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 relative 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

Initial velocity satellite `v_(0) = sqrt(((GM)/(a)))`

When particle is thrown with the velocity v relative to satellite, the resultant velocity of particle will become
`v_(R ) = v_(0) + v`
`= sqrt(((5)/(4))) v_(0) = sqrt(((5)/(4)(GM)/(a)))`
As the particle velocity is greater than the velocity required for circular orbit, hence the particle path deviates from circular path to elliptical path. At position of minimum and maximum distance velocity vectors are perpendicular to instantaneous radius vector. In this elliptical path the minimum distance of particle from earth's centre is a and maximum speed in the path is `v_(R )` and let the maximum distance and minimum speed in the path is r and `v_(1)` respectively.
Now as angular momentum and total energy remain conserved. Applying the law of conservation of angular momentum, we have
`m v_(1) r = m(v_(0) + v) a`
[m = mas of particle]
or `v_(1) = ((v_(0) + v)a)/(r )`
`= (a)/(r ) [sqrt(((5)/(4)(GM)/(a)))]`
`= (1)/(r ) [sqrt(((5)/(4) xx GMa))]`
Applying the law of conservation of energy
`(1)/(2)mv_(1)^(2) - (GMm)/(r ) = (1)/(2)m(v_(0) + v)^(2) - (GMm)/(a)`
or `(1)/(2)m((5)/(4)(GMa)/(r^(2))) - (GMm)/(r )`
`= (1)/(2)m((5)/(4)(GM)/(a)) - (GMm)/(a)`
`(5)/(8) xx (a)/(r^(2)) - (1)/(r ) = (5)/(8) xx (1)/(a) - (1)/(a) = -(3)/(8a)`
or `3r^(2) - 8ar + 5a^(2) = 0`
or `r = a` or `(5a)/(3)`
Thus minimum distance of the particle = a
And maximum distance of the particle `= (5a)/(3)`