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
by )
<<Added by Edit: Also, my undersanding is that acceleration due to gravity is not determined by mass. Mass determines relative weight due to gravity, and which of two masses affects the other mass more and from what distance.>>
Acceleration due to gravity is determined by the sum of the two massesor rather the force is proportional to the product of the two masses (but also inversely with the square of the distance). The mass of the earth is huge in comparison to a given falling object, and and basically determines the acceleration of, say a bowling ball. All objects on the earth's surface fall with the same acceleration because the force is proportional to their mass, and the amount of force needed for a given acceleration is also proportional to their masses. However, that acceleration would be more if the earth's mass were larger (given the same size). In fact the acceleration is proportional to the mass of the central body divided by the square of the radius (or other distance from the center). The moon has about 1/81 the mass of the earth and about 1/4 the radius, so the acceleration at the surface is about 4^2/81 = 1/5 of that of the earth. (a little bit off on the radius, produces a larger amount off on the acceleration, but that's the general idea.)
<<Added by Second Edit: I believe Charlie is correct about the period being the same as the period required for a low orbit. So, why not just calculate the orbit time?>>
That's what I did in my first post, giving an answer of 27.077 minutes, which I later verified using numerical integration of the gravitation (at first an erroneous one not counting the inverse square part, then the corrected one that agreed with the orbit method).

Posted by Charlie
on 20041123 01:41:03 