> orbit at some instant, so that M, the mass within Earth's orbit,
-> suddenly dropped.]

-> Say the Earth's orbit was initially circular. Immediately after the
-> mass loss, the Sun's speed would be the same, but the speed required
-> for a circular orbit would have declined. So the orbit would now be
-> elliptical, with perihelion at the location where the Earth was at the
-> moment of mass loss.

Let's suppose the initial loss of mass is very small, so the orbit
becomes just slightly elliptical. But suppose that, when the planet is
next at perihelion, another small loss of the star's mass occurs. The
result will be to raise the aphelion, leaving the perihelion unmoved,
so the eccentricity of the orbit will increase. A long sequence of
these events, with every mass-loss occurring as the planet passes
through perihelion, will lead to a substantial increase in the
eccentricity of the orbit, even though the loss of mass in each orbital
revolution is very small.

It is not utterly impossible for this to be a realistic scenario. For
example, if the planet is very massive and very close to the star, its
tidal effect on the star could lead to fluctuations in the "solar wind"
which would be synchronized with the planet's revolution around its
elliptical orbit. This could lead to the orbit becoming more eccentric.

dow

ebunn@lfa221051.richmond.edu

Apr11-04, 11:44 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>\n\nIn article <2202379a.0404081758.1d10c2df@posting.google.com>, \nKen S. Tucker <dynamics@vianet.on.ca> wrote:\n\n>Beginning with a circular orbit described conventionally by\n>\n> E=T+V\n>\n>where T=kinetic energy and V = potential energy,\n>we can find 0 = T+V-E\n>\n>I can multiply that by X to get,\n>\n> 0 = X*T + X*V - X*E\n>\n>Usually in orbital equations we expand the kinetic\n>energy to be\n>\n> T=k/r^2\n>\n>where k conserves angular momentum,\n\nI don\'t understand this equation. The kinetic energy\nin an orbit is not inversely proportional to r^2.\n\nConservation of angular momentum says that the tangential component of\nv is inversely proportional to r. So if we compare two moments when\nthe velocity is entirely tangential (i.e., perihelion and aphelion),\nit\'ll be true that v is inversely proportional to r, and hence that T\nis inversely proportional to r^2. But that proportionality fails\nwhenever there\'s a radial component of the velocity (i.e., at all\nmoments other than perihelion and aphelion).\n\nIn any case, I can\'t follow the thread of the rest of your argument.\n\nI\'m virtually sure that, if you aren\'t in a situation where the\nadiabatic approximation or the sudden approximation holds -- that is,\nwhere the time scale of mass loss is neither much longer nor much\nshorter than the orbital period -- there\'s no nice way to solve the\nproblem, and you have to integrate the equation of motion numerically.\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\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>In article <2202379a.0404081758.1d10c2df@posting.google.com>,
Ken S. Tucker <dynamics@vianet.on.ca> wrote:

>Beginning with a circular orbit described conventionally by
>
> E=T+V
>
>where T=kinetic energy and V = potential energy,
>we can find = T+V-E
>
>I can multiply that by X to get,
>
> = X*T + X*V - X*E
>
>Usually in orbital equations we expand the kinetic
>energy to be
>
> T=k/r^2
>
>where k conserves angular momentum,

I don't understand this equation. The kinetic energy
in an orbit is not inversely proportional to r^2.

Conservation of angular momentum says that the tangential component of
v is inversely proportional to r. So if we compare two moments when
the velocity is entirely tangential (i.e., perihelion and aphelion),
it'll be true that v is inversely proportional to r, and hence that T
is inversely proportional to r^2. But that proportionality fails
whenever there's a radial component of the velocity (i.e., at all
moments other than perihelion and aphelion).

In any case, I can't follow the thread of the rest of your argument.

I'm virtually sure that, if you aren't in a situation where the
adiabatic approximation or the sudden approximation holds -- that is,
where the time scale of mass loss is neither much longer nor much
shorter than the orbital period -- there's no nice way to solve the
problem, and you have to integrate the equation of motion numerically.

