Question

find the escape velocity when the escape velocity of planet whose mass is 1 by 6 mass of earth and radius is one by two of earth radius​

Answer 1

Answer:

your answer is as follows

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Step-by-step explanation:

$to \: find : \\ escape \: velocity \: of \: planet \: (Vp) \\ \\ let \: here \\ escape \: velocity \: of \: earth \: be \: Ve \\ and \: that \: of \: planet \: be \: Vp \\ moreover \: let \: the \: mass \: of \: earth \\ be \: Me \: and \: its \: radius \: be \: R \\ and \: that \: of \: mass \: and \: radius \: of \\ planet \: be \: Mp \: and \: R1 \: respectively$

$given : \\ Mp = \frac{1}{6} \times Me \\ \\ Me = 6.Mp \: \: \: \: \: \: (1) \\ \\Rp = \frac{1}{2} \times R \\ \\ R = 2.Rp \: \: \: \: \: \: \: \: \: \: \: (2)$

$we \: know \: that \\ escape \: velocity \: of \: earth = 11.2 \:km \: /s \\ \\ thus \: then \\ Ve = \sqrt{ \frac{2GM}{R} } \\ \\ = \sqrt{ \frac{2.G.6.Mp}{2.Rp} } \: \: \\ \\ 11.2= \sqrt{ \frac{6.GMp}{Rp} } \: \: \: \: \: \: \: (3)$

$similarly \: escape \: velocity \: of \: \\ planet \: would \: be \: given \: by \\ \\ Vp = \sqrt{ \frac{2.GMp}{Rp} } \: \: \: \: \: \: \: \: \: (4) \\ \\ applying \: now \: \: \frac{(3)}{(4)} \\ we \: get \\ \\ \frac{11.2}{Vp} = \frac{ \frac{ \frac{ \sqrt{6GMp} }{ \sqrt{Rp} } }{ \sqrt{2.GMp} } }{ \sqrt{Rp} } \\ \\ = \sqrt{ \frac{6.GMp}{Rp} } \times \sqrt{ \frac{Rp}{2.GMp} } \\ \\ = \frac{11.2}{Vp} = \frac{ \sqrt{6} }{ \sqrt{2} } \\ \\ \frac{11.2}{Vp} = \sqrt{3} \\ \\ Vp = \frac{11.2}{ \sqrt{3} } \: \: km \: /s$