A satellite of mass `M_(S)` is orbitting the earth in a circular orbit of radius `R_(S)`. It starts losing energy slowly at a constant rate C due to friction if `M_(e)` and `R_(e)` denote the mass and radius of the earth respectively show that the satellite falls on the earth in a limit time t given by `t=(GM_(S)M_(e))/(2C)((1)/(R_(e))-(1)/(R_(S)))`

Explanation

Let velocity of satellite in its orbit of radius r be v then we have
`v = sqrt((GM_(e ))/(r ))`
When satellite approaches earth's surface, if its velocity becomes v' then it is given as
`v' = sqrt((GM_(e ))/(R_(e ))`
The total initial energy of satellite at a distance r is
`E_(T_(i))=K_(i)+U_(i)`
`= (1)/(2)mv^(2)-(G M_(e )m)/(r )`
`= -(1)/(2)(GM_(e )m)/(r )`
The total final energy of satellite at a distance `R_(e )` is
`E E_(T_(i)) = K_(f)+U_(f)`
`=(1)/(2)mv'^(2) - (GM_(e )m)/(R_(e ))`
`= (1)/(2) (GM_(e )m)/(R_(e ))`
As satellite is loosing energy at a rate C, if it take a time t in reaching earth, we have
`Ct = E_(T_(f))-E_(T_(i))`
`= (1)/(2)GM_(e )m[(1)/(R_(e ))-(1)/(r )]`
or `t = (GM_(e )m)/(2C)[(1)/(R_(e ))-(1)/(r )]`