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Suppose you drill a tunnel through the earth along a diameter and drop a stone of mass m=13.47 kg down the tunnel. The force of gravity within the volume of Earth is radially inward and equals to:

$F_{g}= -\frac{GMm}{R^{3}}r$

where r is the radial distance from the center of earth. R, G, M are constants with their values shown below. Assuming the density of Earth is constant and no friction or air resistance in the tunnel, the stone will travel all the way until it reaches the other end of the tunnel then travel back to your hand under the effect of the restoring gravitational force in the above equation.

How long (in minutes) is the journey of the stone from the moment you drop it off to the moment when you catch it?

The mass of Earth is M=5.97258*10^24 kg, Radius of Earth is R=6.306*10^6 m, and the gravitational constant is G=6.67384*10^-11 m^3/kg s^2.

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Added Fri, 07 Aug '15
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initial veocity=0

$F_{g}= -\frac{GMm}{R^{3}}r$= -k*r where k=GMm/R3

so accelartion=-GMr/R3

now as we know for an SHM accelaration is directly propotional to distance from equilibrium with a negative sign.

and for SHM a time period will be T=$2\pi\sqrt(\frac{m}{k})$=$2\pi\sqrt(\frac{R^{3}}{GM})$

=2*3.14*((6.306*10^6)3/(6.67384*10^-11*5.97258*10^24))0.5=4981sec=83min 1sec

Edit
Added Fri, 07 Aug '15
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