0
$\begingroup$

Two bodies each of mass $m$ are rotating about their center of mass where the radius is $r$.

Here centripetal force of each body is $\frac{mv^2}{r}$ where $v$ is the linear speed. Now, gravitational force is taken to be $\frac{Gm^2}{(2r)^2}$ and they are then made equal. But i have a doubt.

Since there is center of mass in this case, the center also has mass. So shouldn't there be gravitational force between the center and each of the two individual masses?The teachers and textbooks also don't clarify this point.

$\endgroup$
3
  • 2
    $\begingroup$ physics.stackexchange.com/q/661472/305718 $\endgroup$
    – ACB
    Oct 22, 2021 at 10:41
  • $\begingroup$ How much mass is at the centre? $\endgroup$
    – PM 2Ring
    Jun 30, 2023 at 6:38
  • $\begingroup$ Since there is center of mass in this case, the center also has mass No. Center of mass doesn't mean that there is mass at the $(0,0)$ coordinates, just that this point balances all mass around it, i.e. that $\sum_i m_i \cdot \vec r_i = 0$ at this point. Your reasoning flaw sound like you would say "bottle has center of mass inside it, so it must be full of beer". No, it can be empty, i.e. hollow, but still has center of mass inside empty space of bottle. In general, mass distribution around center of mass can have any form. $\endgroup$ Jun 30, 2023 at 8:02

2 Answers 2

0
$\begingroup$

I will tell you what I feel should be the answer. See, in case two bodies, the gravitational force between them is alright. But the moment you consider their centre of mass, the situation changes entirely. For two individual bodies sitting at some distance from each other will feel the gravitational force. But centre of mass concept is different. This was introduced to solve problems with more than one body (generally). For more than one body, the equations of motion become complicated and so although they are solvable, but require large amount of time. On the other hand, if you consider their centre of mass, then you don't need to deal with individual bodies. You can just consider the motion of the centre of mass. And for the last part of your question, no, the bodies won't feel any gravitational force due to centre of mass. Because when you consider the centre of mass, you don't consider the bodies and vice versa. The idea is that we consider that the whole mass of the system is at the centre of mass of the system. We go over to the centre of mass frame. Individual bodies aren't taken into account at that time.

$\endgroup$
0
$\begingroup$

The two bodies will actually revolve each other in an ellipse-shaped orbit.

The only "real" force that applies to these two masses is actually just the a=Gm/r2 gravity acceleration. The centripetal force is apparent, like an illusion, but does not actually give any real acceleration to the objects.

So the two bodies will keep on orbiting each other, because the gravity attraction that pulls them directly towards each other also have to comingle with the already existing vector motions between the two objects.

For example, if the Earth were to be made suddenly completely still with respect to the sun, then our planet will immediately be pulled in straight line towards the sun and within days the Earth will collide with the sun!

So it is only because the Earth is already moving with respect to the sun from time immemorial that, no matter how long the sun keeps pulling us, the end result is that our two masses will just revolve around each other indefinitely in an ellipse-shaped orbit.

The "center of mass of the two body system" is more like an added analytical concept to this, so we might choose use the concept or not depending whether we feel it is helpful to aid our thinking.

If we choose to use the center of mass framework, then we can start to derive mathematical relations etc. about how does the center of mass will behave with respect to the members, how the center of mass will shift over time as they rotate (or not), etc...

Hope it helps!

$\endgroup$
2
  • 1
    $\begingroup$ The Earth would need to travel near the speed of light to reach the Sun in minutes. If the Earth suddenly lost its orbital speed, it would take ~65 days for it to fall into the Sun. $\endgroup$
    – PM 2Ring
    Oct 22, 2021 at 12:53
  • $\begingroup$ @oops yes, you're right... I was recalling 8 minutes which is actually for lightspeed :) I've updated the numbers, thanks! $\endgroup$
    – user315366
    Oct 22, 2021 at 12:55

Your Answer

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

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