A tunnel is dug along a diameter of the Earth. If $M_{e}$ and $R_{e}$ are the mass and radius respectively of 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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$\frac{{GM}_{e} {mR}_{e}^{3}}{r}$

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

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

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

answer is A.

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Detailed Solution

Let us first calculate the mass of inner solid sphere of radius r.

Mass of inner solid sphere, M

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

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

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