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

Options

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

Answer

B

Explanation

If energy in the form of kinetic energy, which is equal to binding energy, is supplied to the sphere, it leaves the gravitational field of earth.
At a plateform at a height `h`,
escape energy=binding energy of sphere
or `1/2mv'_(e)^(2)=(GMm)/(R+h)`
or `v'_(e)=sqrt((2GM)/(R+h))=sqrt((2GM)/(2R)) ( :' h=R)`
But at the surface of earth,
`v_(e)=sqrt((2GM)/R)`
As given, `v'_(e)=fv_(e)`
Hence, `sqrt((2GM)/(2R))=fsqrt((2GM)/R)`
or `1/(2R)=(f^(2))/R`
`f=1/(sqrt(2))`