# How to show that the gravitational potential energy of a two particle system is -2 times the total kinetic energy, without using the virial theorem?

Consider a two-particle system with identical masses, orbiting in circles about their center of mass. I'm supposed to prove that:

$$U_p = -2U_k$$

With $U_p$ potential energy of the system, and $U_k$ the total kinetic energy of the system.

I'm supposed to solve this pretending I have no knowledge of the virial theorem.

So far this is what I got:

Starting off, in the radial direction, there is no net force on either of the particles. The two forces working on either of the particles are the centripetal force $F_c$ and the gravitational force $F_g$, in such a way that they must balance each other out:

$$F_c = F_g$$

The relation for gravitational potential is given by:

$$dU_p = - F_g dr$$

Rewriting this relation using our previous equation, I wrote:

$$dU_p = - F_c dr$$

(This time with the centripetal rather than the gravitational force.) Integrating on both sides:

$$U_p = - \int F_c dr$$

So far so good, though I don't have any expression for $F_c$ that have only $r$ in it but no other variables that depend on $r$. My first choice was to use $F_c = mr\omega^2$. $\omega$ depends on $r$, and I believe so does $v$ in the subsequent expression. No dice there. Sob!

So I thought, why not work from the second part of the initial relation I'm supposed to find. So I tried calculating the total kinetic energy:

$$U_k = \sum_N \frac{1}{2} m_n v_n^2$$

For an arbitrary particle $n$, with $N$ amount of molecules. Fine, so I have two molecules and $m_1=m_2$. Also, $v_1=v_2$. (Right?)

That gets me to:

$$U_k = 2\cdot\frac{1}{2}mv^2=mv^2$$

And now what? I can't really squeeze out $U_p=-\frac{1}{2} mv^2$.

So, I was hoping to work backwards. Problem is, my expression doesn't even contain any $r$, so what do I have to go on to solve for $F_c$ in this expression:

$$U_p = -\frac{1}{2} mv^2 = - \int F_c dr$$

Any help much appreciated.

• Have you heard of the Virial theorem? Commented Jul 29, 2014 at 19:33
• @ACuriousMind Thanks for bringing it up. I'm supposed to solve this pretending I have no knowledge of the virial theorem. I'll update my question immediately. Commented Jul 29, 2014 at 19:34

Both masses $m$ are at a distance $R$ of the center of mass. The distance between the two masses is then $2R$.

To calculate the potential energy, consider the first mass fixed. The potential energy of the second mass is $\gamma \frac{m^2}{2R}$, with $\gamma$ the gravitational constant. This is also the potential energy of the entire system.

The kinetic energy of the system is $mv^2$. The velocity $v$ can be expressed in function of $m$ and $R$ by requiring the centrifugal force $\frac{mv^2}{R}$ to cancel the centripetal force $\gamma \frac{m^2}{\left( 2R \right)^2}$, what leads to a kinetic energy equalling $\gamma \frac{m^2}{4R}$, i.e. half the potential energy.

Because this is a two particle system, you can do this exactly.

For a given mass $m$ and distance between objects $2r$, you can compute the required angular velocity $\omega$ (to remain in stable orbit) and thus the kinetic energy ($\frac12m(r\omega)^2$).

For that same configuration, you can compute the gravitational energy by seeing how much it takes to get each of the two masses from $r$ to $\infty$

Take the ratio of the two. $m$ and $r$ will drop out of the equation...

• From the equality between gravitational and centripetal force you should get $m\omega^2r=\frac{Gm^2}{4r^2}$. And then it's not much further... Commented Jul 29, 2014 at 20:06