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# 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 the earth, the gravitational acceleration on the surface of that planet is:0.2 g0.4 g2 g4 g

A
0.2 g
B
0.4 g
C
2 g
D
4 g
Solution
Verified by Toppr

#### Acceleration due to gravity g=GMR2Mass of the planet, M=Vd=d×43πR3 where d is the density of the planet⟹ g=43πdGR (for earth)Given : d′=d and R′=0.2R∴ g′=43πd′GR′=43πdG(0.2R)⟹ g′=0.2g

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