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The escape velocity of a sphere of mass ...

The escape velocity of a sphere of mass m is given by (G=Universal gravitational constant,M=Mass of the earth and `R_(e )`=Radius of the earth)

A

`sqrt((2GMm)/(R_(e ))`

B

`sqrt((2GM)/(R_(e ))`

C

`sqrt((GM)/(R_(e ))`

D

`sqrt((2GMm+R_(e ))/(R_(e ))`

Text Solution

Verified by Experts

The correct Answer is:
B
The gravitational potential energy of a body of mass m placed on earth.s surface is given by `U=-(GM_(e )m)/(R_(e ))`
Therefore in order to take a body from the earth.s surface to infinity, the work required is `(GMm_(e )m)/(R_(e ))`. Hence it is evident that if we throw a body of mass m with such a velocity that its kinetic energy is `(GM_(e )m)/(R_(e ))`, then it will move outside the gravitational field of earth. Hence,
`(1)/(2)mv_(e )^(2)=(GM_(e )m)/(R_(e ))" or ", v_(e )=sqrt((2GM_(e ))/(R_(e )))`.
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