The escape velocity form the earth is 11 `kms^(-1)` the esacpe velocity from a planet having twice the radius and the same mean denisty as the earth would be

Options

A. `5.5 km s^(-1)`
B. `11 km s^(-1)`
C. `15.5 km s^(-1)`
D. `22 km s^(-1)`

Answer

D

Explanation

Escape velocity from the earth is
` v_e = sqrt((2GM)/(R ) = sqrt( (2G)/( R) 4/3 pi R^3 rho) = R sqrt((8pi)/(3) G rho)`
As the planet has same density as that of earth
` therefore v_p/v_e = (2R)/( R) = 2 " or " v_p = 2v_e = 2 xx 11 = 22 kms^(-1)`