Suppose that the acceleration of free fall at the surface of a distance planet were found to be equal to that at the surface of the Earth. If the diameter of the planet were twice the diameter of Earth then the ratio of mean density of the planet to that of the Earth would be

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2 :1

1 : 2

4 : 1

1 : 1

answer is D.

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Detailed Solution

$g_{p} = g_{e}; \frac{{GM}_{p}}{{R_{p}}^{2}} = \frac{{GM}_{e}}{{R_{e}}^{2}}; \frac{4}{3} {πR}_{p} d_{p} = \frac{4}{3} {πR}_{e} d_{e}$

or $2 R_{e} d_{p} = R_{e} d_{e}, \frac{d_{p}}{d_{e}} = \frac{1}{2}$

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