3
$\begingroup$

The question and answer are on pg.8-10 of this PDF:

At first, I went through it, thinking nothing of it. But then, I wondered: "What if we picked a final state in which the space junk was NOT at closest approach, but an arbitrary distance away from the center of the moon?" The equation (eq.11) would be exactly the same! What does that mean? Since obviously the distance from the space junk to the moon changes continuously, yet the form of the equation remains the same.

Using conservation of energy: $$\frac{1}{2}mv_i^2-\frac{GMm}{R}=\frac{1}{2}mv_f^2-\frac{GMm}{r}$$ And conservation of angular momentum: $$mRv_i=mrv_f$$ For any $v_f$ and $r$.

Now look. We have two equations and two unknowns, $v_f$ and $r$. This suggests that there is a unique solution for both. If we solve for one and plug that into the other equation, we'll get a unique result (or perhaps end up with a quadratic equation, which doesn't fix the problem). How do we reconcile this?

$\endgroup$
1
  • $\begingroup$ You will end up with a quadratic equation, however one of the solutions probably isn't a physical correct. $\endgroup$
    – fibonatic
    May 22, 2013 at 6:01

1 Answer 1

4
$\begingroup$

The velocity of the orbiting space junk is a vector, with both a radial and a tangential component.

$$\vec{v}_f = \dot{r}_f\hat{r} + r_f\dot{\theta}_f\hat{\theta}$$

(my $r_f$ is your $r$) The equation for conservation of angular momentum involves only the tangential component of velocity, because it comes from the cross product of the radius vector and the velocity.

$$\vec{L}_f = m\vec{r}_f\times\vec{v}_f = mr_f\hat{r}\times\bigl(\dot{r}_f\hat{r} + r_f\dot{\theta}_f\hat{\theta}\bigr) = mr_f(r_f\dot{\theta}_f)\hat{z} = mr_fv_f\hat{z}$$

But the equation for energy conservation involves both components.

$$E_f = \frac{1}{2}mv_f^2 - \frac{GMm}{r_f} = \frac{1}{2}m\dot{r}_f^2 + \frac{1}{2}mr_f^2\dot{\theta}_f^2 - \frac{GMm}{r_f}$$

The combination $r_F\dot{\theta}_f$ corresponds to your $v_f$, but the equation as you've written it in the question is missing the $\frac{1}{2}m\dot{r}_f^2$ term, which corresponds to the energy of motion toward or away from the moon. At apapsis and periapsis (furthest and closest points of the orbit), $\dot{r}_f = 0$ momentarily, so you can ignore this term, as is done in the problem you're asking about. But for other points on the orbit, that term is nonzero, which means you have an extra variable. That prevents you from finding a unique solution without specifying which point in the orbit you're at.

$\endgroup$
1
  • $\begingroup$ Nice explanation. To be cheeky, the pair of equations will indeed yield precisely two solutions: apapsis and periapsis. $\endgroup$
    – Art Brown
    May 22, 2013 at 6:15

Your Answer

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