I am a student with a programming and electronics background attempting to recap on orbital mechanics. I would appreciate anyone with a physics specialisation helping me understand where my ignorance lies on the following:

According to Newtonian calculations the mass of a body orbiting another (larger body) can be calculated with: $$ M = \frac{4\pi^2R^3}{GT^2} $$ The $T$ time period can be calculated by using the mean radius $R_m$: $$ T= \sqrt{R_m^3} $$

My confusion arises when faced with the possibility of an orbit with high eccentricity value:

Using the above calculations only, a high eccentricity means a greater change in speed and thus radius, this then leads to the assumption that the specific mass at different orbital speeds will provide a slightly different mass, or rather will give me a changing "relativistic mass" $M_r$(?).

I'm assuming to then correct this to rest mass, I can use the rest Mass $$ M_0 = \frac{M_r}{\sqrt{(1-(v/c)^2}}$$

This leads to one of two issues with the equations I'm using.

  1. Is the derivative of $M_0/M_r$ so small at most celestial objects speeds that using this correction is pointless?
  2. Is my thinking/solution incorrect?
  • 1
    $\begingroup$ That equation comes is derived under the assumption that the orbit is circular. $\endgroup$ Aug 23 '15 at 23:15
  • $\begingroup$ Aha many thanks!, is there an alternative that factors in eccentricity? $\endgroup$ Aug 23 '15 at 23:18
  • $\begingroup$ Appreciate the edits for clarity! $\endgroup$ Aug 23 '15 at 23:20
  • $\begingroup$ I am not sure. I think a low eccentricity orbit would be well approximated by this equation, just using the average radius. This is completely unrelated to relativistic mass. $\endgroup$ Aug 23 '15 at 23:24
  • 1
    $\begingroup$ In the suggested vis-viva equation, the mass is still the mass of the central body. I guess the question is, what do you really want to do. $\endgroup$ Aug 24 '15 at 0:02

Don't worry about relativistic corrections - they are insignificant for most planetary orbital motion at the level you try to model/understand it.

You want to look at the vis viva equation which is well explained on Wikipedia: $$v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$$

$v$ is the relative speed of the two bodies
$r$ is the distance between the two bodies
$a$ is the semi-major axis (a > 0 for ellipses, a = ∞ or 1/a = 0 for parabolas, and a < 0 for hyperbolas)
$G$ is the gravitational constant
$M$ is the mass of the central body

It allows you to calculate the orbital period and velocity regardless of the eccentricity. Among other things, it shows that the period scales with the semi-major axis of the orbit ($T^2\propto a^3$ which is also Kepler's third law). For a circle, the semimajor axis is equal to the radius; for a more eccentric orbit, there is no "radius" - but there is still a major axis...

See if that gets you going. Ask questions in the comments if this is not sufficiently clear.

  • $\begingroup$ Many thanks!, This will hopefully be enough as I'm deriving everything from a pseudo random based variables of stars. Will let you know, Apologies I can't up vote this due to me current rep. One problem that's bugging my however: I am required to do perform similar equations based on a singularities properties *sigh, would this proportionality still apply? $\endgroup$ Aug 23 '15 at 23:37
  • $\begingroup$ Woot!, Just tested on 250 Billion different star - sun ranges without error. Which is good enough I say :) (Yes GPU calculation is that amazing :P) $\endgroup$ Aug 24 '15 at 0:03
  • $\begingroup$ Just an update: I've reordered things to reference the constant area speed For simulation at least, it helps avoid error. $\endgroup$ Aug 25 '15 at 1:51
  • $\begingroup$ Yes conservation of angular momentum is one constraint you can use on the orbit - which is what "constant area in constant time" really is... $\endgroup$
    – Floris
    Aug 25 '15 at 1:57
  • 1
    $\begingroup$ Just an update: I had to do iterative update based on time and recalculating perihelion with a given variation, after that all good :) $\endgroup$ Oct 15 '15 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.