Two satellites A and B are revolving around the earth circular orbits of radius `r_(1)` and `r_(2)` respectively with `r_(1) lt r_(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. A?t position 2, A is given another impluse so that velocities of A and B at 2 become equal and the move together.

For any elliptical path of the satellite of the time period of revolution is given by Kepler's planetary law as `T^(2)alpha r^(3)` where a is semi major axis of the ellipse which is `(r_(1)+r_(2))/(2)` in this case. Also angular mopmentum 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 along its position vector relative to the earth centre.
If the two have same mass

Options

A. A would have more potential energy than B while on their initial circular paths.
B. A would have more kinetic energy than B while on their initial circular paths.
C. Relative to Earth's centre angular momentum of A when it is in elliptical path would be less than angular momentum of B.
D. During the whole process angular momentum for B would be more than angular momentum of A.

Answer

B, C

Explanation

`U=-(-GMm)/r and K.E.=(GMm)/(2r)`
for same m, if `r_(1)gtr_(2),KE_(1)ltKE_(2)andU_(1)gtU_(2)`
Also, angular momentum of A before reaching the position 2 (during elliptical path) was less than that of B. Since, some impulse is needed (at position 2)for A in direction of its motion, to make its speed and angular momentum equal to that of B.