# What is the reason for assuming angular momentum is conserved for a moon-planet system where the moon is in an elliptical orbit around the planet?

I came across a question in the Princeton Review book for AP Physics 1 and then found the same question in the Collegeboard AP Physics course guide. Both have different answers. (See the questions below)

I understand that the distance can be calulcated using conservation of angular momentum, so that leaves just two answer choices. My question is about the reason that conservation of angular momentum can be used.

I know that the gravitational force on the moon is directed towards the planet, but what does that tell me about the conservation of momentum? Nothing... The direction of the gravitational force is NOT perpendicular to the path or velocity of the moon since it's in an elliptical orbit.

On the other hand, Newton's Third Law is connected to (or perhaps the basis of) the law of conservation of momentum. So shouldn't the answer be A for the Princeton question and D for the Collegeboard question? Am I wrong?

See the questions here:

Princeton Review

Collegeboard

• @MSayanvala The torque exerted by a force is given by $\tau = \vec{r}\times\vec{F}$. Because the force $\vec{F}$ is directed along the position vector $\vec{r}$ this cross-product vanishes and hence there is no torque. Commented Mar 6, 2020 at 9:31
If your frame of reference is placed on the center of mass of the planet, there is no torque exerted on the moon ($$\mathbb{r}\parallel\mathbb{F}$$), so angular momentum is conserved. At the points A and B the velocity of the moon is orthogonal to its position vector, so the magnitude of the angular momentum vector is $$|L|=m r_A v_A=mr_ B v_B.$$ At this point you just solve for $$r_B$$.