Two satellites `A` and `B` are revolving around the earth in circular orbits of radius `r_(1)` and `r_(2)` respectively with `r_(1)ltr_(2)`. Plane of motion of the two are same. At position `1`, `A` is given an impulse in the direction of velocity by firing a rocket so that it follows an elliptical path to meet `B` at position `2` as shown. focal lengths of the elliptical path are `r_(1)` and `r_(2)` respectively. at position `2`, `A` is given another impulse so that velocities of `A` and `B` at `2` become equal and the two move together. for any elliptical path of the satellite time period of revolution is given Kepler's planetry law as `T^(2)alphar^(3)` where a is semi-major axis of the ellipse which is `(r_(1)+r_(2))/2` in this case. also angular momentum of any satellite revolving around the earth will remain a constant about earth's centre as force of gravity on the satellite which keeps it in elliptical path is given its position vector relative to the earth centre.
When `A` is given its first impulse at that moment

Options

A. `A,B` and the earth centre are in same straight line
B. `B` is a head of `A` angularly
C. `B` is behind of `A` angularly
D. none of the above

Answer

B

Explanation

Total Energy `E=1/2mV_(1)^(2)-(GMm)/(r_(1))`
Substuitute `V_(1)` and `r_(1)` and simplify, `E=-(GMm)/(2a)`