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7.Gravitation
hard
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.
A
$1:4$
B
$1:5$
C
$1:2$
D
$3:2$
(AIIMS-2019)
Solution
The acceleration due to gravity at the surface of earth is,
$g=\frac{G M}{R^{2}}$ $\ldots( I )$
And the mass is,
$M=V \rho$
$M=\frac{4 \pi R^{3}}{3} \rho$
It is given that,
$\rho_{e}=\rho_{p} \text { and } G_{p}=2 G_{e}$
Substitute the values in equation $(I).$
$\frac{g_{e}}{g_{p}}=\frac{\frac{G_{e}\left(\frac{4 \pi R_{e}^{3}}{3} \rho_{e}\right)}{R_{e}^{2}}}{\frac{G_{p}\left(\frac{4 \pi R_{p}^{3}}{3} \rho_{p}\right)}{R_{p}^{2}}}$
$1=\frac{G_{e} R_{e}^{3} \times R_{p}^{2}}{2 G_{e} R_{p}^{3} \times R_{e}^{2}}$
$\frac{R_{p}}{R_{e}}=\frac{1}{2}$
Standard 11
Physics
