Question

The escape velocity on the surface of the earth is 11.2 kms^-1. If mass and radius of a planet is 4 and 2 times of the earth respectively, then what is the escape velocity from the planet?

Options

• 11.2 kms^-1

• 1.112 kms^-1

• 15.8 kms^-1

• 22.4 kms^-1

Answer 1

15.8 kms^-1

Explanation:

Escape velocity from the surface of earth is given by

`"V"_"es(e)" = sqrt((2"Gm"_"e")/"R"_"e")`                  ...(i)

Here: `"V"_"es(e)"` = 11.2 km/s, M_p = 4 M_e

R_p = 2R_e

`"V"_"es(e)"=sqrt((2"Gm"_"p")/"R"_"p")=sqrt((2"G"xx4"M"_"e")/(2"R"_"e"))`                ...(ii)

Dividing eq. (ii) by eq. (i), we get

So, `"V"_"es(p)"=sqrt2xx"V"_"es(e)"`

= 1.414 × 11.2 

= 15.8 m/s