The ratio of binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around of the earth at a height h above the earth's surface is (R = radius of the earth).

Options

A. `(2(R+h))/(R )`
B. `(R+h)/(2R)`
C. `(R+h)/(R )`
D. `(R )/(R+h)`

Answer

A

Explanation

Binding energy of the satellite on the surface of the earth is
`E_(1)=(GMm)/(R )" "` …..(1)
When it is at a height h above the surface of the earth, its
`P.E.=-(GMm)/(R+h)`
and `K.E.=(1)/(2)mv_(o)^(2)=(1)/(2)m((GM)/(R+h))=(GMm)/(2(R+h))`
where `v_(o)` = orbital velocity
`therefore` Total energy `E_(2)=(GMm)/(2(R+h))-(GMm)/(R+h)`
`therefore E_(2)=-(GMm)/(2(R+h))`
`therefore` Binding energy `=+(GMm)/(2(R+h))" "` ......(2)
`therefore (E_(1))/(E_(2))=(GMm)/(R )xx(2(R+h))/(GMm)=(2(R+h))/(R )" "` .......(3)