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Q.

The diameter of a planet is 4 times that of the earth and its mean density is equal to that of earth. If the acceleration due to gravity on the earth surface is $9.8 {ms}^{2}$ , the acceleration due to gravity on the planet surface is

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a

19.6 {ms}^{- 2}

b

4.9 {ms}^{- 2}

c

9.8 {ms}^{- 2}

d

39.2 {ms}^{- 2}

answer is D.

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Detailed Solution

$D_{P} = 4 D_{e} \Rightarrow R_{P} = 4 R_{e} and ρ_{P} = ρ_{e}$

$g = \frac{G M}{R^{2}}$

$M = ρ X V$ , where $ρ$ is the density and $V$ is the volume.

$g = \frac{G \frac{4}{3} {πR}^{3} ρ}{R^{2}}$

$g = \frac{4}{3} π G ρ R$

$g \propto R$

$\Rightarrow \frac{g_{p}}{g_{e}} = \frac{R_{p}}{R_{e}}$

$\Rightarrow g_{p} = g_{e} (\frac{R_{p}}{R_{e}})$

$= 9.8 \times 4 = 39.2 {ms}^{- 2}$

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The diameter of a planet is 4 times that of the earth and its mean density is equal to that of earth. If the acceleration due to gravity on the earth surface is9.8 ms2 , the acceleration due to gravity on the planet surface is