Science > Physics > Q about formula for eccentricity under Newtonian Gravity.
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Science > Physics |
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"" |
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11 Dec 2003 09:15:54 AM |
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Q about formula for eccentricity under Newtonian Gravity. |
I found this on a web search, the result was stated without the
details of derivation:
For any conical orbit of a small test particle in a Newtonian
gravitational field around a central mass m, the eccentricity (e) is
given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total
energy (kinetic plus potential), h = rv_t is the angular momentum, v
is the total speed, v_t is the tangential component of the speed, and
r is the radial distance from the center of the mass. (Note v_t is v
subscript t).
Can someone please explain exactly why the above is true! Thanks.
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| User: "Franz Heymann" |
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| Title: Re: Q about formula for eccentricity under Newtonian Gravity. |
12 Dec 2003 05:32:44 AM |
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<tech_why@yahoo.com> wrote in message
news:964fe272.0312110715.ad4291d@posting.google.com...
I found this on a web search, the result was stated without the
details of derivation:
For any conical orbit of a small test particle in a Newtonian
gravitational field around a central mass m, the eccentricity (e) is
given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total
energy (kinetic plus potential), h = rv_t is the angular momentum, v
is the total speed, v_t is the tangential component of the speed, and
r is the radial distance from the center of the mass. (Note v_t is v
subscript t).
Can someone please explain exactly why the above is true! Thanks.
Solve Newton's Equations of motion for such orbits and you will see that it
comes out of the wash.
This is a standard first year classical mechanics problem, dealt with in any
worth while text book on the subject
Franz Heymann
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| User: "Greg Neill" |
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| Title: Re: Q about formula for eccentricity under Newtonian Gravity. |
11 Dec 2003 01:55:51 PM |
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<tech_why@yahoo.com> wrote in message
news:964fe272.0312110715.ad4291d@posting.google.com...
I found this on a web search, the result was stated without the
details of derivation:
For any conical orbit of a small test particle in a Newtonian
gravitational field around a central mass m, the eccentricity (e) is
given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total
energy (kinetic plus potential), h = rv_t is the angular momentum, v
is the total speed, v_t is the tangential component of the speed, and
r is the radial distance from the center of the mass. (Note v_t is v
subscript t).
Can someone please explain exactly why the above is true! Thanks.
Your m in the above is actually mu = G*M, the gravitational
parameter. It's possible that you've chosen a canonical
system of units where G = 1.
You will find a derivation of this expression in "Fundamentals
of Astrodynamics" by Bate et al. The text is only about ten
bucks on Amazon; it's a great buy.
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| User: "Harry Conover" |
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| Title: Re: Q about formula for eccentricity under Newtonian Gravity. |
11 Dec 2003 02:10:47 PM |
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wrote in message news:<964fe272.0312110715.ad4291d@posting.google.com>...
I found this on a web search, the result was stated without the
details of derivation:
For any conical orbit of a small test particle in a Newtonian
gravitational field around a central mass m, the eccentricity (e) is
given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total
energy (kinetic plus potential), h = rv_t is the angular momentum, v
is the total speed, v_t is the tangential component of the speed, and
r is the radial distance from the center of the mass. (Note v_t is v
subscript t).
Can someone please explain exactly why the above is true! Thanks.
I'd suggest you reference Becker's book "Introduction to Theoretical
Mechanics", Section 10-5 in which he addresses "Equation of the Orbit
by the Energy Method". While the equations of the derivation are a
bit too much to post on Usenet, essentially Becker begins with
expressions for the total energy and angular-momentum :
T + V = W
mr^2*Theta(dot) = J
After a bit of integration and routine substitutions, Becker imports a
previously derived relationship derived earlier in chapter 10:
r = 1/((1/ep) - (1/p)Cos(Theta))
and obtains:
e = sqrt((2WJ^2)/mk^2) + 1)
This is a remarkably similar to form as the equation that you posted,
and by expanding to equivalent terms as that you posted, I suspect
that they will or should turn out to be equivalent. I wasn't
sufficiently ambitious to perform this final step.
Harry C.
p.s. My copy of Becker's text is dated 1954 (Lord, how time flies!),
so you may have a difficult time locating a copy. If so, I'd be happy
to mail you a copy of this interesting chapter. If you need this,
please post a followup and I'll contact you via email (my posting
email address is not my real email address because of spam avoidance).
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| User: "" |
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| Title: Re: Q about formula for eccentricity under Newtonian Gravity. |
12 Dec 2003 08:53:20 AM |
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(Harry Conover) wrote in message news:<7ce4e226.0312111210.781b42a4@posting.google.com>...
tech_why@yahoo.com wrote in message news:<964fe272.0312110715.ad4291d@posting.google.com>...
