Calculation of orbital speed for given eccentricityby theskipirate Tags: calculation, eccentricity, orbital, speed 

#1
Nov1109, 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 



#2
Nov1209, 02:22 AM

P: 774

v^{2} = 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(1e)) or highest point (which is Q = a(1+e)) on the orbit. And, of course, a = (q + Q)/2 



#3
Nov1209, 06:58 AM

Mentor
P: 14,434

The six classical orbital elements used to describe an elliptical orbit are
Alternately, some use the mean anomaly M_{0} at some epoch time t_{0} 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.
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 zxz 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 righthand 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}{1e}} \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 (1e^2)} \\ \vec h &= h\hat z_{\text{orb}} \\ \vec e &= e\hat x_{\text{orb}} \\ \vec v &= \frac 1 {a(1e^2)} \vec h \times (\vec e  \hat r) \endaligned[/tex] 


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