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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.)
part 1: different answer | Comment 4 of 22 |
(In reply to Completely Analytical Solution (Physics + Differential Calculus) by np_rt)

Somewhere I learned that a fall like this through the center of a planet and back again would take the same amount of time as one orbit around the planet in a circular orbit skimming the surface.  One quarter of such oscillation would take one quarter of the time, as the trip going would be the same length as the trip coming back, and within each trip, the time getting to the center would be symmetrical with the time leaving the center.

The formula for the number of radians per second covered in a circular orbit is sqrt(G*(M+m)/a^3) where a is the radius of the orbit, G is the gravitional constant of 6.673*10^-8 cm^3/(g*s^2). (agrees with np_rt's value when units converted). So, with necessary conversions, one full orbit of 2*pi radians would take 2*pi/(sqrt((6.673/10^8)*7.38*10^25/(3480*100000/2)^3) *60) = 108.3088 minutes.  One quarter of that would be 27.077 minutes, the amount of time to reach the center in the through-the-center "orbit".



Edited on November 22, 2004, 10:00 pm
  Posted by Charlie on 2004-11-22 20:31:17

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