Question 251: Gravitation

Question

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 ________ g.

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

Gravitation

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 ________ g.

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