Acceleration due to gravity on the surface of earth is ' $g$ '. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is $\ldots\ldots$ g.

Question

Aicharya · Accepted Answer

$\because$ acceleration due to gravity on surface is given by
$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^2}$
Now since diameter is reduced to $1 / 3^{\text {rd }}$, radius also reduces to $1 / 3^{\text {rd }}$, keeping mass constant
New value of acceleration due to gravity on Earth's surface is
$\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\left(\frac{\mathrm{R}_{\mathrm{e}}}{3}\right)^2}=9 \frac{\mathrm{GMe}}{\mathrm{R}_{\mathrm{e}}^2}=9 \mathrm{~g}$

Acceleration due to gravity on the surface of earth is ' $g$ '. If the diameter of earth is reduced to one…

Solution