Escape velocity of a body from the surface of earth is 11.2km/sec. from the earth surface. If the mass of earth becomes double of its present mass and radius becomes half of its present radius then escape velocity will become

Options

A. 5.6 km/s
B. 11.2 km/s
C. 44.8 km/s
D. 22.4 km/s

Answer

D

Explanation

The escape velocity of a body from the earth.s surface is `v_e = sqrt((2GM)/(R )) = 11.2km//s `
where M and R are the mass and radius of the earth respectively. When M becomes 2M and R becomes R/2, then
` v._e = sqrt((2G(2M))/((R//2)) ) = 2 sqrt((2GM)/(R )) = 2 xx 11.2 km//s = 22.4 km//s`