Earth radius R = 6.371 times 106m Gravity at surface, gR = 9.81 m/s2 Ignore rota

Question

Earth radius R = 6.371 times 106m Gravity at surface, gR = 9.81 m/s2 Ignore rotation of earth, friction, etc. Assume unit mass for particle: m = 1 Assume earth has mass M and uniform density, rho. A straight, smooth hole is drilled eccentrically through the earth from A to B. The lateral distance between the hole and the center of the earth is d (see figure). At time t = 0, a particle P is dropped into the hole from point A, passes through the smooth hole (no sir resistance or friction between the particle and the sides of the hole) and accelerates under the force of gravity until it reaches point B, where it stops momentarily before falling in the opposite direction back to point A. Determine the time tau for the round trip from A back to A. For the particle location rho shown in the figure, the only mass contributing to the gravity at that point is that part of the earth's mass at a smaller radius than the particle radius tau (shown shaded in the figure): Mass = (4pir3/3)rho. Points will be awarded for the derivation of the appropriate equations, with only a small part of the score for the correct numerical value of the solution.

Explanation / Answer

relevant equations are

effective mass of the earth
M=4/3r3
. I ignore the R-r mass assuming that it can be neglected. This results in the ODE:

r¨+(4/3G)r=0


with

2=4/3G


and since
=2T

hance T =(4/3G)/2

=2.66 sec

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