Advertisement Remove all ads

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? - Science and Technology 1

Numerical

Solve the problem.

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?

Advertisement Remove all ads

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, escape velocity of the planet is
\[v_{esc}' = \sqrt{\frac{2G M_p}{R_p}} = \sqrt{\frac{8}{2}}\sqrt{\frac{2G M_e}{R_e}} = \sqrt{4} \times 11 . 2 = 22 . 4 km/s\]

Solution 2

Given : Mass of the planet, MP = 8 ME
Radius of the planet, RP = 2 RE
Escape velocity for the Earth, Ves cE = 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

Concept: Launching of Satellite
  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×