The escape velocity on a planet with radius double that of earth and mean density equal to that of earth will be (escape velocity on earth $= 11.2 K m / s)$

11 K m / s

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22 K m / s

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5.5 K m / s

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15.5 K m / s

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Solution

The correct option is B $22 K m / s$
Escape velocity $v_{e} = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{2 G (\frac{4 π d R^{3}}{3})}{R}} = R \sqrt{\frac{8 π d G}{3}}$
If the radius is radius new radius $r = 2 R$
So, $V_{p} = r \sqrt{\frac{8 G π d}{3})} = 2 R \sqrt{\frac{8 π d G}{3}} = 2 v_{e} = 2 (11) = 22 k m / s$

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The escape velocity on a planet with radius double that of earth and mean density equal to that of earth will be (escape velocity on earth =11.2Km/s)

The correct option is B 22Km/sEscape velocity ve=√2GMR=  ⎷2G(4πdR33)R=R√8πdG3If the radius is radius new radius r=2RSo, Vp=r√8Gπd3)=2R√8πdG3=2ve=2(11)=22 km/s

The escape velocity on a planet with radius double that of earth and mean density equal to that of earth will be (escape velocity on earth $= 11.2 K m / s)$