-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

Ken S. Tucker

Apr14-04, 08:36 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\n\nebunn@lfa221051.richmond.edu wrote in message news:<c56n6q\\$ho7\\$1@lfa222122.richmond.edu>...\ n>In article <2202379a.0404081758.1d10c2df@posting.google.com>, \n>Ken S. Tucker <dynamics@vianet.on.ca> wrote:\n>\n>>Beginning with a circular orbit described conventionally by\n>>\n>> E=T+V\n>>\n>>where T=kinetic energy and V = potential energy,\n>>we can find 0 = T+V-E\n>>\n>>I can multiply that by X to get,\n>>\n>> 0 = X*T + X*V - X*E\n>>\n>>Usually in orbital equations we expand the kinetic\n>>energy to be\n>>\n>> T=k/r^2\n>>\n>>where k conserves angular momentum,\n>\n>I don\'t understand this equation. The kinetic energy\n>in an orbit is not inversely proportional to r^2.\n\nThank you, my reference is Blanco & McCuskey\'s\n"BASIC PHYSICS of the SOLAR SYSTEM"\nfollowing their Eq.(4-40) they write,\n\n(1/2)*m*h^2*u^2 - (mu)*m*u - E = 0\n\nwhere m = mass of orbiting particle, (mu) = GM\nwith M the central Mass, and u = 1/r, r=radius.\nThe "h" is a constant based on Kepler\'s 2nd law\nto "sweep" equal areas in equal times.\n\nSo I abbreviated their 1st term above to\n\nT = k/r^2\n\nby setting m=1 and k= (1/2) h^2.\n\nI also backed up to their Eq.(4-33)\n\nE = (1/2)*m*v^2 - GMm/r\n\nto confirm the meaning of "T" as to be interpretated\nas kinetic energy.\nI sure hope you have access to that book, (I see I made\na mistake on the sign of the potential in my previous post,\nsorry about that), but are we ok with the orbital equations\nreferenced in this post?\n\n>Conservation of angular momentum says that the tangential component of\n>v is inversely proportional to r. So if we compare two moments when\n>the velocity is entirely tangential (i.e., perihelion and aphelion),\n>it\'ll be true that v is inversely proportional to r, and hence that T\n>is inversely proportional to r^2. But that proportionality fails\n>whenever there\'s a radial component of the velocity (i.e., at all\n>moments other than perihelion and aphelion).\n\nYes, understood, more below...\n\n>In any case, I can\'t follow the thread of the rest of your argument.\n>I\'m virtually sure that, if you aren\'t in a situation where the\n>adiabatic approximation or the sudden approximation holds -- that is,\n>where the time scale of mass loss is neither much longer nor much\n>shorter than the orbital period -- there\'s no nice way to solve the\n>problem, and you have to integrate the equation of motion numerically.\n>-Ted\n\nUsing Newtonian Mechanics, a decrement of the central\nMass will cause an orbital perturbation. Is this same orbital\nperturbation equivlent to leaving M constant and applying\nreaction thrust?\n\nIf so the reduction of M in the 1st scenario, can be duplicated\nby leaving M constant and applying a reaction thrust in the\ndirection of the mass\'s motion. In that 2nd scenario the\neccentricity should increase until the orbit becomes parabolic.\nPending Ted\'s approval of the standard orbital equations\nabove, (IMO), we might regard the 2nd scenario.\nRegards\nKen S. Tucker\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>ebunn@lfa221051.richmond.edu wrote in message news:<c56n6q$ho7$1@lfa222122.richmond.edu>...
>In article <2202379a.0404081758.1d10c2df@posting.google.com>,
>Ken S. Tucker <dynamics@vianet.on.ca> wrote:
>
>>Beginning with a circular orbit described conventionally by
>>
>> E=T+V
>>
>>where T=kinetic energy and V = potential energy,
>>we can find = T+V-E
>>
>>I can multiply that by X to get,
>>
>> = X*T + X*V - X*E
>>
>>Usually in orbital equations we expand the kinetic
>>energy to be
>>
>> T=k/r^2
>>
>>where k conserves angular momentum,
>
>I don't understand this equation. The kinetic energy
>in an orbit is not inversely proportional to r^2.

Thank you, my reference is Blanco & McCuskey's
"BASIC PHYSICS of the SOLAR SYSTEM"
following their Eq.(4-40) they write,

(1/2)*m*h^2*u^2 - (\mu)*m*u - E =

where m = mass of orbiting particle, (\mu) = GM
with M the central Mass, and u = 1/r, r=radius.
The "h" is a constant based on Kepler's 2nd law
to "sweep" equal areas in equal times.

So I abbreviated their 1st term above to

T = k/r^2

by setting m=1 and k= (1/2) h^2.

