The radius of a planet is `1/4` of earth’s radius and its acceleration due to gravity is double that of earth’s acceleration due to gravity. How many times will the escape velocity at the planet’s surface be as compared to its value on earth’s surface

Options

A. `(1)/(sqrt(2))`
B. `sqrt(2)`
C. 2
D. `2sqrt(2)`

Answer

A

Explanation

`R_(p)=(R_(e))/(4)` and `g_(p)=2g_(e )` and `V_(e )=sqrt(2g_(e )R_(e ))`
`therefore (V_(P))/(V_(e ))=sqrt((2xx g_(P)R_(P))/(2xx g_(e)R_(e)))=sqrt((2g_(e)R_(e))/(4g_(e)R_(e)))=(1)/(sqrt(2))`