0
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ Have you heard of the Virial theorem? $\endgroup$
    – ACuriousMind
    Jul 29, 2014 at 19:33
  • 1
    $\begingroup$ @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. $\endgroup$
    – user55789
    Jul 29, 2014 at 19:34

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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...

$\endgroup$
1
  • $\begingroup$ 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... $\endgroup$
    – Floris
    Jul 29, 2014 at 20:06

Your Answer

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

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