A hole is drilled all the way through the center of the Earth, from one side to

Question

A hole is drilled all the way through the center of the Earth, from one side to the other, and a particle of mass m is dropped in. Recall that the gravitational attraction of the Earth for a particle a distance r from the center of the Earth really arises entirely from that portion of the matter of the Earth internal to the position of the particle, and the external shells exert no force at all. a) Assuming that the Earth has uniform density, find the force exerted on the particle as a function of r. b) How long does it take to get to the other end? c) What happens then? Comment intelligently, d) (XC )Actually this result is true for any chord, not just one through the center of the Earth. Show this for XC.

Explanation / Answer

a)

assuming uniform density, rho = mass / volume = M / (4 pi Re^3 / 3)

rho = 3 M / (4 pi Re^3)

for radius r :

mass of underneath earth, M' = (3 M / 4 pi Re^3)(4 pi r^3 / 3 ) = M r^3 / Re^3

F = G M' m / r^2 = G ( M r^3 / Re^3 ) m / r^2

F = (G M m / Re^3) r

b) F is directly propotional to r .

hence mass will perofrm Simple Harmonic motion.

F = k x   

where k = (G M m / Re^3)

and time period of SHM, T = 2pi sqrt(m / k )

T = 2pi sqrt(m / (G M m / Re^3)) = 2pi sqrt(Re^3 / G M )

so time taken to reach the onther end, t = T/2 = pi sqrt(Re^3 / G M )

c) Reaching the other end , mass will stop and then starts moving to the centre of earth and
then other end.

so mass will perform SHM like this.

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