I also backed up to their Eq.(4-33)

E = (1/2)*m*v^2 - GMm/r

to confirm the meaning of "T" as to be interpretated
as kinetic energy.
I sure hope you have access to that book, (I see I made
a mistake on the sign of the potential in my previous post,
sorry about that), but are we ok with the orbital equations
referenced in this post?

>Conservation of angular momentum says that the tangential component of
>v is inversely proportional to r. So if we compare two moments when
>the velocity is entirely tangential (i.e., perihelion and aphelion),
>it'll be true that v is inversely proportional to r, and hence that T
>is inversely proportional to r^2. But that proportionality fails
>whenever there's a radial component of the velocity (i.e., at all
>moments other than perihelion and aphelion).

Yes, understood, more below...

>In any case, I can't follow the thread of the rest of your argument.
>I'm virtually sure that, if you aren't in a situation where the
>adiabatic approximation or the sudden approximation holds -- that is,
>where the time scale of mass loss is neither much longer nor much
>shorter than the orbital period -- there's no nice way to solve the
>problem, and you have to integrate the equation of motion numerically.
>-Ted

Using Newtonian Mechanics, a decrement of the central
Mass will cause an orbital perturbation. Is this same orbital
perturbation equivlent to leaving M constant and applying
reaction thrust?

If so the reduction of M in the 1st scenario, can be duplicated
by leaving M constant and applying a reaction thrust in the
direction of the mass's motion. In that 2nd scenario the
eccentricity should increase until the orbit becomes parabolic.
Pending Ted's approval of the standard orbital equations
above, (IMO), we might regard the 2nd scenario.
Regards
Ken S. Tucker

David Williams

Apr15-04, 04:03 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\n-> Using Newtonian Mechanics, a decrement of the central\n-> Mass will cause an orbital perturbation. Is this same orbital\n-> perturbation equivlent to leaving M constant and applying\n-> reaction thrust?\n\n-> If so the reduction of M in the 1st scenario, can be duplicated\n-> by leaving M constant and applying a reaction thrust in the\n-> direction of the mass\'s motion. In that 2nd scenario the\n-> eccentricity should increase until the orbit becomes parabolic.\n-> Pending Ted\'s approval of the standard orbital equations\n-> above, (IMO), we might regard the 2nd scenario.\n-> Regards\n-> Ken S. Tucker\n\nObviously, it would be *possible*, in a situation where the central\nstar\'s mass is constant, to use rocket power to drive a planet in the\npath that it would have followed if the star\'s mass decreased and the\nplanet moved inertially. However, the *velocity* of the rocket-driven\nplanet at any point along this path would *not* be the same as the\nvelocity of the inertial planet at the same point. Basically, the\nrocket-driven planet would be orbiting a star that is more massive than\nthe one that the inertial planet is orbiting, some time after the\nstellar mass has started decreasing.\n\nSo the analogy between the two planets, one driven by a rocket and the\nother in free orbit around a star that is losing mass, can *not* be\nexact.\n\nI doubt that it is a useful consideration, in any way.\n\ndow\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>-> Using Newtonian Mechanics, a decrement of the central
-> Mass will cause an orbital perturbation. Is this same orbital
-> perturbation equivlent to leaving M constant and applying
-> reaction thrust?

-> If so the reduction of M in the 1st scenario, can be duplicated
-> by leaving M constant and applying a reaction thrust in the
-> direction of the mass's motion. In that 2nd scenario the
-> eccentricity should increase until the orbit becomes parabolic.
-> Pending Ted's approval of the standard orbital equations
-> above, (IMO), we might regard the 2nd scenario.
-> Regards
-> Ken S. Tucker

Obviously, it would be *possible*, in a situation where the central
star's mass is constant, to use rocket power to drive a planet in the
path that it would have followed if the star's mass decreased and the
planet moved inertially. However, the *velocity* of the rocket-driven
planet at any point along this path would *not* be the same as the
velocity of the inertial planet at the same point. Basically, the
rocket-driven planet would be orbiting a star that is more massive than
the one that the inertial planet is orbiting, some time after the
stellar mass has started decreasing.

So the analogy between the two planets, one driven by a rocket and the
other in free orbit around a star that is losing mass, can *not* be
exact.

