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.
Answer (integer)
9
Solution
<p>$\because$ acceleration due to gravity on surface is given by</p>
<p>$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^2}$</p>
<p>Now since diameter is reduced to $1 / 3^{\text {rd }}$, radius also reduces to $1 / 3^{\text {rd }}$, keeping mass constant</p>
<p>New value of acceleration due to gravity on Earth's surface is</p>
<p>$$g^{\prime}=\frac{\mathrm{GM}}{\left(\frac{\mathrm{R}_{\mathrm{e}}}{3}\right)^2}=9 \frac{\mathrm{GMe}}{\mathrm{R}_{\mathrm{e}}^2}=9 \mathrm{~g}$$</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
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