A planet has same density as that of earth an | vedclass

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Question

The acceleration due to gravity at the surface of earth is,

$g=\frac{G M}{R^{2}}$ $\ldots( I )$

And the mass is,

$M=V \rho$

$M=\frac{4 \pi

Vedclass · Accepted Answer

$1:2$

Vedclass · Answer

The acceleration due to gravity at the surface of earth is,

$g=\frac{G M}{R^{2}}$ $\ldots( I )$

And the mass is,

$M=V ho$

$M=\frac{4 \pi R^{3}}{3} ho$

It is given that,

$ho_{e}=ho_{p} 	ext { and } G_{p}=2 G_{e}$

Substitute the values in equation $(I).$

$\frac{g_{e}}{g_{p}}=\frac{\frac{G_{e}\left(\frac{4 \pi R_{e}^{3}}{3} ho_{e}ight)}{R_{e}^{2}}}{\frac{G_{p}\left(\frac{4 \pi R_{p}^{3}}{3} ho_{p}ight)}{R_{p}^{2}}}$

$1=\frac{G_{e} R_{e}^{3} 	imes R_{p}^{2}}{2 G_{e} R_{p}^{3} 	imes R_{e}^{2}}$

$\frac{R_{p}}{R_{e}}=\frac{1}{2}$

A planet has same density as that of earth and universal gravitational constant $G$ is twice that of earth, the ratio of acceleration due to gravity, is.

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