I doubt that it is a useful consideration, in any way.

dow

ebunn@lfa221051.richmond.edu

Apr16-04, 02:28 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>In article <2202379a.0404131031.7794e31a@posting.google.com>, \nKen S. Tucker <dynamics@vianet.on.ca> wrote:\n\n>Thank you, my reference is Blanco & McCuskey\'s\n>"BASIC PHYSICS of the SOLAR SYSTEM"\n>following their Eq.(4-40) they write,\n>\n> (1/2)*m*h^2*u^2 - (mu)*m*u - E = 0\n>\n>where m = mass of orbiting particle, (mu) = GM\n>with M the central Mass, and u = 1/r, r=radius.\n> The "h" is a constant based on Kepler\'s 2nd law\n>to "sweep" equal areas in equal times.\n\nThis is a correct equation to describe the perihelion and aphelion\nof a Keplerian orbit. It certainly doesn\'t describe an arbitrary\npoint on the orbit though.\n\nThe easiest way to see this is to note that this is a quadratic\nequation for u, with everything else being constant, so it has at most\ntwo solutions. Those two solutions are the values of u at perihelion\nand aphelion. But clearly the equation can\'t be true at any other\npoint in the orbit.\n\n>So I abbreviated their 1st term above to\n>\n> T = k/r^2\n>\n>by setting m=1 and k= (1/2) h^2.\n>\n>I also backed up to their Eq.(4-33)\n>\n> E = (1/2)*m*v^2 - GMm/r\n>\n>to confirm the meaning of "T" as to be interpretated\n>as kinetic energy.\n\nThat\'s fine. But it\'s still all only true at those two specific\npoints in the orbit, not in between. But in your previous argument,\nas far as I could tell anyway, you were assuming that this equation\nheld at all times, not just at two specific instants.\n\n>Using Newtonian Mechanics, a decrement of the central\n>Mass will cause an orbital perturbation. Is this same orbital\n>perturbation equivlent to leaving M constant and applying\n>reaction thrust?\n\nI can\'t interpret this question in any sensible way. It seems to me\nto mean "Can we attach a rocket to the mass m and fire it in such a\nway that it follows the same orbit as it would if M were decreasing?"\nThe answer to that question is obviously yes, if we allow arbitrary\nrocket firings. I can\'t imagine that that\'s what you really mean to\nask, but I can\'t figure out any other interpretation.\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\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>In article <2202379a.0404131031.7794e31a@posting.google.com>,
Ken S. Tucker <dynamics@vianet.on.ca> wrote:

>Thank you, my reference is Blanco & McCuskey's
>"BASIC PHYSICS of the SOLAR SYSTEM"
>following their Eq.(4-40) they write,
>
> (1/2)*m*h^2*u^2 - (\mu)*m*u - E =
>
>where m = mass of orbiting particle, (\mu) = GM
>with M the central Mass, and u = 1/r, r=radius.
> The "h" is a constant based on Kepler's 2nd law
>to "sweep" equal areas in equal times.

This is a correct equation to describe the perihelion and aphelion
of a Keplerian orbit. It certainly doesn't describe an arbitrary
point on the orbit though.

The easiest way to see this is to note that this is a quadratic
equation for u, with everything else being constant, so it has at most
two solutions. Those two solutions are the values of u at perihelion
and aphelion. But clearly the equation can't be true at any other
point in the orbit.

>So I abbreviated their 1st term above to
>
> T = k/r^2
>
>by setting m=1 and k= (1/2) h^2.
>
>I also backed up to their Eq.(4-33)
>
> E = (1/2)*m*v^2 - GMm/r
>
>to confirm the meaning of "T" as to be interpretated
>as kinetic energy.

That's fine. But it's still all only true at those two specific
points in the orbit, not in between. But in your previous argument,
as far as I could tell anyway, you were assuming that this equation
held at all times, not just at two specific instants.

>Using Newtonian Mechanics, a decrement of the central
>Mass will cause an orbital perturbation. Is this same orbital
>perturbation equivlent to leaving M constant and applying
>reaction thrust?

I can't interpret this question in any sensible way. It seems to me
to mean "Can we attach a rocket to the mass m and fire it in such a
way that it follows the same orbit as it would if M were decreasing?"
The answer to that question is obviously yes, if we allow arbitrary
rocket firings. I can't imagine that that's what you really mean to
ask, but I can't figure out any other interpretation.

