-1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – jacob1729
    Jun 18, 2018 at 15:41
  • 1
    $\begingroup$ The first and third equation contradict each other because, well, they are not equivalent. $\endgroup$ Jun 18, 2018 at 15:57
  • 1
    $\begingroup$ Related: Satellite in Elliptical orbit. $\endgroup$ Jun 18, 2018 at 18:40

1 Answer 1

4
$\begingroup$

Your first two equations are correct, the third one is wrong.

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:

$\frac{v_1^2}{r_1}-\ddot{r}=\frac{GM}{r_1^2}$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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