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

C

Explanation

Given,
`g =10 m//s ^(2)`
`R = 6400 km = 6.4 xx 10 ^(6) m`
Escape velocity of earth, `v _(c) = sqrt(2gR)`
`= sqrt( 2xx10 6.4 xx 10 ^(6))`
` = sqrt(128xx 10 ^(6))`
` = 11.3 xx 10 ^(3) m//s `
Since given,
`g ' = 2g = 20 m//s^(2)`
` R ' = (R)/(2) = 3.2 xx 10 ^(6) m`
Then `R ' = sqrt ( 2g ^(1) R ^(1))`
`= sqrt(2xx 20 xx 3.2 xx 10 ^(6))`
` = sqrt(128 xx 10 ^(6))`
`=11.3 xx 10 ^(3) m//s `
` therefore v _(e) ^(1) ~~12 km//s.`