If the radius of the Earth shrinks to 1/n of | vedclass

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(B) The acceleration due to gravity on the surface of the Earth is given by $g_e = \frac{G M_e}{R_e^2}$.Given that the new radius $R'_e = \frac{R_e}{n

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$n^2 g_e$

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(B) The acceleration due to gravity on the surface of the Earth is given by $g_e = \frac{G M_e}{R_e^2}$.Given that the new radius $R'_e = \frac{R_e}{n}$ and the mass $M_e$ remains constant.The new acceleration due to gravity $g'_e$ is given by $g'_e = \frac{G M_e}{(R'_e)^2}$.Substituting the value of $R'_e$,we get $g'_e = \frac{G M_e}{(R_e/n)^2} = \frac{G M_e n^2}{R_e^2}$.Since $g_e = \frac{G M_e}{R_e^2}$,we can write $g'_e = n^2 g_e$.

If the radius of the Earth shrinks to $1/n$ of its present value while its mass remains constant,what will be the value of $g'_e$ on its surface?

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