Does the mass change for a body with eccentric orbit?

Question

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?

Answer 1

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: $$v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$$

where:
$v$ 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.