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)

<<So r=R*cos(C*t) where C=sqrt(GM/R^3)

G=6.67*10^(-11) m^3/(kg*s)
M=7.38*10^22 kg
R=3480000 m
C=0.000342/s

The time for r=0 occurs when cos(C*t)=0. The first time that occurs is C*t=pi/2.

t=pi/(2C)
 =4596.19 s (1 hour, 16 minutes, and 36 seconds)>>

But it is the diameter of the moon that's 3480000 m, not the radius.  Your R is twice what it should be, making C too small by a factor of sqrt(8), and therefore t too large by that same factor.

4596.19 s / sqrt(8) = 1625 s, or  27.08 minutes, in agreement with both my methods of calculation.

Posted by Charlie on 2004-11-23 05:04:17