Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A particle $P$ of mass $m$ moves under the repulsive inverse cube field $\vec{F}=\frac{m\gamma}{r^3}\vec{e_r}$ ($\vec{e_r}$ is a unit vector along a position vector $\vec{r}$).
Initially $P$ is at a great distance from $O$ and is moving with speed $U$. Find conservation of energy equation.

I found that potential energy is $V=-\int\frac{m\gamma}{r^3}dr=\frac{m\gamma}{2r^2}$.
Kinetic energy is $T=\frac{1}{2}mv^2$.
So, conservation of energy equation is

Now, I have to find $E$. From initial conditions, $T=\frac{1}{2}mU^2$. What is happening with potential energy? Is it going to be zero since $r$ is very large, so $E$ will be $\frac{1}{2}mU^2$?

share|cite|improve this question
It has to be assumed that $PE=0$ at a very large distance. To define potential energy you need a reference potential energy of an arbitrary point. – udiboy1209 Jul 21 '13 at 13:49
Yes, that is what it's assumed to be. – Ali Jul 21 '13 at 13:51
up vote 0 down vote accepted

It's a quite common assumption. When the particle is at infinity(or fairly distant from the origin), the potential is assumed to be zero. Other examples can be found in numerous scattering problems.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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