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A tunnel is dug along a diameter of the earth. If M_e and R. are the mass and radius, respectively, of the earth, then the force on a particle of mass m placed in the tunnel at a distance r from the centre is

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$\frac{{GM}_{e} m}{R_{e}^{2}} r$

$\frac{{GM}_{e} m}{R_{e}^{3}} r$

$\frac{{GM}_{e} m}{R_{e}^{3} r}$

$\frac{{GM}_{e} {mR}_{e}^{3}}{r}$

answer is A.

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Detailed Solution

Let us first calculate the mass of the inner solid sphere of radius r, Mass of the inner solid sphere is

$M^{'} = \frac{M_{e}}{\frac{4}{3} {πR}_{e}^{3}} \times \frac{4}{3} {πr}^{3} = \frac{r^{3} M_{e}}{R_{e}^{3}}$

$Now, g = \frac{{GM}_{e} r^{3}}{R_{e}^{3}} \times \frac{1}{r^{2}} or g = \frac{{GM}_{e} r}{R_{e}^{3}}$

Force on the particle of mass $m = mg = \frac{{GM}_{e} mr}{R_{e}^{3}}$

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