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

The corresponding situation is shown in figure

Initial velocity of statellite `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 hence the particle path deviates from circular path to ellipticle path. At postions of minimu and maximum distance velocity vector are perpendicular to instantaneous radius vector. In this elliptical path the minimumm 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
`mV_(1) r =m(V_(0)+V)a` [m = mass of particle]
or `V_(1)=((V_(0)+V)a)/(r )`
`= (a)/(r )[sqrt(((5)/(4)(GM)/(a)))]`
`=(1)/(r )[sqrt(((5)/(4)xxGMa))]`
Applying the law of conservation of energy
`(1)/(2)mvV_(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)-8a r+5a^(2)=0`
or `r = a` or `(5a)/(3)`
Thus minimum distance of the particle = a
And maximum distance of the particle `= (5a)/(3)`