Question

Escape velocity for Earth is 11.2 kms - 1. What will be the escape velocity for a planet whose mass and radius are twice those of the Earth?

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \\ escape \: velocity \: of \: earth \: (V1) = 11.2 \: \frac{km}{s} \\ \\ moreover \: let \: mass \: of \: earth \: be \: M \: and \: its \: \\ radius \: be \: R \\ moreover \: for \: a \: certain \: planet \: \\ let \: its \: mass \: be \: M1 \: and \: its \: radius \: be \: R1$

$so \: we \: know \: that \\ escape \: velocity \: = \sqrt{ \frac{2G M}{R} } \\ \\ so \: for \: a \: certain \: planet \\ R1 = 2.R \\ M1 - 2.M \\ \\ so \: escape \: velocity \: of \: earth \: ie \\ V1 = \sqrt{ \frac{2G M}{R} } \\ \\ 11.2 = \sqrt{ \frac{2G M}{R} } \: \: \: \: \: \: (1)$

$similarly \: \\ escape \: velocity \: of \: planet \: ie \\ V2 = \sqrt{ \frac{2G . M1}{R1} } \\ \\ = \sqrt{ \frac{2G \ 2 M}{2R} } \\ \\ = \sqrt{ \frac{2G . M}{R} } \\ \\ = 11.2 \: \: \frac{km}{s} \: \: \: \: \: \: \: from \: (1)$

$hence \: escape \: velocity \: of \: planet \\ whose \: mass \: and \: radius \: are \: twice \: of \: \\ that \: of \: earth \: \: is \: \: 11.2 \: \: km/s$