Consider two satellites A and B of equal mass moving in the same circular orbit of radius r around the earth but in the opposite sense and therefore, on a collision course

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The total mechanical energy $E_{A} + E_{B}$ of the two satellite plus-earth system before collision is $- \frac{{GM}_{e} m}{r}$

The combined satellite will fall on to the earth

The combined satellite revolves around the earth in elliptical orbit.

If the collision is completely inelastic the total mechanical energy immediately after collision is $- \frac{2 {GM}_{e} m}{r}$

answer is A, B, D.

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Detailed Solution

(a) $E_{A} = - \frac{1}{2} \frac{{GM}_{e} m}{r}; E_{B} = - \frac{1}{2} \frac{{GM}_{e} m}{r}$
$E_{Q} + E_{B} = - \frac{{GM}_{e} m}{r}$
(b) Applying Law of conservation of momentum along the axis perpendicular to the gravitational force i.e., along the tangent, we get
$mv + (- mv) = 2 mv'$
$\Rightarrow v' = 0$
Energy of the system = Gravitational energy
$E = \frac{2 {GM}_{e} m}{r}$
Since velocity = 0, satellite will fall vertically down.