# Radial and Angular Velocities of a Mass in an Elliptical Orbit

1. Dec 2, 2013

### ecastro

Given:
According to my reference (An Introduction to Modern Astrophysics), I should start with this equation:

$r = \frac{a\left(1 - e^2\right)}{1+ e \cos\left(\theta\right)}$

And by using Kepler's Second Law, I should be able to derive the expression for $\mathbf{v}_r$ and $\mathbf{v}_\theta$.

Attempt:

Since $\mathbf{v}_r = \frac{dr}{dt}\mathbf{\hat{r}}$, then,

$\frac{dr}{d\theta}\frac{d\theta}{dt} = \frac{d}{d\theta} \left[\frac{a\left(1 - e^2\right)}{1+ e \cos\left(\theta\right)}\right]\frac{d\theta}{dt} \mathbf{\hat{r}}$, and

$\frac{d}{d\theta} \left[\frac{a\left(1 - e^2\right)}{1+ e \cos\left(\theta\right)}\right] = \frac{ae\sin\left(\theta\right)\left(1 - e^2\right)}{\left[1+ e \cos\left(\theta\right)\right]^{2}}$.

According to Kepler's Second Law,

$\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt} \\ \frac{1}{2}\frac{L}{\mu} = \frac{1}{2}r^2\frac{d\theta}{dt} \\ \frac{L}{\mu r^2} = \frac{d\theta}{dt} \\ \frac{d\theta}{dt} = \frac{L}{\mu r}\frac{1}{r} = \frac{v}{r} = \frac{2\pi}{P}\\$

Thus,

$\mathbf{v}_r = \frac{2\pi ae\sin\left(\theta\right)\left(1 - e^2\right)}{P\left[1+ e \cos\left(\theta\right)\right]^{2}} \mathbf{\hat{r}}.$

The same goes for $\mathbf{v}_\theta$,

$\mathbf{v}_\theta = r \frac{d\theta}{dt} \mathbf{\hat{\theta}} = \frac{2\pi a\left(1 - e^2\right)}{P\left[1+ e \cos\left(\theta\right)\right]} \mathbf{\hat{\theta}}.$

My final answers should only be functions of $P, a, e,$ and $\theta$ only. Please check if my calculations are wrong.

Now, I need to square both and add them up to be able to acquire this,

$v = G\left(m_1 + m_2\right)\left(\frac{2}{r} - \frac{1}{a}\right)$

Squaring and adding the two velocities,

$v_r^2 = \frac{4\pi^2 a^2e^2\sin^2\left(\theta\right)\left(1 - e^2\right)^2}{P^2\left[1+ e \cos\left(\theta\right)\right]^{4}}$

$v_\theta^2 = \frac{4\pi^2 a^2\left(1 - e^2\right)^2}{P^2\left[1+ e \cos\left(\theta\right)\right]^2}$

Their sum should be something like this,

$v_r^2 + v_\theta^2 = \frac{4 \pi^2 a^2 \left(1 - e^2\right)^2}{P^2\left[1+ e \cos\left(\theta\right)\right]^2}\left\{\frac{e^2\sin^2\left(\theta\right)}{\left[1+ e \cos\left(\theta\right)\right]^2} + 1\right\}$

If I multiply $\frac{a}{a}$, I will be able to bring out the gravitational constant $G$,

$\frac{G\left(m_1 + m_2\right)\left(1 - e^2\right)^2}{a\left[1+ e \cos\left(\theta\right)\right]^2} \left\{\frac{e^2\sin^2\left(\theta\right)}{\left[1+ e \cos\left(\theta\right)\right]^2} + 1\right\}$.

This is where I get stuck. I tried combining the two latter terms and having a trigonometric identity of $e^2\sin^2\left(\theta\right) + e^2\cos^2\left(\theta\right) = e^2$, but it seems I don't get anywhere doing it. Could you help me or give me tips on what should I do?