-2
$\begingroup$

If the centre of mass is taken as the origin, then the gravitational potential energy of two bodies is \begin{equation} V=-\frac{Gm_{1}m_{2}}{(r_{1}+r_{2})} \end{equation} where $r_{1}$ and $r_{2}$ are the distances from the center of mass of the system to the bodies $m_{1}$ and $m_{2}$ respectively.

Is this right? Correct me if I'm wrong.

$\endgroup$
  • $\begingroup$ If you place the origin instead at one of the bodies, is the separation distance the same as $r_1+r_2$? No. $\endgroup$ – Captain Morgan May 25 '18 at 8:56
  • $\begingroup$ @CaptainMorgan But the problem I'm trying to solve requires the centre of mass of the system to be treated as the origin. $\endgroup$ – Viswa May 25 '18 at 9:10
  • $\begingroup$ That's fine, I'm just trying to clarify that because the separation distance is unequal in the two case, then one is wrong. Physics doesn't change when you shift reference frames. Check $r_{12}$ in the CM frame and $r_2$ in a $r_1$ centered frame and you'll see the force of gravity is then equal $\endgroup$ – Captain Morgan May 25 '18 at 15:39
  • $\begingroup$ @CaptainMorgan No, the separation distance is the same. The CM is in between the two bodies along the line joining them. The body $m_{1}$ is to the left of CM at $-r_{1}$ (as mentioned in the question, $r_{1}$ is distance between the CM and $m_{1}$) and $m_{2}$ is at $r_{2}$, to the right of CM. So, the distance between the two bodies is $r_{1}+r_{2}$. If the origin is placed at $m_{1}$ (let this be primed coordinate system), then $r'_{1}=0$ and let $m_{2}$ be at $r'_{2}$. The distance between the two bodies in this frame is just $r'_{2}$. Also, $r'_{2}=r_{1}+r_{2}$. $\endgroup$ – Viswa May 26 '18 at 11:51
  • $\begingroup$ You're defining $m_1$'s position as $-r_1$? Well, I suppose then yes $r'_{2}=r_{1}+r_{2}$. However, I must advise you to not do such a thing. In all of my years, I've never before encountered a position defined negatively. This was our disconnect. It's conventional to define a position as $r_i$ and if the vector is negative in all of its components, we leave it like that $\endgroup$ – Captain Morgan May 26 '18 at 14:56
0
$\begingroup$

This is almost correct. The formula is $ V=-\frac{Gm_{1}m_{2}}{r_{12}} $ where $r_{12} $ is the distance between the centers of mass of the two particles.

$\endgroup$
  • 1
    $\begingroup$ Why do you say that it's 'almost' correct? Isn't $r_{12}=r_{1}+r_{2}$, in CM frame? $\endgroup$ – Viswa May 25 '18 at 9:24
  • $\begingroup$ You are right. For two bodies it is the same. $\endgroup$ – my2cts May 25 '18 at 11:51
  • $\begingroup$ $r_{12}=|r_1-r_2|$ $\endgroup$ – Captain Morgan May 25 '18 at 17:44
0
$\begingroup$

Consider $m_1$ at $\mathbf{r}_1=\sum r_i \mathbf{e}_i$. Consider then a second mass $m_2$ at $\mathbf{r}_2=\sum r_i \mathbf{e}_i$. This (arbitrary) choice of coordinates is such that our origin is the CM for the two. The separation between the two is, $$r_{12}\equiv|\mathbf{r}_1-\mathbf{r}_2|$$ Let us now move our coordinate system to where the origin is at $m_1$. Shifting all positions by subtracting $\mathbf{r}_1$, $$\mathbf{r'}_1=\mathbf{r}_1-\mathbf{r}_1= \mathbf{0}$$ $$\mathbf{r'}_2=\mathbf{r}_2-\mathbf{r}_1$$ In this reference, the separation distance is simply the magnitude of $\mathbf{r'}_2$. In both cases we see that the separation distance is the same regardless of where we place the origin of our coordinate system. This is good because the laws of physics should agree regardless of the chosen inertial frame. That is to specifically say, $V_{12}$ is the same in both cases as it should. If it weren't, the bodies would fall towards each other differently depending on where you look at them. That doesn't make sense.

$\endgroup$

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.