First slide

Gravitational field intensity

Question

The density of newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be

Moderate

A

2 R
B

4 R
C

$\frac{1}{4} R$
D

$\frac{1}{2} R$

Solution

Given that, d_planet = 2 d_earth
and g_planet = g_earth
$∴ \frac{{GM}_{p}}{R_{p}^{2}} = \frac{{GM}_{e}}{R_{e}^{2}} or \frac{M_{p}}{R_{p}^{2}} = \frac{M_{e}}{R_{e}^{2}}$
or $\frac{\frac{4}{3} {πR}_{p}^{3} d_{p}}{R_{p}^{2}} = \frac{\frac{4}{3} {πR}_{e}^{3} d_{e}}{R_{e}^{2}}$
or $R_{p} \times d_{p} = R_{e} \times d_{e}$
or $R_{p} = \frac{R_{e} d_{e}}{d_{p}} = \frac{R_{e}}{2} = \frac{R}{2}$

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