Consider 2 bodies of mass $m_1$ and $m_2$ at infinite separation. As these bodies approach each other, their gravitational potential energy is converted to kinetic energy and we let the bodies collide, the kinetic energy may be released in the form of heat. Let the bodies settle down and reach a stable state after colliding with each other. Let the system have lost energy equal to $E$ in the collision process. Let the mass of the combined system after the collision be $M$.

My question is, Does $m_1+m_2=M$? If not, does $m_1+m_2=M+\dfrac{E}{c^2}$? In other words, where does the energy $E$ show up from? Is it initially a part of $m_1$ and $m_2$? If we weigh the objects before and after the collision, will we see a difference in mass between the two states?


1 Answer 1


Yes, potential energy is related to mass as $E=mc^2$. The correct option in your question is $$m_1+m_2=M-\dfrac{E}{c^2}$$ with the minus sign, because the gravitational potential energy is negative ($M$ is smaller than $m_1+m_2$).

  • $\begingroup$ So the final mass is bigger? $\endgroup$
    – Mauricio
    Nov 30, 2017 at 18:44
  • $\begingroup$ @Mauricio The final mass is M. Please read the last line of my answer. $\endgroup$
    – safesphere
    Nov 30, 2017 at 20:06
  • $\begingroup$ @saphesphere sorry my mistake. But a real question, it seem counterintuitive, but does your result mean that the closer the masses start from one another, the smaller the final mass? $\endgroup$
    – Mauricio
    Nov 30, 2017 at 20:54
  • $\begingroup$ @Mauricio No, the "final" mass depends only on the "final" distance between them. Please note that $m_1$ and $m_2$ in the question are defined at infinity. If they are closer to each other, then tbeir masses would be smaller. What would indeed be smaller when the masses start closer is the amount of the heat energy released upon the impact. This all is simple energy conservation. The sum of two masses at a distance equals the sum of two masses up close plus the amount of energy released (divided by $c^2$). The closer the masses get, the smaller they are and the bigger energy released is. $\endgroup$
    – safesphere
    Nov 30, 2017 at 21:40
  • $\begingroup$ @safesphere And this energy becomes a part of the rest mass of the object, correct? $\endgroup$
    – Aniansh
    Dec 3, 2017 at 4:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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