# Pigs in space and circular orbit

1. May 12, 2013

### swevener

1. The problem statement, all variables and given/known data
Two [STRIKE]particles[/STRIKE] pigs of mass $m$ and $M$ undergo uniform circular motion about each other at a separation of $R$ under the influence of an attractive force $F$. The angular velocity is $\omega$ radians per second. Show that $$R = \frac{F}{\omega^2}\left( \frac{1}{m} + \frac{1}{M} \right).$$

2. Relevant equations
$F = ma$

3. The attempt at a solution
I'm having trouble reconciling my physical reasoning (which is probably wrong) with the given answer. Okay so since the pigs are orbiting each other I think that means their tangential velocities are equal. And we've got a central force $F$ keeping them a distance $R$ apart.

The book hasn't covered center of mass yet, but I don't care; I want to find the center of mass. (It's likely this is where I'm going astray, but this is what makes sense to me, so I'm going to do it and hope you'll tell me why I'm wrong.) Let's put the origin at $M$ for the time being, so the center of mass is $mR/(m+M)$ along the line connecting $m$ and $M$. That makes the distance of $m$ from the center of mass $R - mR/(m+M) = MR/(m+M)$. Now move the origin to the center of mass and define $r_M = mR/(m+M)$ and $r_m = MR/(m+M)$, noting that if $m = M$, $r_m = r_M = R/2$, so that seems okay.

With respect to my possibly inconvenient origin, $m$ and $M$ have the same tangential velocity $v$ but will have different angular velocities, $v = r_m \omega_m = r_M \omega_M$. But with respect to the midpoint between them, I think they'd have the same angular velocity $v = (R/2) \omega$. With that relation we can get $$\omega_M = \frac{R}{2 r_M} \omega = \frac{m + M}{2m} \omega$$ and $$\omega_m = \frac{R}{2 r_m} \omega = \frac{m + M}{2M} \omega.$$ (Edit: Looking at this again, these equations would mean $\omega_M \ge \omega_m$, which doesn't make sense when $r_M \le r_m$. )

Now the accelerations. For $M$, $$\mathbf{a}_M = -\frac{F}{M} \hat{r} = -r_M \omega_M^2 \hat{r} \rightarrow \frac{F}{M} - r_M \omega_M^2 = 0.$$ For $m$, $$\mathbf{a}_m = -\frac{F}{m} \hat{r} = -r_m \omega_m^2 \hat{r} \rightarrow \frac{F}{m} - r_m \omega_m^2 = 0.$$

Since 0 + 0 = 0, $$\frac{F}{M} + \frac{F}{m} - r_M \omega_M^2 - r_m \omega_m^2 = 0.$$ At this point, if I just let $r_M \omega_M^2 = r_m \omega_m^2 = v^2 / (R/2) = R \omega^2 / 2$, the answer pops right out. That doesn't make much sense to me, though. Why would the same force acting on two unequal masses yield the same acceleration? So I'll keep going. \begin{align}F \left( \frac{m + M}{mM} \right) &= r_M \omega_M^2 + r_m \omega_m^2 \\ &= r_M \left( \frac{m + M}{2m} \omega \right)^2 + r_m \left( \frac{m + M}{2M} \omega \right)^2 \\ \frac{F}{mM} &= r_M \frac{m + M}{4m^2} \omega^2 + r_m \frac{m + M}{4M^2} \omega^2 \\ &= \left( \frac{mR}{m+M} \frac{m+M}{4m^2} + \frac{MR}{m+M} \frac{m+M}{4M^2} \right) \omega^2 \\ &= \left( \frac{1}{m} + \frac{1}{M} \right) \frac{R \omega^2}{4} \\ F &= \frac{R \omega^2 (m + M)}{4}.\end{align} Thus $$R = \frac{4F}{\omega^2 (m+M)}.$$So why am I wrong?

Last edited: May 12, 2013
2. May 12, 2013

### swevener

Never mind. I got it. (See the edit.)