# Rocket orbitting Moon [closed]

The Figure shows six paths by which a rocket orbiting a moon might move from point a to point b. Rank the paths according to the net work done on the rocket by the gravitational force from the moon, greatest first. I was not able to find the work done. As i think that in the elliptical path, the work done shouldn't be 0, the work done in every path would be different.

• The potential energy difference between a and b is path independent. Sep 26 at 6:41
• Hello there, and welcome to the Physics Stack Exchange! Homework and "check my work" questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. Please read this post on asking homework questions and this post for "check my work" questions. Sep 26 at 7:49

Think about how the force from gravity is directed. It's in the radial direction seen from the moon. Since it is radially then when moving perpendicular to the radial direction there is no work done. This means that both the work $$W=F_g*h$$ done and the potential energy $$E=mgh$$ is path independent and depends only on the net difference in height $$h$$. I now leave you to answer the question of the ranking in a and b which should not be a struggle with the previous being stated.

• But the work done by it on point a and b wouldn't be zero because it wouldn't be perpendicular to the radial direction! Sep 26 at 8:13
• I'm unsure If I understand your comment. If you mean that the work between a and b is non-zero regardless of path (assuming that the height relative to moon is different which it seems in the figure) then I agree. But what only matters for the work and potential energy is the difference in height relative to the moon.
– ludz
Sep 26 at 8:46
• The work done would be zero for the elliptical if the height difference between a and b is zero.
– ludz
Sep 26 at 11:39
• I may guess what can confuse you and that is "The elliptical is not moving perpendicular to the radial direction of the moon" and that is true but the movement of the elliptical will have two components, one moving perpendicular and one moving radially which when added will give the trajectory that you see. The perpendicular component does no work but the radially component does work and this work is determined by the net difference of height relative to the moon from the initial position a.
– ludz
Sep 26 at 11:43
• Thank you so much for clearing my doubt Sep 26 at 12:54