JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2025
- Easy
- Numerical
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.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
\(\because\) 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
\(\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\left(\frac{\mathrm{R}_{\mathrm{e}}}{3}\right)^2}=9 \frac{\mathrm{GMe}}{\mathrm{R}_{\mathrm{e}}^2}=9 \mathrm{~g}\)
\(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^2}\)
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
\(\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\left(\frac{\mathrm{R}_{\mathrm{e}}}{3}\right)^2}=9 \frac{\mathrm{GMe}}{\mathrm{R}_{\mathrm{e}}^2}=9 \mathrm{~g}\)
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