-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

Ken S. Tucker

Apr16-04, 02:28 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>david.williams@bayman.org (David Williams) wrote in message news:<1084722573.1572.1084696684@bayman.org>...\n>-> Using Newtonian Mechanics, a decrement of the central\n>-> Mass will cause an orbital perturbation. Is this same orbital\n>-> perturbation equivlent to leaving M constant and applying\n>-> reaction thrust?\n>>ken\n\n>Obviously, it would be *possible*, in a situation where the central\n>star\'s mass is constant, to use rocket power to drive a planet in the\n>path that it would have followed if the star\'s mass decreased and the\n>planet moved inertially. However, the *velocity* of the rocket-driven\n>planet at any point along this path would *not* be the same as the\n>velocity of the inertial planet at the same point. Basically, the\n>rocket-driven planet would be orbiting a star that is more massive than\n>the one that the inertial planet is orbiting, some time after the\n>stellar mass has started decreasing.\n>\n>So the analogy between the two planets, one driven by a rocket and the\n>other in free orbit around a star that is losing mass, can *not* be\n>exact.\n>\n>I doubt that it is a useful consideration, in any way.\n> dow\n\nThanks for the clarification Dave...\n\nOk, one of course could replace the thrusters by a\nlocal g-field perturbation, to eliminate the inertial\nobjection, but that suggests we move to GR which\nis better able to handle g-fields that vary in time, which\nI think the problem requires.\n\nThe important point David alludes to (IMHO) with\nhis "inertial" objection is the requirement that the solution\nfollows a geodesic in a time varying g-field.\n\nI do not have a lot of experience on time varying\ng-fields, as the star loses mass, so advice is welcome.\nFor example, it looks like the geodesic equation,\n\nDU^u/ds =0\n\n((0=dU^u/ds + GAMMA^u_ab U^a U^b))\n\n(D is absolute derivative, U^u = dx^u/ds)\nwill need to be true throughout the orbital\nevolution. (?)\n\nThe GAMMA term lends itself easily to time\ndependant partials of the metrics like g_00,0 or\ng_11,0 that is not so clear in Newton Mechanics\nthat I can see (?).\n\nThanks in advance\nKen S. Tucker\n\nPS: snippable, might be interesting to find a weak\nfield problem more easily solved by GR than\nNewton Mechanics!\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>david.williams@bayman.org (David Williams) wrote in message news:<1084722573.1572.1084696684@bayman.org>...
>-> Using Newtonian Mechanics, a decrement of the central
>-> Mass will cause an orbital perturbation. Is this same orbital
>-> perturbation equivlent to leaving M constant and applying
>-> reaction thrust?
>>ken

>Obviously, it would be *possible*, in a situation where the central
>star's mass is constant, to use rocket power to drive a planet in the
>path that it would have followed if the star's mass decreased and the
>planet moved inertially. However, the *velocity* of the rocket-driven
>planet at any point along this path would *not* be the same as the
>velocity of the inertial planet at the same point. Basically, the
>rocket-driven planet would be orbiting a star that is more massive than
>the one that the inertial planet is orbiting, some time after the
>stellar mass has started decreasing.
>
>So the analogy between the two planets, one driven by a rocket and the
>other in free orbit around a star that is losing mass, can *not* be
>exact.
>
>I doubt that it is a useful consideration, in any way.
> dow

Thanks for the clarification Dave...

Ok, one of course could replace the thrusters by a
local g-field perturbation, to eliminate the inertial
objection, but that suggests we move to GR which
is better able to handle g-fields that vary in time, which
I think the problem requires.

The important point David alludes to (IMHO) with
his "inertial" objection is the requirement that the solution
follows a geodesic in a time varying g-field.

I do not have a lot of experience on time varying
g-fields, as the star loses mass, so advice is welcome.
For example, it looks like the geodesic equation,

DU^u/ds =0((0=dU^u/ds + \GAMMA^u_{ab} U^a U^b))

(D is absolute derivative, U^u = dx^u/ds)
will need to be true throughout the orbital
evolution. (?)

The \GAMMA term lends itself easily to time
dependant partials of the metrics like g_{00},0 or
g_{11},0 that is not so clear in Newton Mechanics
that I can see (?).

Thanks in advance
Ken S. Tucker

PS: snippable, might be interesting to find a weak
field problem more easily solved by GR than
Newton Mechanics!

