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(B) The gravitational potential energy $U$ of a system of two masses $M$ and $m$ separated by a distance $R$ is given by $U = -\frac{GMm}{R}$.Here,$M

Vedclass · Accepted Answer

$6.67 	imes 10^{-10}\, J$

Vedclass · Answer

(B) The gravitational potential energy $U$ of a system of two masses $M$ and $m$ separated by a distance $R$ is given by $U = -\frac{GMm}{R}$.Here,$M = 100\, kg$,$m = 10\, g = 0.01\, kg$,$R = 10\, cm = 0.1\, m$,and $G = 6.67 	imes 10^{-11}\, N m^2/kg^2$.Substituting these values into the formula:$U = -\frac{6.67 	imes 10^{-11} 	imes 100 	imes 0.01}{0.1}$$U = -\frac{6.67 	imes 10^{-11} 	imes 1}{0.1} = -6.67 	imes 10^{-10}\, J$.The work required to take the particle to infinity is equal to the change in potential energy,which is $W = U_{\infty} - U_{surface} = 0 - (-6.67 	imes 10^{-10}\, J) = 6.67 	imes 10^{-10}\, J$.

$A$ particle of mass $10\, g$ is kept on the surface of a uniform sphere of mass $100\, kg$ and radius $10\, cm$. Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $G = 6.67 \times 10^{-11}\, N m^2/kg^2$).

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