The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is `fv_(e)`, where `v_(e)` is its escape velocity from the surface of the earth. Find the value of `f`.

Options

A. `(1)/(sqrt(2))`
B. `sqrt(2)`
C. `(1)/(3)`
D. `(1)/(2)`

Answer

A

Explanation

The escape velocity of a body projected from the surface of the earth is `v_(e)=sqrt((2GM)/(R ))`
The platform is at a height R from the surface of the earth.
`therefore` If the body of mass m is on the platform, its P.E.
`=-(GMm)/(R+h)=-(GMm)/(2R)`
`therefore` Its binding energy `= (GMm)/(2R)`
`therefore` Its escape energy is `(1)/(2)m(v_(e ))^(2)`
For escaping, `(1)/(2)m(v_(e))^(2)=(GMm)/(2R)`
`therefore v'_(e)=sqrt((GM)/(R ))`
But it is given that `v'_(e)=fv_(e)`
`therefore sqrt((GM)/(R ))=f sqrt((2GM)/(R ))`
`therefore 1 = f sqrt(2)` or `f= (1)/(sqrt(2))`