A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of radius of earth R will be

Question

A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of radius of earth R will be (A) R / 4 (B) R / 2 (C) R / 3 (D) R / 8

Tardigrade Admin · Accepted Answer

It is given that, the density of planet is twice the density of earth and both of them have same acceleration due to gravity.
if

g

and

g^{'}

are the acceleration due to gravity of earth and planet, then

g = g^{'}

∴ \frac{G}{R ^{2}} [\frac{4}{3} π R^{3} ρ] = \frac{G}{R ^{'2}} \times \frac{4}{3} π R^{'3} (2 ρ)

\Rightarrow R^{'} = \frac{R}{2}

Q. A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of radius of earth $R$ will be

$R /4$

$R /2$

$R /3$

$R /8$

Q. A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of radius of earth R will be

R/4

R/2

R/3

R/8

It is given that, the density of planet is twice the density of earth and both of them have same acceleration due to gravity. if g and g′ are the acceleration due to gravity of earth and planet, then g=g′ ∴R2G​[34​πR3ρ]=R′2G​×34​πR′3(2ρ) ⇒R′=2R​

Q. A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of radius of earth $R$ will be

$R /4$

$R /2$

$R /3$

$R /8$