0
$\begingroup$

I know there are similar questions with answers like this one in physics stackexchange but before I read these I had a completely different argument with a different conclusion which I thought was correct but now I'm suspecting that there is something wrong with the argument. I just want to know what exactly is wrong with it.

In a two body problem with say particles of masses $m_1$ and $m_2$ interacting under a central force $f(r)\hat{r}$ where $\mathbf{r} = \mathbf r_2 - \mathbf r_1$ the equations of motion of the two particles are $$m_1\ddot{\mathbf r}_1 = f(r)\hat{r}$$ $$m_2 \ddot{\mathbf r}_2= -f(r)\hat{r}$$ enter image description here

From these two equations I have $$\ddot{\mathbf{r}}_2 - \ddot{\mathbf{r}}_1 = -f(\mathbf{r}) \hat{r} \left(\frac{1}{m_1} + \frac{1}{m_2} \right)$$ or $$\mu \ddot{\mathbf r} = - f(r)\hat{r}$$ where $\mu = \frac{m_1m_2}{m_1 + m_2}$ Now by just looking at the last equation it seems that there is a single particle under a central force and $\vec{r}$ is just the position vector of that particle.

enter image description here

But I also know that $\vec{r}$ is the separation vector between particles of masses $m_1$ and $m_2$. So it seems to me that when I find the trajectory of $\mu$ (when I find $\vec{r}(t)$) it would also be the trajectory of $m_2$ with $m_1$ at the origin. But it seems my reasoning was wrong because it turns out that the center of mass is at the origin which is further reinforced by the fact that the angular momentum and energy of $\mu$ is the angular momentum and energy of the two body system in the center of mass frame. So there must be something I was missing. But I can't figure out what it is.

$\endgroup$
  • 1
    $\begingroup$ In a gravitational two-body problem with negative energy, both bodies follow similar elliptic orbits with the same orbital period around their common barycenter. Also the relative position of one body with respect to the other follows an elliptic orbit.en.wikipedia.org/wiki/Elliptic_orbit $\endgroup$ – Wolphram jonny Jul 6 at 20:56
1
$\begingroup$

I figured what was wrong with my argument a while back. First of all J. Thomas was right, I do implicitly put the center of mass at the origin. I guess the way the one body problem equation was derived in my book confused me. If I put my origin at the center of mass, then the position vectors of $m_1$ and $m_2$ relative to the center of mass are $-\frac{m_2}{m_1 + m_2} \mathbf{r}$ and $\frac{m_1}{m_1 + m_2} \mathbf{r}$ respectively. So the equations of motion for $m_1$ and $m_2$ in the cm frame are $$m_1 \frac{d^2}{dt^2}\left( -\frac{m_2}{m_1 + m_2} \mathbf{r} \right) = f(r)$$ or $$\mu \ddot{\mathbf{r}} = - f(r)$$ and $$m_2 \frac{d^2}{dt^2}\left( \frac{m_1}{m_1 + m_2} \mathbf{r} \right) = - f(r)$$ or $$\mu \ddot{\mathbf{r}} = - f(r)$$

But I'm still solving for the separation vector $\mathbf{r}$ and not the actual positions of $m_1$ or $m_2$ with respect to the center of mass. My mistake was forgetting that the tail of $\mathbf{r}$ was an accelerating point. I thought that because I solved for $\mathbf{r}$ with it's tail fixed at my origin, it's motion must be the same $\mathbf{r}$ as seen if $m_1$ was my origin. But the rate of change of vectors when viewed in rotating frames are different from rate of change of vectors when viewed in inertial frames. So the viewed trajectory would be different. If $\mathbf{\Omega}$ is the angular velocity of $m_1$ at an instant with respect to the center of mass, then the rate of change of the separation vector as viewed with $m_1$ at the center is $$\left( \frac{d\mathbf{r}}{dt} \right)_{m_1} = \left( \frac{d\mathbf{r}}{dt} \right)_{cm} - \mathbf{\Omega} \times \mathbf{r} $$ So the trajectories are different. The actual trajectories of $m_1$ and $m_2$ are ofcourse given by $-\frac{m_2}{m_1 + m_2} \mathbf{r}$ and $\frac{m_1}{m_1 + m_2} \mathbf{r}$ respectively with the origin (center of mass) at the focus. But if $m_2 \gg m_1$ like in the Earth-Sun case, then the trajectory of $m_1$ with respect to cm is approximately $\mathbf{r}(t)$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

You started by implicitly putting the center of mass at the origin.

If you put $m_1$ at the origin you have changed the setup.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How did I put cm at origin implicitly? $\endgroup$ – Brain Stroke Patient Jul 6 at 13:13
  • $\begingroup$ @Brain The vector $\vec{r}$ with $r=0$ is the centre of mass. $\endgroup$ – PM 2Ring Jul 6 at 13:24
  • $\begingroup$ @PM2Ring $\mathbf{r} = 0$ only when $m_1$ and $m_2$ are at the same point and I agree that then that's also where the center of mass is, but in general for any value of $\mathbf{r}$ has it's tail at $m_1$ and it's head at $m_2$, no? And the center of mass lies somewhere in between? $\endgroup$ – Brain Stroke Patient Jul 6 at 13:33
  • $\begingroup$ @Brain Not really. You've converted a 2 body system to an equivalent 1 body system. See en.wikipedia.org/wiki/Reduced_mass $\endgroup$ – PM 2Ring Jul 6 at 13:50
  • $\begingroup$ But isn't the $\mathbf{r}$ in the one body problem equation mathematically the same as the separation vector between $m_1$ and $m_2$? $\endgroup$ – Brain Stroke Patient Jul 6 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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