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If two particles of equal mass $m$ separated by a distance $r$ are released from rest and start on a collision course, attracting each other through gravity (so their gravitational potential energy is being converted into kinetic energy as they accelerate towards each other), which of these is the correct representation of conservation of energy (when they collide each particle has speed $v$)?

$$mv^2 = 2\frac{Gm^2}{r}\tag{1}$$

or

$$mv^2 = \frac{Gm^2}{r}\tag{2}$$

One one hand, each particle should have a potential energy of $E_p = -\frac{Gm^2}{r}$, which would support the first equation, but on the other hand it seems reasonable that the decrease in potential energy for one particle would increase the kinetic energy of both particles, since they are both moving towards each other, which would support the second equation.

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  • $\begingroup$ Comment to the post (v2): The total mechanical energy in the COM frame is constant but not necessarily zero. $\endgroup$
    – Qmechanic ♦
    Commented Oct 27, 2017 at 15:21

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The potential energy of the system is $G\frac{m_1m_2}{R}$. If the two objects are of equal mass, then the kinetic energy is shared between them. $$G\frac{m_1m_2}{R}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ $m_1=m_2$ and $v_1 = v_2$ so our expression is now: $$G\frac{m^2}{R}= mv^2$$

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  • $\begingroup$ So it's the energy of the whole system, and not each individual particle. I've always learnt that the potential energy of one mass is $\frac{GMm}{r}$, but I guess that's wrong and it's actually the shared potential energy of the entire system. Thank you. $\endgroup$
    – Pancake_Senpai
    Commented Oct 27, 2017 at 14:50
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    $\begingroup$ @Pancake_Senpai: if the particles are separated by a distance $r$ then each is $\tfrac{1}{2}r$ from their common centre of mass. So if you bring the particles together each has moved only $\tfrac{1}{2}r$. So the PE of each particle is half of what you think it is. $\endgroup$
    – John Rennie
    Commented Oct 27, 2017 at 15:06

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