A uniform sphere A with radius R exerts a for | vedclass

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Question

(C) Let the mass of the original sphere $A$ be $M$. The force $F$ on particle $B$ of mass $m$ at distance $2R$ is $F = \frac{GMm}{(2R)^2} = \frac{GMm}

Vedclass · Accepted Answer

$F^{\prime} = \frac{7}{9} F$

Vedclass · Answer

(C) Let the mass of the original sphere $A$ be $M$. The force $F$ on particle $B$ of mass $m$ at distance $2R$ is $F = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$.The mass of the removed spherical portion of radius $r = R/2$ is $M^{\prime} = \rho \cdot \frac{4}{3} \pi (R/2)^3 = \frac{M}{\frac{4}{3} \pi R^3} \cdot \frac{4}{3} \pi \frac{R^3}{8} = \frac{M}{8}$.The distance of the centre of the removed portion from the particle $B$ is $d = 2R - R/2 = 3R/2$.The force exerted by the removed portion on $B$ is $F_{removed} = \frac{G M^{\prime} m}{d^2} = \frac{G (M/8) m}{(3R/2)^2} = \frac{GMm}{8 \cdot (9R^2/4)} = \frac{GMm}{18R^2}$.Since $F = \frac{GMm}{4R^2}$,we have $\frac{GMm}{R^2} = 4F$. Thus,$F_{removed} = \frac{4F}{18} = \frac{2}{9} F$.The new force $F^{\prime}$ is $F - F_{removed} = F - \frac{2}{9} F = \frac{7}{9} F$.

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