Gravitation

JEE Main level, single correct, medium

8 of 223

Question 8 of 223

A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing WW on earth will weigh on that planet:

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Answer: (B) 213W{2}^{\dfrac{1}{3}}W

The formula for acceleration due to gravity is
g=GMR2g=\dfrac{GM}{{R}^{2}}
Let the density of earth and planet be ρ\rho, and the radius of the planet be R′R'. Let RR be the radius of the earth and gg be the acceleration due to gravity of earth and g′g' be the acceleration due to gravity of planet.
So, the masses are
M=43πR3ρ...(i)M=\dfrac{4}{3}\pi {R}^{3}\rho ...(i)
M′=43π(R′)3ρ...(ii)M'=\dfrac{4}{3}\pi {\left(R'\right)}^{3}\rho ...(ii)
So, g′g=GM′R′2GMR2=M′M(RR′)2...(iii)\dfrac{g'}{{g}^{}}=\dfrac{\dfrac{GM'}{R{'}^{2}}}{\dfrac{GM}{{R}^{2}}}=\dfrac{M'}{M}{\left(\dfrac{R}{R'}\right)}^{2}...(iii)
From equation (i) and (ii) we get R′=213RR'={2}^{\dfrac{1}{3}}R
Substituting the value in equation (iii),
g′g=(21)2−23=213g\dfrac{g'}{g}=\left({2}^{1}\right){2}^{\dfrac{-2}{3}}={2}^{\dfrac{1}{3}}g
Weight is given by W′=mg′=213WW'=mg'={2}^{\dfrac{1}{3}}W

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