# Conservation of energy constant

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
$\frac{1}{2}mv^2+\frac{m\gamma}{2r^2}=E$.

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 $\frac{1}{2}mU^2$?

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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

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.

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