Ken S. Tucker

Apr18-04, 03:51 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>ebunn@lfa221051.richmond.edu wrote in message news:<c5mc6t\\$1da\\$1@lfa222122.richmond.edu>...\ n>In article <2202379a.0404131031.7794e31a@posting.google.com>, \n>Ken S. Tucker <dynamics@vianet.on.ca> wrote:\n>\n>>Thank you, my reference is Blanco & McCuskey\'s\n>>"BASIC PHYSICS of the SOLAR SYSTEM"\n>>following their Eq.(4-40) they write,\n>> (1/2)*m*h^2*u^2 - (mu)*m*u - E = 0\n>>where m = mass of orbiting particle, (mu) = GM\n>>with M the central Mass, and u = 1/r, r=radius.\n>> The "h" is a constant based on Kepler\'s 2nd law\n>>to "sweep" equal areas in equal times.\n\n>This is a correct equation to describe the perihelion and aphelion\n>of a Keplerian orbit. It certainly doesn\'t describe an arbitrary\n>point on the orbit though.\n>The easiest way to see this is to note that this is a quadratic\n>equation for u, with everything else being constant, so it has at most\n>two solutions. Those two solutions are the values of u at perihelion\n>and aphelion. But clearly the equation can\'t be true at any other\n>point in the orbit.\n\nThanks, the notation of (u=1/r) is sometimes\ndynamic and sometimes point specific as you\nexplained. ...\n\n>>I also backed up to their Eq.(4-33)\n>> E = (1/2)*m*v^2 - GMm/r\n>>to confirm the meaning of "T" as to be interpretated\n>>as kinetic energy.\n\n>That\'s fine. But it\'s still all only true at those two specific\n>points in the orbit, not in between. But in your previous argument,\n>as far as I could tell anyway, you were assuming that this equation\n>held at all times, not just at two specific instants.\n\nActually, I think Eq.(4-33) would be true\nanywhere in the orbit, because now two dynamic\nvariable, "v" and "r" are relating.\n\n>>Using Newtonian Mechanics, a decrement of the central\n>>Mass will cause an orbital perturbation. Is this same orbital\n>>perturbation equivlent to leaving M constant and applying\n>>reaction thrust?\n>\n>I can\'t interpret this question in any sensible way. It seems to me\n>to mean "Can we attach a rocket to the mass m and fire it in such a\n>way that it follows the same orbit as it would if M were decreasing?"\n>The answer to that question is obviously yes, if we allow arbitrary\n>rocket firings. I can\'t imagine that that\'s what you really mean to\n>ask, but I can\'t figure out any other interpretation.\n>-Ted\n\nThat\'s it Ted. Instead of decrementing M in the\npotential Phi = M/r we could decrement Phi by\nincrementing r, to change the graviatational potential,\nwhile leaving M constant.\nBeginning with E = T - Phi, a reducing central Mass,\nappears equivalent to a reducing Phi. So I reasoned a\nreducing Phi retains the portionality by increasing E and T\nby thrust reaction.\n\nHere\'s a few problems I have:\nAs the central Mass decreases we all agree\nthe orbit varies, it is spiral and not elliptical, hence\nall the elliptical orbital theory is doubtful. We try to\nretain elliptical orbital thoery by making dM/dt small,\nthen we pretend to call a spiral orbit one close to an\nelliptical orbit. Then using elliptical orbital equations\nwe simplify out the true spiral orbital perturbation,\n....that to me is scary.\n\nWe really can\'t use the simple lame definition of\neccentricity determined from a ratio of of semi-\nmajor and minor axes when these are dynamic.\nThere are means to calculate eccentricity at\nany point in the orbit given the paramenters at\nthat point in spacetime,\n(please see my ref, following Eq.(4-39).\n\nThanks again,\nKen S. Tucker\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>ebunn@lfa221051.richmond.edu wrote in message news:<c5mc6t$1da$1@lfa222122.richmond.edu>...
>In article <2202379a.0404131031.7794e31a@posting.google.com>,
>Ken S. Tucker <dynamics@vianet.on.ca> wrote:
>
>>Thank you, my reference is Blanco & McCuskey's
>>"BASIC PHYSICS of the SOLAR SYSTEM"
>>following their Eq.(4-40) they write,
>> (1/2)*m*h^2*u^2 - (\mu)*m*u - E =
>>where m = mass of orbiting particle, (\mu) = GM
>>with M the central Mass, and u = 1/r, r=radius.
>> The "h" is a constant based on Kepler's 2nd law
>>to "sweep" equal areas in equal times.

