A satellite is revolving round the earth in a circular orbit of radius `r` and velocity `upsilon_(0)`. A particle is projected from the satellite in forward direction with realative velocity `upsilon = (sqrt(5//4) - 1) upsilon_(0)`. Calculate its minimum and maximum distances from earth's centre during subsequent motion of the particle.

Explanation

`upsilon_(0) = sqrt((GM)/(r)) =` orbital speed of satellite ..(i)
where, `M =` mass of earth.
Absoulte velocity of particle would be
`upsilon_(p) = upsilon + upsilon_(0) = sqrt((5)/(4)) upsilon_(0) = sqrt(1.25) upsilon_(0)` ..(ii)
Since, `upsilon_(p)` lies between orbital velocity and ascape velocity, path of the particle would be can ellipe with 'r' being the minimum distance.
Let `r'` be the maximum distance and `upsilon'_(p)` its velocity at that moment.
`V_(p) = sqrt((5)/(4)) V_(0)`

Then, from conservation of angular momentum and conservation of mechanical energy , we get
`m upsilon_(p) = m upsilon'_(p) r'` ..(iii)
and `(1)/(2) m upsilon_(p)^(2) - (GMm)/(r) = (1)/(2) m upsilon'_(p)^(2) - (GMm)/(r')` ..(iv)
Solving the above Eqs. (i), (ii), (iii) and (iv), we get
`r' = (5r)/(3)` and `r`
Hence, the maximum and minimum distance are `(5 r)/(3)` and 'r' respectively.