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(C) The gravitational force $F$ between two spheres of mass $m$ and radius $R$ whose centers are separated by a distance $d = 2R$ is given by Newton's

Vedclass · Accepted Answer

$R^4$

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(C) The gravitational force $F$ between two spheres of mass $m$ and radius $R$ whose centers are separated by a distance $d = 2R$ is given by Newton's Law of Gravitation: $F = \frac{G m^2}{(2R)^2}$.Since the spheres are solid copper,their mass $m$ is given by $m = 	ext{Volume} 	imes 	ext{Density} = (\frac{4}{3} \pi R^3) ho$,where $ho$ is the density of copper.Substituting the expression for $m$ into the force equation:$F = \frac{G (\frac{4}{3} \pi R^3 ho)^2}{4R^2} = \frac{G \cdot \frac{16}{9} \pi^2 R^6 ho^2}{4R^2}$.Simplifying the expression:$F = \frac{4}{9} G \pi^2 ho^2 R^4$.Since $G$,$\pi$,and $ho$ are constants,we find that $F \propto R^4$.

Two identical solid copper spheres of radius $R$ are placed in contact with each other. The gravitational attraction between them is proportional to

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