Would a change in the mass or radius of the moon have any effect on its speed?

When asked this question in an assignment, I stated that it did not. My reasoning being that a change in the mass or radius of the moon would in no way change the acceleration or velocity of the moon. The moon would travel along the same path, the only difference being that the gravitational force being exerted by/on the Earth and Moon would be of a lesser magnitude. As the mass of the Moon decreases by ½, does the gravitational force.



The original gravitational force being exerted is double that of the new gravitational force

2(G MiMii/r2)=G Mi(1/2Mii)/r2

I got this answer wrong and I am unsure why. Was I wrong in my approach? Are there other forces that I should have taken into account when coming to my conclusion?

Edit:The mass of the moon simply vanishes.


closed as off-topic by Peter Shor , Brandon Enright, tpg2114, Qmechanic Dec 3 '13 at 0:58

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    $\begingroup$ Because mass is conserved, there is no physical way to change the mass of the moon without removing part of it. And if you remove part of it, whether the speed changes depends on how you remove part of it. So you're asking about an impossible scenario. On the other hand, a change in the radius would not have any effect on its speed. $\endgroup$ – Peter Shor Dec 2 '13 at 21:30

If you remove a bit of moon and leave the rest with the same velocity, it will continue to follow the same orbit. Since you are carrying away some of the mass, as you carry it away it will exert a gravitational force on the moon, which could change the velocity. You are correct that cutting the mass in half will cut the gravitational force in half, but so will the inertia be cut in half, so the trajectory does not change.

  • $\begingroup$ would the velocity change if .5 of the moons mass simply vanished? $\endgroup$ – Aris Dec 3 '13 at 0:23
  • $\begingroup$ Peter Shor's comment is well taken. There is no physical way to make it just vanish. If you define "just vanishing" as leaving the rest in its previous state of motion, you have said the velocity does not change. If you define "just vanishing" as leaving the rest with the same kinetic energy, it will accelerate. I do not believe there is a reasonable answer. $\endgroup$ – Ross Millikan Dec 3 '13 at 0:28
  • $\begingroup$ Thank you for the input. Perhaps I need to look over the question again, maybe my understanding of the question is where my error was. $\endgroup$ – Aris Dec 3 '13 at 0:32

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