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?

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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, M

Radius of the planet, R

Thus, escape velocity of the planet is

Mass of the planet, M

_{p}= 8 `× ` Mass of the Earth (M_{e})Radius of the planet, R

_{p}= 2 `× `Radius of the Earth (R_{e})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 M_{E}

Radius of the planet, R_{P} = 2 R_{E}

Escape velocity for the Earth, V_{es} _{cE} = 11.2 km/s

Escape velocity for the planet, V_{escP} = ?

`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?

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