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?


Kinetic Energy isn't the same in different reference frames. This isn't really surprising: A test mass has 0J of Kinetic Energy in its rest frame, but by switching to a frame in which it is moving, can be made to have arbitrarily high energy.

Now imagine a system of orbiting, gravitationally bound bodies. Their Kinetic plus Gravitational Potential energy is negative, else it wouldn't be bound. But this has to be calculated in the reference frame of the system's center of mass! Why? Because you could take that same system, and make it have higher kinetic energy just by changing frames, but it must still be gravitationally bound regardless of what frame you are in.

This is what's gone wrong in your calculation.

To remedy it: Rather than saying the system is bound if the energy of the system less than zero, you should say it is bound if the energy is less than the kinetic energy of the systems centre of mass as described by your new reference frame.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.