# Explain solution to me(Gravitation)

1. Jan 4, 2012

### Gavroy

first thing, that you might wanna know is, that i am not from an english-speaking country, so please, excuse my grammar...;-)
1. The problem statement, all variables and given/known data
hi

i have a problem here and i already know the solution, but the thing is, that i do not understand it.

let's see:
the first question was: the moon rotates around the earth and now some evil person stops the rotation. so the moon is still that far away from the earth as he was before but now his velocity is 0.

The question now was: how long does it take until the moon crashes into the earth?

And the solution was, that you use kepler's third law and set: T²/(r/2)³=t²/r³

i should explain that " t " is the time the moon would ordinarily need to go one time around the earth and r is the distance between the earth and the moon

and T/2 is the solution. That seems okay to me, because you only want to know how much time it takes to go one way to the earth and not back again. but why do they set r/2 in the denominator?

i always thought that there you should use the aphelion, which is the farthest distance between the central object and the travelling object. but in this case, the starting position should be the farthest position, as the moon would not go farther from the earth than this initial distance, or am i wrong?

2. Jan 4, 2012

### D H

Staff Emeritus
Your friend is wrong. There's a missing term in his solution.

First, look at T²/(r/2)³=t²/r³. The solution to this is T²=t²/8, or T=t/(2√2)=(√2/4)t.

That is the period of an orbit with a semi-major axis half that of the Moon's semi-major axis. One way to look at this is to imagine what happens in the limit as the Moon's velocity approaches zero. As the Moon's velocity decreases, the Moon's orbit would become more eliptical, with perigee approaching zero in the limit v→0. Since the semi-major axis is the average of perigee and apogee, this means that the semi-major axis would approach r/2 as v→0. Hence the r/2.

However, $T=\frac{\sqrt 2}4t$ is the period of this limiting orbit. This is not the quantity to be solved for. The question asks how long it would take for the Moon to fall. Going from apogee to perigee is half of an orbit, so the correct solution is
$$T=\frac{\sqrt 2}{8}t$$

3. Jan 4, 2012

### Gavroy

wow, thank you...