>This is a correct equation to describe the perihelion and aphelion
>of a Keplerian orbit. It certainly doesn't describe an arbitrary
>point on the orbit though.
>The easiest way to see this is to note that this is a quadratic
>equation for u, with everything else being constant, so it has at most
>two solutions. Those two solutions are the values of u at perihelion
>and aphelion. But clearly the equation can't be true at any other
>point in the orbit.

Thanks, the notation of (u=1/r) is sometimes
dynamic and sometimes point specific as you
explained. ...

>>I also backed up to their Eq.(4-33)>> E = (1/2)*m*v^2 - GMm/r
>>to confirm the meaning of "T" as to be interpretated
>>as kinetic energy.

>That's fine. But it's still all only true at those two specific
>points in the orbit, not in between. But in your previous argument,
>as far as I could tell anyway, you were assuming that this equation
>held at all times, not just at two specific instants.

Actually, I think Eq.(4-33) would be true
anywhere in the orbit, because now two dynamic
variable, "v" and "r" are relating.

>>Using Newtonian Mechanics, a decrement of the central
>>Mass will cause an orbital perturbation. Is this same orbital
>>perturbation equivlent to leaving M constant and applying
>>reaction thrust?
>
>I can't interpret this question in any sensible way. It seems to me
>to mean "Can we attach a rocket to the mass m and fire it in such a
>way that it follows the same orbit as it would if M were decreasing?"
>The answer to that question is obviously yes, if we allow arbitrary
>rocket firings. I can't imagine that that's what you really mean to
>ask, but I can't figure out any other interpretation.
>-Ted

That's it Ted. Instead of decrementing M in the
potential \Phi = M/r we could decrement \Phi by
incrementing r, to change the graviatational potential,
while leaving M constant.
Beginning with E = T - \Phi, a reducing central Mass,
appears equivalent to a reducing \Phi. So I reasoned a
reducing \Phi retains the portionality by increasing E and T
by thrust reaction.

Here's a few problems I have:
As the central Mass decreases we all agree
the orbit varies, it is spiral and not elliptical, hence
all the elliptical orbital theory is doubtful. We try to
retain elliptical orbital thoery by making dM/dt small,
then we pretend to call a spiral orbit one close to an
elliptical orbit. Then using elliptical orbital equations
we simplify out the true spiral orbital perturbation,
....that to me is scary.

We really can't use the simple lame definition of
eccentricity determined from a ratio of of semi-
major and minor axes when these are dynamic.
There are means to calculate eccentricity at
any point in the orbit given the paramenters at
that point in spacetime,
(please see my ref, following Eq.(4-39).

Thanks again,
Ken S. Tucker

David Williams

Apr19-04, 01:55 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>-> Using Newtonian Mechanics, a decrement of the central\n-> Mass will cause an orbital perturbation. Is this same orbital\n-> perturbation equivlent to leaving M constant and applying\n-> reaction thrust?\n\n-> If so the reduction of M in the 1st scenario, can be duplicated\n-> by leaving M constant and applying a reaction thrust in the\n-> direction of the mass\'s motion. In that 2nd scenario the\n-> eccentricity should increase until the orbit becomes parabolic.\n-> Pending Ted\'s approval of the standard orbital equations\n-> above, (IMO), we might regard the 2nd scenario.\n-> Regards\n-> Ken S. Tucker\n\nYour rocket will escape if its thrust is constant. It might not escape\nif the thrust declines asymptotically toward zero over time. Which\nrocket, the constant-thrust one or the declining-thrust one, would more\naccurately model the motions of an inertially-moving planet of a\ndeclining-mass star? Unless you can convincingly argue in favour of a\nthrust-time function that predicts escape (constant thrust being one\nsuch function), your analogy predicts nothing.\n\ndow\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>-> Using Newtonian Mechanics, a decrement of the central
-> Mass will cause an orbital perturbation. Is this same orbital
-> perturbation equivlent to leaving M constant and applying
-> reaction thrust?

