An earth-like planet has a radius equal to double the earth’s radius. The acceleration due to gravity on its surface will be

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g

\frac{g}{2}

\frac{g}{4}

g^{2}

answer is C.

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Detailed Solution

The acceleration due to gravity on its surface will be

\frac{g}{4}

.
Acceleration due to the gravity of the earth is

g = \frac{GM}{R^{2}}

.
Earth-like planet has a radius double the earth that is,

R^{'} = 2 R

. So acceleration due to gravity on this planet is,

{\Rightarrow g}^{*} = \frac{GM}{{(2 R)}^{2}}

{\Rightarrow g}^{*} = \frac{GM}{4 R^{2}}

\Rightarrow g^{*} = \frac{g}{4}

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