Gravitation

JEE Main level, single correct, medium

33 of 223

Question 33 of 223

If the radius of earth shrinks by 22% while its mass remains same. The acceleration due to gravity on the earth's surface will approximately

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Answer: (D) increase by 44%

The acceleration due to gravity on the surface of the earth is given by, g=GMR2g=\dfrac{GM}{{R}^{2}}.
The radius shrunk by 22%. Therefore, ∆RR×100=−2\dfrac{∆R}{R}\times 100=-2%.
As the mass is constant and the change in the radius of the earth is relatively small, we can write
∆gg=−2∆RR⇒∆gg×100=4\dfrac{∆g}{g}=-2\dfrac{∆R}{R} \Rightarrow \dfrac{∆g}{g}\times 100=4%
Hence, the acceleration due to gravity increases by 44%.

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