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Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is $g$ and that on the surface of the new planet is $g'$, then -
A
$g' = 3g$
B
$g' = 9g$
C
$g' = \dfrac{g}{9}$
D
$g' = 27g$
Detailed Solution
$g = \dfrac{GM}{R^2}$, and the mass is $M = \dfrac{4}{3}\pi R^3\rho$
$g = \dfrac{G \cdot \frac{4}{3}\pi R^3\rho}{R^2} = \dfrac{4}{3}\pi G\rho R$
So for the same density, $g \propto R$.
The new planet has a radius 3 times that of the earth.
$\dfrac{g'}{g} = \dfrac{3R}{R} = 3$
$\Rightarrow g' = 3g$
$g = \dfrac{G \cdot \frac{4}{3}\pi R^3\rho}{R^2} = \dfrac{4}{3}\pi G\rho R$
So for the same density, $g \propto R$.
The new planet has a radius 3 times that of the earth.
$\dfrac{g'}{g} = \dfrac{3R}{R} = 3$
$\Rightarrow g' = 3g$