- #1

etotheipi

- Homework Statement
- Two masses $$m_{A}, m_{B}$$ of radii $$r_{A}, r_{B}$$ are released from rest at infinity. Find their final relative velocity at the point of collision.

- Relevant Equations
- $$GPE = -\frac{Gm_{A}m_{B}}{r}$$

There have been some other threads on similar problems but none address one specific point I'm confused about.

The change in GPE of a body is the negative of the work done on that body by a gravitational field between two points; by this logic, since the same (but opposite) gravitational forces have acted on both A and B from infinity to a final separation of r

So I thought that the final relative velocity of the masses should be $$v_{rel}= |v_{A}| + |v_{B}| = \sqrt{\frac{2Gm_{B}}{r_{A}+r_{B}}} + \sqrt{\frac{2Gm_{A}}{r_{A}+r_{B}}}$$ However, the solution given equates the sum of the final kinetic energies to one GPE, as in the following $$\frac{Gm_{A}m_{B}}{r_{A} + r_{B}} = \frac{1}{2}mv_{A}^{2} + \frac{1}{2}mv_{B}^{2}$$ which when followed through by considering momentum (that part of the argument is fine) yields a different relative velocity of $$\sqrt{\frac{2G(m_{A}+m_{B})}{r_{A}+r_{B}}}$$I don't understand how we can equate the drop in GPE of only one body to the gain in KE of both bodies, since surely this same amount of work is done on each separately and is not shared - I was wondering if someone could help me out. Thanks a bunch!

The change in GPE of a body is the negative of the work done on that body by a gravitational field between two points; by this logic, since the same (but opposite) gravitational forces have acted on both A and B from infinity to a final separation of r

_{A}+ r_{B}, I assumed that**both**undergo a change in GPE of $$\Delta GPE = -\frac{Gm_{A}m_{B}}{r_{A} + r_{B}}$$ the negative of which should be then equal to the work done on each body and consequently also the change in kinetic energy.So I thought that the final relative velocity of the masses should be $$v_{rel}= |v_{A}| + |v_{B}| = \sqrt{\frac{2Gm_{B}}{r_{A}+r_{B}}} + \sqrt{\frac{2Gm_{A}}{r_{A}+r_{B}}}$$ However, the solution given equates the sum of the final kinetic energies to one GPE, as in the following $$\frac{Gm_{A}m_{B}}{r_{A} + r_{B}} = \frac{1}{2}mv_{A}^{2} + \frac{1}{2}mv_{B}^{2}$$ which when followed through by considering momentum (that part of the argument is fine) yields a different relative velocity of $$\sqrt{\frac{2G(m_{A}+m_{B})}{r_{A}+r_{B}}}$$I don't understand how we can equate the drop in GPE of only one body to the gain in KE of both bodies, since surely this same amount of work is done on each separately and is not shared - I was wondering if someone could help me out. Thanks a bunch!

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