If the acceleration due to gravity g doubles and the radius of earth becomes half that of the present value, then the value of escape velocity is (Assume, `g = 10 m//s^2 ` and radius of earth, R = 6400 km)

Options

A. `12 km//s`
B. `16 sqrt2 km//s`
C. `8 sqrt2 km//s`
D. `4 sqrt2 km//s `

Answer

A

Explanation

Given, radius of earth, ` R_e = 6400 km = 6.4 xx 10^6 m `
and acceleration of gravity, `g = 10 m//s^2`
`therefore ` Escape velocity of the surface of earth, `v_e = sqrt(2gR_e) = sqrt( 2 xx 10 xx 6.4 xx 10^6) = 11.3 xx 10^3 m//s`
Again, when `g' = 2g`
`R' = R/2`
Now, escape velocity is given by
` therefore v'_e = sqrt(2g'R') = sqrt( 2 xx 2g xx R/2) = sqrt(2 g R ) = v_e`
` = 11.3 xx 10^3 m//s = 11.3 km//s ~~ 12 km//s `