First slide

Gravitational field intensity

Question

The density of a 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

4R
B

$\frac{R}{4}$
C

$\frac{R}{2}$
D

2R

Solution

$g = \frac{GM}{R^{2}} = \frac{4}{3} πGρR$
Now according to question g_planet = g_earth
So, $\frac{4}{3} {πGρ}_{planet} R_{planet} = \frac{4}{3} {πGρ}_{earth} R_{earth}$
$\Rightarrow R_{planet} = \frac{1}{2} R_{earth} (∵ ρ_{planet} = 2 ρ_{earth})$

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