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I am trying to calculate an instantaneous merger of two rotating spheres into one. Two spheres each rotating around their own axis of rotation (which are generally not aligned) and moving relative to each other. Imagine two entirely inelastically colliding planets.

While I can calculate their individual spin angular momentum and the orbital angular momentum, it is unclear to me how to combine this into one (spin) angular momentum for the resulting sphere. It seems to me I cannot simply add the four vectors together... or can I?

After merging the two into a single body maintaining the initial momentum, can I calculate the released energy (in the inelastic collision) simply as the difference in kinetic energy, using the same reference frame?

I realize this should be pretty trivial, but I have thus far not been able to neither derive nor locate and answer.

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It seems like you know what the answer is, but you just don't know how to prove it. You are right though.

To make things simpler, just view things in the center of mass frame. Then the total momentum is zero, and, like you said, the total angular momentum is just the sum of the orbital momentum of each planet plus the sum of the spin angular momentum of each planet.

After the collision, the linear velocity of the final planet must be zero by conservation of momentum. Thus the final planet will have no orbital angular momentum. However, we know that the angular momentum must be conserved, so the planet must be spinning about sum axis, and you must have $\vec{L} = I \vec{\omega}$. Since you know what $\vec{L}$ is and you presumably know $I$, you know what $\omega$ must be.

So basically you had it right.

To get the released energy, you need to take into account both the kinetic and potential energy. I will call the sum of these two the total energy. Then the total energy of the final state will be less than the total energy of the initial state. So the energy released (the energy that will cause the final planet to be hot) will be the intial total energy minus the final total energy.

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  • $\begingroup$ Thanks for taking the time to answer my question :-) Good to get a confirmation as well as the basic explanation. I think I was primarily confusing myself with orbital vs spin angular momentum, since I have previously always dealt with one or the other. Then, or course, I also forgot the potential energy for the heating, which I have little excuse for, except the late hour. Thanks for the help. $\endgroup$ – Tom Taic Apr 8 '14 at 12:06

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