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

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Question

(A) Outside the planet (at the surface),the gravitational force on the man is given by Newton's law of gravitation: $F_{	ext{old}} = \frac{GMm}{R^2}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Outside the planet (at the surface),the gravitational force on the man is given by Newton's law of gravitation: $F_{	ext{old}} = \frac{GMm}{R^2}$.Inside the planet,the man is inside the spherical shell of mass $\frac{2M}{3}$. According to the shell theorem,the gravitational field due to a spherical shell at any point inside it is zero. Thus,the shell exerts no force on the man.The man is at a distance $R$ from the point mass $\frac{M}{3}$ located at the center. The gravitational force exerted by this point mass is $F_{	ext{new}} = \frac{G(\frac{M}{3})m}{R^2} = \frac{GMm}{3R^2}$.The change in the force of gravity experienced by the man is $\Delta F = F_{	ext{old}} - F_{	ext{new}} = \frac{GMm}{R^2} - \frac{GMm}{3R^2} = \frac{2GMm}{3R^2}$.

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