Correct Answer - B
`E=m/2xx(6GM_(e))/(5r)-(GM_(e)m)/(2r)=-2/3(GM_(e)m)/r`
which is the total energy of the earth satellite system.
So, semi major axis of the elliptical orbit is `a =(5r)/4`
Speed of the satellite at the apogee position is
`v_(A)=(v_(P)xxr)/(2a-r)=2/3sqrt((6GM_(e))/(5r))`
For orbit to change to a circle of radius `3r//2=(2a-r)` the rocket has to be fired when the satellite is at the apogee position.
New orbital speed is ` v_(0)=sqrt((GM_(e))/(3r//w))=sqrt((2FGM_(e))/(3r))`
Required change in the orbital speed is
`/_v=v_(A)-v_(0)=0.085sqrt((GM_(e))/R)`.