Satellites and gravitation A satellite with mass $m$ orbits a planet of mass $M$ in a circular path with radius $r$ and velocity $v$. Due to some internal technical failure, the satellite breaks into two, similar parts with mass $m/2$ each. In the satellite's frame of reference, it appears the two parts move radially, in opposite directions, along the line connecting the original satellite and the planet's center, each with velocity $v_0/2$. I am expected to show that right after the technical failure, each of the two parts has a total energy equal to $-3GM/(16r)$ and angular momentum equal to $(m/2)√(GMr)$, wrt the planet's center.
The total energy of each of the two parts should be, I believe: $E_{tot} = mv_0^2/16 - GmM/(2r)$. Now, isn't angular momentum preserved despite the failure? However, why isn't the angular momentum zero if the two parts are moving in opposite directions?
 A: There is something strange with the problem. In particular, when the satellite divides, the consevation of lineal momentum implies both parts moving each with the same tangential velocity (same direction) in the orbital direction but half mass, so $(mv/2+mv/2=mv)$ (orbital velocity). The angular momentum cannot be zero because of two reasons, first, as you say, the total angular momentum of each piece of the system needs to be conserved with respect to the center of the planet, simply because the gravitational force is radial $(\mathbf{r}\times F\hat{r}=0)$ and the technical failure could only be an internal force (no net torque)(the radial velocities do not contribute, so they do not matter in the momentum). So the angular momentum they ask you to demonstrate is simply $\mathbf{r}\times m\mathbf{v}/2$. Since each of the parts ends up with half of the original momentum (you can demonstrate this is what they give you). Now for $v_{0}$/2, I will assume that the text has a mistake, and that $v/2$ is what they mean. For the energy simply E=$1/2(m/2){(v)}^2-GMm/2r+1/2(1/2m)(v/2)^{2}=-3GMm/16r$, as you  can easily verify. 
