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Moon Experiment (Posted on 2004-11-22) Difficulty: 5 of 5
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)

See The Solution Submitted by Brian Smith    
Rating: 2.4000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: part 1: another method; another answer (wrong assumption) | Comment 6 of 22 |
(In reply to part 1: another method; another answer by Charlie)

Acceleration due to gravity is not proportional to the cube of the distance from the center of the moon.

I had stated in my solution that the acceleration is proportional to the distance from the center of the moon. I had stated that it was based on symmetry, which isn't wrong but hard to see.

For those who have learned E&M (Electromagnetism), this is exactly the same as Coulomb's Law and Gauss's Law. Coulomb's law says that electric force is proportional to 1/r^2  for point charges. Gauss's Law (although universally correct) is useful only for symmetrical shapes. It is actually used to prove something similar to the given situation.

If you have a sphere of charge and you have a point charge inside the sphere of radius R at a distance r. Only the charge contained in the smaller sphere (radius r) exerts a net force equivalent to that of a point charge carrying a charge equivalent to the smaller sphere. In other words, all charges contained between r and R don't count.

By defining gravitational field lines (similar to electric field lines), you can define a "Gauss's Law" for gravity and use that to prove the same thing.


Edited on November 22, 2004, 11:25 pm
  Posted by np_rt on 2004-11-22 23:24:06

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