If the density of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration on the surface of the planet is

Question

If the density of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration on the surface of the planet is (A) 0.2 g (B) 0.4 g (C) 2 g (D) 4 g

Tardigrade Admin · Accepted Answer

We know that, g=(G M/R2)=(G((4/3) π R3) ρ/R2)=(4/3) π G R ρ (g prime/g)=(R prime/R)=(0.2 R/R)=0.2 ∴ g prime=0.2 g

Q. If the density of a small planet is the same as that of earth, while the radius of the planet is $0.2$ times that of the earth, the gravitational acceleration on the surface of the planet is

$0.2 g$

$0.4 g$

$2 g$

$4 g$

We know that,
$g = \frac{GM}{R ^{2}} = \frac{G ( \frac{4}{3} π R ^{3} ) ρ}{R ^{2}} = \frac{4}{3} π GRρ$
$\frac{g ^{'}}{g} = \frac{R ^{'}}{R} = \frac{0.2 R}{R} = 0.2$
$∴ g^{'} = 0.2 g$

Q. If the density of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration on the surface of the planet is

0.2g

0.4g

2g

4g