I found this on a web search, the result was stated without the
details of derivation:
For any conical orbit of a small test particle in a Newtonian
gravitational field around a central mass m, the eccentricity (e) is
given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total
energy (kinetic plus potential), h = rv_t is the angular momentum, v
is the total speed, v_t is the tangential component of the speed, and
r is the radial distance from the center of the mass. (Note v_t is v
subscript t).
Can someone please explain exactly why the above is true! Thanks.
I'd suggest you reference Becker's book "Introduction to Theoretical
Mechanics", Section 10-5 in which he addresses "Equation of the Orbit
by the Energy Method". While the equations of the derivation are a
bit too much to post on Usenet, essentially Becker begins with
expressions for the total energy and angular-momentum :
T + V = W
mr^2*Theta(dot) = J
After a bit of integration and routine substitutions, Becker imports a
previously derived relationship derived earlier in chapter 10:
r = 1/((1/ep) - (1/p)Cos(Theta))
and obtains:
e = sqrt((2WJ^2)/mk^2) + 1)
This is a remarkably similar to form as the equation that you posted,
and by expanding to equivalent terms as that you posted, I suspect
that they will or should turn out to be equivalent. I wasn't
sufficiently ambitious to perform this final step.
Harry C.
p.s. My copy of Becker's text is dated 1954 (Lord, how time flies!),
so you may have a difficult time locating a copy. If so, I'd be happy
to mail you a copy of this interesting chapter. If you need this,
please post a followup and I'll contact you via email (my posting
email address is not my real email address because of spam avoidance).
Thanks. If you have a scanner you could email me the document at
tech_why@yahoo.com. Thanks.
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| User: "Igor" |
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| Title: Re: Q about formula for eccentricity under Newtonian Gravity. |
11 Dec 2003 04:05:19 PM |
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wrote in message news:<964fe272.0312110715.ad4291d@posting.google.com>...
I found this on a web search, the result was stated without the
details of derivation:
For any conical orbit of a small test particle in a Newtonian
gravitational field around a central mass m, the eccentricity (e) is
given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total
energy (kinetic plus potential), h = rv_t is the angular momentum, v
is the total speed, v_t is the tangential component of the speed, and
r is the radial distance from the center of the mass. (Note v_t is v
subscript t).
Can someone please explain exactly why the above is true! Thanks.
This is quite adequately derived in most theoretical mechanics
textbooks (at either the undergrad or grad level). The one I
recommend is Herbert Goldstein. It's the most thorough treatment of
orbital motion I'm aware of.
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| User: "Rob Johnson" |
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| Title: Re: Q about formula for eccentricity under Newtonian Gravity. |
11 Dec 2003 09:26:13 PM |
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In article <964fe272.0312110715.ad4291d@posting.google.com>,
tech_why@yahoo.com wrote:
I found this on a web search, the result was stated without the
details of derivation:
For any conical orbit of a small test particle in a Newtonian
gravitational field around a central mass m, the eccentricity (e) is
given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total
energy (kinetic plus potential), h = rv_t is the angular momentum, v
is the total speed, v_t is the tangential component of the speed, and
r is the radial distance from the center of the mass. (Note v_t is v
subscript t).
Can someone please explain exactly why the above is true! Thanks.
In my article at http://www.whim.org/nebula/math/planets.html, I derive
formulas, which with a small bit of work, verify this formula. In that
article, I name a couple of orbital constants that can be derived from
the initial position, p, and velocity, v, of the satellite along with
the gravitational constant, G, and the mass of the primary, M.
1
k = - |p x v| [3a]
1 2
1 2 GM
k = - |v| - --- [6a]
2 2 |p|
k_1 is the rate of area swept out per unit time; it is also your h/2.
k_2 is the energy per unit mass of the satellite; it is also your E.
Later in the article, while showing that orbits are conic sections, I
derive the formula of the orbit as
1
r = ---------------- [10]
b + c cos(@-k_3)
where b = GM/(4 k_1^2) and c^2 = b^2 + k_2/(2 k_1^2). I match that
formula with the formula in terms of the semimajor axis, a, and
eccentricity, e.
a(1-e^2)
r = ------------ [10a]
1 + e cos(@)
Looking at these formulas,
c
e = -
b
k_2
= sqrt( 1 + ----------- )
2 k_1^2 b^2
8 k_2 k_1^2
= sqrt( 1 + ----------- )
(GM)^2
2 E h^2
= sqrt( 1 + ------- )
(GM)^2
which is your equation using m = GM. This implies that if E < 0, the
orbit is elliptical and if E > 0, the orbit is hyperbolic.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
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