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Maharashtra State BoardSSC (English Medium) 10th Standard

If Mass of a Planet is Eight Times the Mass of the Earth and Its Radius is Twice the Radius of the Earth, What Will Be the Escape Velocity for that Planet?

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Question

Solve the problem.

If the mass of a planet is eight times the mass of the Earth and its radius is twice the radius of the Earth, what will be the escape velocity for that planet?

Numerical
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Solution 1

Escape velocity for Earth is

\[v_{esc} = \sqrt{\frac{2G M_e}{R_e}} = 11 . 2 km/s\]
Given:
Mass of the planet, Mp = 8 `× ` Mass of the Earth (Me) 
Radius of the planet, Rp = 2 `× `Radius of the Earth (Re)
Thus, the escape velocity of the planet is
\[v_{esc}' = \sqrt{\frac{2G M_p}{R_p}} = \sqrt{\frac{8}{2}}\sqrt{\frac{2G M_e}{R_e}} = \sqrt{4} \times 11 . 2 = 22 . 4 km/s\]
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Solution 2

Given:

Mass of the planet, MP = 8 ME
Radius of the planet, RP = 2 RE
Escape velocity for the Earth, VescE = 11.2 km/s
Escape velocity for the planet, VescP =? 

`V_("escP") = sqrt((2"GM"_"p")/("R"_"p")) `

=`sqrt((2"G"(8"M"_"E"))/(2"R"_"E"))`

`=sqrt(8/2 xx (2"GM"_"E")/("R"_"E")`

`= sqrt(8/2) xx sqrt((2"GM"_"E")/"R"_"E")`

`= sqrt(4) xx "V"_("escE")`    

∵ `"V"_("escE") = sqrt((2"GM"_"E") /("R"_"E") `

`= 2 xx 11.2`

`"V"_("escP")` = 22.4 Km/s

Escape velocity for the planet = 22.4 km/s

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Chapter 10: Space Missions - Answer the Following [Page 48]

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