1.

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

Answer»

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

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


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