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_(f)) = K_(f) + U_(r )`
`= (1)/(2)mv^(2) - (GM_(e)m)/(R_(e))`
`= -(1)/(2)(GM_(e)m)/(r )`
The total final energy of satellite at a distance `R_(e)` is
`E_(T_(f)) = K_(f) + U_(r )`
`= (1)/(2) mv^('2) - (GM_(e)m)/(R_(e))`
`= (1)/(2)(GM_(e)m)/(R_(e))`
As satellite is loosing energy at rate C, if it takes a time t in reaching earth, we have
`Ct = E_(T_(i)) - E_(T_(f))`
`= (1)/(2)GM_(e)m[(1)/(R_(e)) - (1)/(r )]`
`rArr t = (GM_(e)m)/(2C) [(1)/(R_(e)) - (1)/(r )]`