Assume the moon is a perfect sphere and a straight tunnel has been drilled through the center. How long would it take a 1kg ball dropped from one end of the tunnel to reach the center? Ignore all resistances.
If a second 1kg ball is dropped 10 seconds after the first one, when and where in the tunnel would they first meet?
Idealized Moon Stats:
 Diameter: 3480 kilometers
 Mass: 7.38x10^22 kilograms (uniform density)
(In reply to
Completely Analytical Solution (Physics + Differential Calculus) by np_rt)
I started to work out this problem, and then I read np_rt's solution which I found to be excellent; so I stopped working on it. I didn't double check the calculations, but I recall from school that such a ball dropped through a transplanetary hole will undergo simple harmonic motion to "China" and back. The meeting on the far side of the moon 5 seconds after the ball starts falling back towards the center makes perfect sense. The only way you get simple harmonic motion is by having x'' = kx. ie acceleration proportional to r. But this reasoning is based on the fact that I happen to remember something from school. Just like Charlie remembers that this problem in some way equates to the problem of an object in orbit at the surface of the planet. I don't remember that; I must have been sick that day.
But np_rt worked it out from first principles, which is A number 1 in my book. So unless someone can point out where np_rt made a mistake (I haven't found one), then I'm going to use my lifeline and bet on np_rt's solution.
And that's my story.

Posted by Larry
on 20041123 03:45:55 