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Q.
If density of a planet is double that of the earth and the radius $1.5$ times that of the earth, the acceleration due to gravity on the surface of the planet is
Gravitation
Solution:
Acceleration due to gravity on the surface of a planet is
given by, $g=\frac{G M}{R^{2}}$
$M \rightarrow$ Mass of the planet
$R \rightarrow$ Radius of the planet
Also, $M=\frac{4}{3} \pi R^{3} \times \rho$
$\Rightarrow g=\frac{G}{R^{2}} \times \frac{4}{3} \pi R^{3} \rho=\frac{4}{3} \rho G \pi R$
$\rho \rightarrow$ Density of the planet.
$\Rightarrow $ Acceleration due to gravity $\alpha \rho R$
$\Rightarrow \frac{g_{\text {planet }}}{g_{\text {earth }}}=\frac{2 \rho_{e} \times 1.5 R_{e}}{\rho_{e} \times R_{e}}=3$
$\Rightarrow $ Acceleration due to gravity on the surface of planet is 3 times that on the surface of earth.