A man of mass m starts falling towards a plan | vedclass

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(A) Before entering the hole (at the surface),the force of gravity on the man is $F_1 = \frac{GMm}{R^2}$.As the man enters the hole,he is inside the s

Vedclass · Accepted Answer

$\frac{2}{3} \frac{GMm}{R^2}$

Vedclass · Answer

(A) Before entering the hole (at the surface),the force of gravity on the man is $F_1 = \frac{GMm}{R^2}$.As the man enters the hole,he is inside the spherical shell of mass $2M/3$. According to the shell theorem,the gravitational field inside a uniform spherical shell is zero.Therefore,the force of gravity experienced by the man inside the hole is only due to the point mass $M/3$ at the centre: $F_2 = \frac{G(M/3)m}{R^2} = \frac{GMm}{3R^2}$.The change in the force of gravity is $\Delta F = F_1 - F_2 = \frac{GMm}{R^2} - \frac{GMm}{3R^2} = \frac{2GMm}{3R^2}$.

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