-> If so the reduction of M in the 1st scenario, can be duplicated
-> by leaving M constant and applying a reaction thrust in the
-> direction of the mass's motion. In that 2nd scenario the
-> eccentricity should increase until the orbit becomes parabolic.
-> Pending Ted's approval of the standard orbital equations
-> above, (IMO), we might regard the 2nd scenario.
-> Regards
-> Ken S. Tucker

Your rocket will escape if its thrust is constant. It might not escape
if the thrust declines asymptotically toward zero over time. Which
rocket, the constant-thrust one or the declining-thrust one, would more
accurately model the motions of an inertially-moving planet of a
declining-mass star? Unless you can convincingly argue in favour of a
thrust-time function that predicts escape (constant thrust being one
such function), your analogy predicts nothing.

dow

Stephen Riley

Apr20-04, 02:34 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>In message <2202379a.0404171100.33e0ee30@posting.google.com >, Ken S.\nTucker <dynamics@vianet.on.ca> writes\n\n>Here\'s a few problems I have:\n> As the central Mass decreases we all agree\n>the orbit varies, it is spiral and not elliptical, hence\n>all the elliptical orbital theory is doubtful.\n\nI\'ve come at this from another angle and integrated the equations of\nmotion numerically. In 3D rather than just one degree of freedom Ted\nmentioned would be adequate. Anyway the upshot was that for small\ncontinuous mass losses (a small fraction of the mass lost per orbit),\neccentricity did indeed remain constant, and the semi-major axis\nlengthen. I ran this until about 1/3 the original mass remained and\nstill no change in eccentricity. Actually there likely is a small change\nin eccentricity but this can be made arbitrarily small by reducing the\nrate at which mass is lost, IMO. But for greater mass losses per orbit,\nthe satellite spirals outwards.\n\nMy simulation, an applet using an adaptive step 4th order Runge-Kutta,\nwas a toy but good enough here I think :\n\nhttp://www.sriley.co.uk/RocketVar\n\nIf anyone should want to run this applet, I should say it\'s a bit of a\nmemory hog, so it might be a good idea not to be in the middle of an\nunsaved edit. And secondly it takes a fair bit of CPU.\n\nIt\'s easy to simulate this kind of thing, why not do that and tackle\nGoldstein later...\n\n\n\n>We try to\n>retain elliptical orbital thoery by making dM/dt small,\n>then we pretend to call a spiral orbit one close to an\n>elliptical orbit. Then using elliptical orbital equations\n>we simplify out the true spiral orbital perturbation,\n>...that to me is scary.\n>\n>We really can\'t use the simple lame definition of\n>eccentricity determined from a ratio of of semi-\n>major and minor axes when these are dynamic.\n> There are means to calculate eccentricity at\n>any point in the orbit given the paramenters at\n>that point in spacetime,\n>(please see my ref, following Eq.(4-39).\n>\n>Thanks again,\n>Ken S. Tucker\n>\n\nRegards\n\n--\nStephen Riley\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>In message <2202379a.0404171100.33e0ee30@posting.google.com>, Ken S.
Tucker <dynamics@vianet.on.ca> writes

>Here's a few problems I have:
> As the central Mass decreases we all agree
>the orbit varies, it is spiral and not elliptical, hence
>all the elliptical orbital theory is doubtful.

I've come at this from another angle and integrated the equations of
motion numerically. In 3D rather than just one degree of freedom Ted
mentioned would be adequate. Anyway the upshot was that for small
continuous mass losses (a small fraction of the mass lost per orbit),
eccentricity did indeed remain constant, and the semi-major axis
lengthen. I ran this until about 1/3 the original mass remained and
still no change in eccentricity. Actually there likely is a small change
in eccentricity but this can be made arbitrarily small by reducing the
rate at which mass is lost, IMO. But for greater mass losses per orbit,
the satellite spirals outwards.

My simulation, an applet using an adaptive step 4th order Runge-Kutta,
was a toy but good enough here I think :

http://www.sriley.co.uk/RocketVar

If anyone should want to run this applet, I should say it's a bit of a
memory hog, so it might be a good idea not to be in the middle of an
unsaved edit. And secondly it takes a fair bit of CPU.

It's easy to simulate this kind of thing, why not do that and tackle
Goldstein later...

>We try to
>retain elliptical orbital thoery by making dM/dt small,
>then we pretend to call a spiral orbit one close to an
>elliptical orbit. Then using elliptical orbital equations
>we simplify out the true spiral orbital perturbation,
>...that to me is scary.
>
>We really can't use the simple lame definition of
>eccentricity determined from a ratio of of semi-
>major and minor axes when these are dynamic.
> There are means to calculate eccentricity at
>any point in the orbit given the paramenters at
>that point in spacetime,
>(please see my ref, following Eq.(4-39).
>
>Thanks again,
>Ken S. Tucker
>

Regards

--
Stephen Riley