# 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?

#### 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
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