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.
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 of `B` would be more than angular momentum of `A`

Answer

B, C

Explanation

Acceleration due to gravity near the surface of shell can be assumed to be uniform `(h lt lt R)`
`g=(G(2M))/((2R)^(2))=(GM)/(2R^(2))` from `h=1/2g t^(2)`
`t=sqrt((2h)/g)=2sqrt((hR^(2))/(GM))`