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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?

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I think you need some relation between $v$ and $v_0$ (are they equal?) Without it, $v_0$ could be very small and the kinetic energy of separation will be zero. Once you have that relation, the two terms can combine. – Ross Millikan Jan 15 '14 at 22:21
Alright, how about the following relation: $(1/2)mv^2 = 2(1/2)(m/2)(v_0/2)^2$? – peripatein Jan 15 '14 at 22:55
The two pieces presumably keep the orbital velocity, but add radial components of $\pm v_0$. You cannot use the orbital KE to assess the separation velocity. Think of an explosion making the $v_0$-it doesn't relate to the orbital velocity. – Ross Millikan Jan 15 '14 at 23:16
Hint: The two parts are moving in radially opposite directions - in the satellite frame of reference. The satellite frame is non-inertial. Angular momentum depends on motion tangent to the radius. That help? – Simon Bridge Jan 15 '14 at 23:18
@peripatein your relation can be simplified to $v_0=2v$, which would seem like a very, very bad structural failure. This will actually put both parts on an parabolic escape trajectory with periapsis of $\frac{r}{2}$. – fibonatic Sep 26 '14 at 3:13

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.

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Why are you adding to the total energy of one of the parts its kinetic energy prior to the failure, when the question asks for the total energy past the separation? – peripatein Jan 16 '14 at 0:25
I am assuming that the resulting satellite parts have two velocity components, orbital and radial one equal to $v$ and the other equal to $v/2$. The orbital term follows from linear momentum conservation just before and just after technical failure. – LagraMen0123 Jan 16 '14 at 0:34
The angular momentum is conserved. Each piece has half the original angular momentum, because it has half the mass and the same tangential velocity. The new velocity is along the radius, so doesn't change the angular momentum of each piece. Each piece has the same higher energy, so the orbits will have the same shape, but the perigees will be offset. – Ross Millikan Jun 20 '14 at 4:09

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