The density of a newly discovered planet is twice that of the 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 $R$ and the radius of the planet is $R^{^{'}}$ . Then what is the value of $\frac{R}{R^{^{'}}}$ ?

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Solution

Given acceleration due to gravity on the 2 planets are equal and density on new planet is twice as that of earth.
$\frac{G M_{e}}{R^{2}} = \frac{G M_{p}}{R^{' 2}}$
$\frac{G \frac{4}{3} π R^{3} ρ}{R^{2}} = \frac{G \frac{4}{3} π R^{' 3} \times 2 ρ}{R^{' 2}}$
$⟹ R = 2 R^{'}$

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The density of a newly discovered planet is twice that of the 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 and the radius of the planet is R′. Then what is the value of RR′?

Given acceleration due to gravity on the 2 planets are equal and density on new planet is twice as that of earth.GMeR2=GMpR′2 G43πR3ρR2=G43πR′3×2ρR′2 ⟹R=2R′

Given acceleration due to gravity on the 2 planets are equal and density on new planet is twice as that of earth.
$\frac{G M_{e}}{R^{2}} = \frac{G M_{p}}{R^{' 2}}$
$\frac{G \frac{4}{3} π R^{3} ρ}{R^{2}} = \frac{G \frac{4}{3} π R^{' 3} \times 2 ρ}{R^{' 2}}$
$⟹ R = 2 R^{'}$