Two particles separated a distance r, each of mass m, are being launched at opposite directions with the same speed v. If we’re in the reference frame of the center of mass of the particles, what speed is needed by each of the particles so that they escape each other’s gravitational pull?
What I decided to do is go into the reference frame of one of the particles and have that one be stationary and made the speed of the other particle 2v.
Setting up conservation of energy equations, I get:
$-\frac{Gm^2}{r} + \frac{1}{2}m(2v)^2 = 0+0$
$-\frac{Gm^2}{r} + 2mv^2 = 0$
$...$
However, the solution says to stay in the reference frame of the center of the two particles, which yields:
$-\frac{Gm^2}{r} + 2(\frac{1}{2}mv^2) = 0+0$
which obviously yields different results. Could you please explain?