# Angular momentum, potential gravitational energy, rotation

Suppose a satellite of mass $m$ is orbiting a much heavier object of mass $M$ due to gravitational attraction in an elliptical orbit. The distance between the objects at which they are the closest is $r_1$ at which point the speed of $m$ is $v_1$. At the farthest point apart they are $r_2$ and $v_2$. This means that $v_1\perp r_1$ and $v_2\perp r_2$.

Given this situation, the three following equations come to my mind:

• Due to conservation of angular momentum $$mv_1r_1=mv_2r_2.$$(as $v_1\perp r_1$ and $v_2\perp r_2$.)

• Due to energy conservation $$\frac{1}{2}mv_1^2-G\frac{Mm}{r_1}=\frac{1}{2}mv_2^2-G\frac{Mm}{r_2}.$$

• Due to the fact that $v_1\perp r_1$ and $v_2\perp r_2$, the relationship between radius, linear velocity and centripetal acceleration could (?) be applied: $a=\frac{v^2}{r}$, where $a=\frac{GM}{r^2}$, thus: $$v_1^2r_1=GM=v_2^2r_2$$

As they give very different relationships between distances and velocities, my question is, which of them are actually applicable to the situation.

• Do any of these equations directly contradict each other? It seems plausible that not all of $v_1,v_2,r_1,r_2$ are independent quantities. Jun 18, 2018 at 15:41
• The first and third equation contradict each other because, well, they are not equivalent. Jun 18, 2018 at 15:57
• Related: Satellite in Elliptical orbit. Jun 18, 2018 at 18:40

The third equation you are trying to use the formula for centripetal acceleration $a=\frac{v^2}{r}$. However, this requires not only that the acceleration be purely radial (which it is, we're dealing with a central force) but also that the radial acceleration is zero. In general the acceleration in plane polar co-ordinates $(r,\theta)$ is:
$\vec{a} = (\ddot{r}-r\dot{\theta}^2)\hat{r} + (2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\theta}$
At the apopapse/periapse we have $\dot{r}=0$ not $\ddot{r}=0$, and $\vec{F}=m\vec{a}=-\frac{GM}{r^2}\hat{r}$ as you've identified. This means the correct version of your third equation should be:
$\frac{v_1^2}{r_1}-\ddot{r}=\frac{GM}{r_1^2}$.