A satellite of mass M is in a circular orbit of radius R about the center of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speed of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be:

Let the mass of the satellite be M and the mass of the meteorite be M. Let the speed of the satellite and the meteorite just before the collision be v. Since the collision is completely inelastic, the two bodies stick together after the collision. The combined mass is 2M. Let the velocity of the combined body after the collision be v'.

By conservation of momentum:
Mv + M(-v) = 2Mv'
0 = 2Mv'
v' = 0

The velocity of the combined body after the collision is zero. However, this is incorrect. The meteorite is falling towards the Earth, it doesn't have the same velocity vector as the satellite. Let's assume the satellite is orbiting with velocity v in the positive x-direction. Let's consider the meteorite falling vertically with velocity v. Before collision, the total momentum is:
P_x = Mv
P_y = Mv
After collision, the combined mass 2M will have a velocity v' at some angle θ:
P_x = 2Mv'cos(θ) = Mv
P_y = 2Mv'sin(θ) = Mv
This gives:
v'cos(θ) = v/2
v'sin(θ) = v/2
Squaring and adding these two equations:
v'^2(cos^2(θ) + sin^2(θ)) = v^2/4 + v^2/4 = v^2/2
v'^2 = v^2/2
v' = v/√2
The velocity is not zero but v/√2, implying a change in both speed and direction. The combined body will now be in an elliptical orbit. The total energy of the satellite before collision is given by:
E = KE + PE = 1/2Mv^2 - GMm/R = -GMm/2R
After collision, the velocity is reduced. The kinetic energy changes, and thus, the energy of the system changes. The new orbit will have a different total energy and therefore will be an elliptical orbit.
Therefore, the subsequent motion of the combined body will be in an elliptical orbit.