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(A) Let the density of the sphere be $ho$.Mass of the original sphere $M = ho \cdot \frac{4}{3}\pi R^3$.Volume of the removed sphere $V_{	ext{rem

Vedclass · Accepted Answer

$\frac{7GM^2}{576R^2}$

Vedclass · Answer

(A) Let the density of the sphere be $ho$.Mass of the original sphere $M = ho \cdot \frac{4}{3}\pi R^3$.Volume of the removed sphere $V_{	ext{removed}} = \frac{4}{3}\pi (\frac{R}{2})^3 = \frac{1}{8} (\frac{4}{3}\pi R^3)$.Mass of the removed sphere $m = ho \cdot V_{	ext{removed}} = \frac{M}{8}$.Mass of the remaining part of the sphere $M' = M - m = M - \frac{M}{8} = \frac{7M}{8}$.The gravitational force between the remaining part of the sphere and the removed sphere is given by Newton's law of gravitation:$F = \frac{G M' m}{r^2}$Here,$r = 3R$ is the distance between the centers of the two spheres.$F = \frac{G (\frac{7M}{8}) (\frac{M}{8})}{(3R)^2}$$F = \frac{G (\frac{7M^2}{64})}{9R^2}$$F = \frac{7GM^2}{576R^2}$

From a sphere of mass $M$ and radius $R$,a smaller sphere of radius $\frac{R}{2}$ is carved out. For the configuration shown in the figure where the distance between the center of the original sphere and the center of the removed sphere is $3R$,the gravitational force between the two spheres is:

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