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Calculation of orbital speed for given eccentricity

by theskipirate
Tags: calculation, eccentricity, orbital, speed
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theskipirate
#1
Nov11-09, 03:59 AM
P: 1
Hi,

I am trying to work out, for a given eccentricity,a known planet mass and a known value for the semi major axis is there a formula to work out the orbital speed of a planet around a central star?

I have looked around online with not much success and was wondering if anyone could point me in the right direction?

Cheers
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qraal
#2
Nov12-09, 02:22 AM
P: 775
Quote Quote by theskipirate View Post
Hi,

I am trying to work out, for a given eccentricity,a known planet mass and a known value for the semi major axis is there a formula to work out the orbital speed of a planet around a central star?

I have looked around online with not much success and was wondering if anyone could point me in the right direction?

Cheers
The average orbital speed is simply √(GM/a). I suspect that what you want is the orbital speed at a given position on the orbital ellipse. You can't have looked too far on the Net if you haven't found that yet. But it usually isn't formulated with the eccentricity in the equation.

v2 = GM( 2/r - 1/a )

...that's it. Knowing the eccentricity isn't needed in this particular version. You do need to know it if you want to know the speed at the lowest (which is q = a(1-e)) or highest point (which is Q = a(1+e)) on the orbit. And, of course, a = (q + Q)/2
D H
#3
Nov12-09, 06:58 AM
Mentor
P: 15,067
Quote Quote by qraal View Post
v2 = GM( 2/r - 1/a )
That's the vis-viva equation. It is a consequence of conservation of energy. So, how to determine the radius?

The six classical orbital elements used to describe an elliptical orbit are
  • a, the semi-major axis length
  • e, the orbital eccentricity
  • i, the inclination of the orbital plane
  • Ω, the right ascension of the ascending node
  • ω, the argument of periapsis
  • tp, the time of periapsis passage

Alternately, some use the mean anomaly M0 at some epoch time t0 in lieu of the time of periapsis passage. Specifying the time of periapsis passage is a special case of this more general form. The mean anomaly is zero at periapsis.

The problem at hand: Given a set of orbital element, how do you compute the current position and velocity? Those three angles, i, Ω, and ω, are only needed if you need to know the position and velocity vectors. The problem is a bit easier if all you want to know are the orbital radius r and the magnitude of the velocity vector v at some point in time.

First, some scalars.
  • [tex]\mu\equiv GM[/tex] The value of the gravitational constant G is known to only four decimal places. The product G*M is known to many more places than that. Scientists know the Sun's gravitational parameter to over 10 places, the Earth's to nearly 9 places. Astronomers almost always use μ in lieu of G*M.

  • [tex]\dot M=\sqrt{\frac{\mu}{a^3}}[/tex] Mean motion.

  • [tex]M(t) = M_0 + (t-t_0)\dot M[/tex] The mean anomaly at the time in question.

  • [tex]M=E-e\sin E[/tex] The eccentric anomaly at the time in question. This equation (Kepler's equation) however gives the mean anomaly in terms of the eccentric anomaly. Newton's method works very well for small values of e. It can fail for large values of e and a bad initial guess. Google "Kepler's equation" and you will find lots of references on how to solve this.
  • [tex]r=a(1-e\cos E)[/tex] The orbital radius at the time in question. Combine this with the vis-viva and you have the magnitude of the orbital velocity.

A bit more work is needed if you want the position and velocity vectors. Now those three angles come into play. Ω, i, and ω form an Euler rotation sequence (zxz) from the inertial frame in which those angles are referenced to the orbital reference frame. (Google Euler angles; this is a specific example of the standard astronomical z-x-z Euler rotation.) The orbital reference frame has [tex]\hat x_{\text{orb}}[/tex] pointing toward periapsis, [tex]\hat z_{\text{orb}}[/tex] pointing along the orbital angular momentum vector, and [tex]\hat y_{\text{orb}}=\hat z_{\text{orb}} \times \hat x_{\text{orb}}[/tex] completing a right-hand system.

One more angle, the true anomaly, is needed to determine the position and velocity vectors. This is related to the eccentric anomaly via

[tex]\tan\left(\frac {\theta} 2\right) = \sqrt{\frac{1+e}{1-e}} \tan\left(\frac E 2\right)[/tex]

With this,

[tex]\aligned
\hat r &= \cos \theta \hat x_{\text{orb}} + \sin \theta \hat y_{\text{orb}} \\
\vec r &= r \hat r \\
h &= \sqrt{\mu a (1-e^2)} \\
\vec h &= h\hat z_{\text{orb}} \\
\vec e &= e\hat x_{\text{orb}} \\
\vec v &= \frac 1 {a(1-e^2)} \vec h \times (\vec e - \hat r)
\endaligned[/tex]


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