0
$\begingroup$

A satellite is in a circular orbit a distance $h$ above the surface of the earth with speed $v_0$. It suffers a head-on collision with some debris which reduces its speed to $kv_0$, where $k$ is a constant in the range $0<k<1$, but does not change its direction. Calculate the eccentricity of the new orbit.

We assume that the satellite is being acted on by an attractive inverse square force.

Now before the collision we have $$\frac{d^2u}{d\theta^2}+u=\frac{\gamma}{L^2}\implies\frac{1}{R} = \frac{\gamma}{R^2 v_0^2} \implies \gamma = Rv_0^2$$

After the collision, $\gamma$ is unchanged, but the angular momentum is. We shall denote that by $L'$. how do I progress from here? Because I don't know how the magnitude of the angular component of the velocity has changed. What is L'?

Would it be correct to say after the collision $L' = kRv_0$? I feel this is not correct.

$\endgroup$

1 Answer 1

0
$\begingroup$

If you assume the velocity immediately after the collision is in the same direction it was before, it is correct to say $L'=kRv_0$. At that moment the radius vector and the velocity vector are perpendicular. If the direction has changed, you need to project the new velocity on the perpendicular to the radius vector to get the new angular momentum.

$\endgroup$
3
  • $\begingroup$ Ah right, I see. Thanks Ross. $\endgroup$
    – user197848
    Oct 10, 2015 at 19:54
  • $\begingroup$ Ross, I get the eccentricity to be $k^2 - 1$ this is going to be less than 0. Or do I take the absolute value of this? $\endgroup$
    – user197848
    Oct 10, 2015 at 21:00
  • $\begingroup$ No, it shouldn't be less than zero. If $k$ is a tiny bit less than $1$, the eccentricity should be small because the orbit will still be almost circular. I would suspect you dropped a sign and it should be $1-k^2$ (or perhaps the square root of that). $\endgroup$ Oct 10, 2015 at 21:21

You must log in to answer this question.