The ratio of escape velocity at earth `(v_(e))` to the escape velocity at a planet `(v_(y))` whose radius and density are twice

Options

A. `1:4`
B. `1:sqrt2`
C. `1:2`
D. `1:2sqrt2`

Answer

D

Explanation

As escape velocity
` v = sqrt((2GM)/(R )) = sqrt((2G)/(R ) . (4pi R^3)/(3) rho) = R sqrt((8pi G)/(3) rho)`
` therefore v_e/v_P = R_e/R_P xx sqrt(rho_e/rho_P) = 1/2 xx sqrt(1/2) = (1)/(2sqrt2)`