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A hole is drilled from one end to other end of earth and an object of mass 'm' is dropped down the hole. The gravitational force acting on the object as a function of distance (r) of the object from the center of earth is
(Mass of earth =M, Radius of earth = R and consider uniform density of earth)

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$\vec{F} = \frac{GMm}{R^{2}} \hat{r}$

$\vec{F} = \frac{GMm}{R^{2}} \vec{r}$

$\vec{F} = \frac{GMm}{R^{3}} \hat{r}$

$\vec{F} = \frac{GMm}{R^{3}} \vec{r}$

answer is D.

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Detailed Solution

Force on the particle is mg'

where

$g' = g (1 - \frac{d}{R}) = g (\frac{R - d}{R}) = \frac{g r}{R} r i s t h e d i s \tan c e f r o m t h e c e n t e r$

$F o r c e = m g' = m \frac{g r}{R} = \frac{G m M}{R^{3}} \overset{⇀}{r}$

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