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Here is a problem from the 2009 $F=ma$ contest on orbital mechanics:

Two stars, one of mass $M$, the other of mass $3M$, orbit their common center of mass. When the stars are collinear with the center of mass, the distance between the two stars is $d$. Find the period of orbit for the star of mass $3M$.

First of all, I am a bit clueless ab out how to solve this problem, partially because I don't understand the statement of the problem:

  1. What does it actually mean for two bodies to "orbit their center of mass". Does that mean that the two bodies move in ellipses, and the center of mass is a foci of each ellipse?

2.Suppose we are given two masses with non-collinear initial velocity vectors. Assuming the only relevant force is the gravitational attraction between the two masses, does it follow that the bodies orbit their center of mass? In other words, under what conditions do two bodies orbit their center of mass?

I would appreciate the following two three things:

  1. You answer my above questions.

  2. You provide a solution to the problem.

  3. You mention any interesting generalizations, related problems, or things I should study to solve problems like this.

Note: I am preparing for the F=ma exam, so any help is truly appreciated.

Edit: See the diagram here: http://www.aapt.org/physicsteam/2010/upload/2009_F-ma.pdf

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  • $\begingroup$ there are 2 "concentric" ellipses, lighter star orbits on a bigger ellipse and heavier star orbits a smaller one. $\endgroup$ – Asphir Dom Jan 11 '15 at 2:24
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What does it actually mean for two bodies to "orbit their center of mass". Does that mean that the two bodies move in ellipses, and the center of mass is a foci of each ellipse?

It is also called a barycenter. Any two (or more) objects in orbit around each other all orbit the barycenter. When working with 2 objects, the center of mass is the barycenter. I think you are confusing the center of mass for each object with the center of mass for the system.

Suppose we are given two masses with non-collinear initial velocity vectors. Assuming the only relevant force is the gravitational attraction between the two masses, does it follow that the bodies orbit their center of mass? In other words, under what conditions do two bodies orbit their center of mass?

The 2 object must always orbit their center of mass of the system.

As for solving the question, it appears there is not enough information provided to determine the period.

Added (corrected):
The period can be found from: $T=2\pi \sqrt {\frac{r^3}{G(M_1+M_2)}}$
The masses are M and 3M and the radii of the orbits is $\frac{1}{4}d + \frac{3}{4}d$.
Substituting the values gives:
$T=2\pi \sqrt {\frac{d^3}{4GM}}$
$T=\pi \sqrt {\frac{d^3}{GM}}$
Which is A.

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  • $\begingroup$ Nice, exactly what I was going to write. If you knew the separation of the bodies, would you be able to solve the problem? Or would you also need to know their velocities? $\endgroup$ – HDE 226868 Jan 11 '15 at 2:24
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    $\begingroup$ You do need to know the distance. I edited the question to reflect this. $\endgroup$ – Joshua Benabou Jan 11 '15 at 2:37
  • $\begingroup$ LDC3: Not if the intial velocity vectors are 0 relative to the center of mass! And not if the initial velocity vectors are collinear with the vector pointing to the center of mass! $\endgroup$ – Joshua Benabou Jan 11 '15 at 2:39
  • $\begingroup$ Also, I read briefly the link about the barycenter. It appears I was right that each body moves in an ellipse with the center of mass of the system (barycenter) at one of the foci. This is an interesting thing and I'm wondering how might one prove this. $\endgroup$ – Joshua Benabou Jan 11 '15 at 2:45
  • $\begingroup$ @JoshuaBenabou If the initial velocities of the objects are zero, then they are attracted to each other and collide. The same thing happens when their velocities are pointed at the center of mass, they collide. I think you mean perpendicular. Also, the center of mass for the 2 objects is always collinear with the 2 objects. $\endgroup$ – LDC3 Jan 11 '15 at 2:49
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I have found the answer to my own question, and I've cleared up the confusion.

Let us first recall:

1.Two bodies moving in space only under the influence of the gravitational attraction between them move in ellipses, with the center of mass of the system at a foci of each ellipse. This is a very nontrivial statement but I will not prove it here.

  1. Two bodies orbiting their center of mass are always collinear with the center of mass. It follows that the gravitational attractive force exerted on each body is a central force (i.e it points towards the center of mass).

The problem statement says "the two bodies are seperated by a distance $d$". The wording is a bit ambiguous, but I believe what is meant to be said is "the two bodies are seperated by a fixed distance $d$ throught their orbits".

Thus we can suppose that the two bodies move in concentric circles with center at the center of mass of the system.

Now, it is tempting to apply Kepler's Third Law here to the orbit of the body with mass $3M$ (as LDC3 did). However the derivation of keplers third law assumes that the mass which provides the centripetal force on the orbiting body is at the same distance from the orbiting body as the mass which provides the gravitational force. In other words, the central force exerted on each body is not originating from the center of mass of the system.

The correct way to proceed is to note the gravitational force on the mass $3M$ is $\frac{G(M)(3M)}{d^2}$ but the centripetal force is $\frac{(3M)v^2}{r}$ where $r$ is the distance to the center of mass, which can be determined to be $d/4$. Using $v=\frac{2\pi r}{T}$ with $T$ the period, we can equate the centripetal force to the gravitational force and arrive at $T=\pi\sqrt{\frac{d^3}{GM}}$. This is in fact an answer choice :).

What we essentialy did was generalize Kepler's Third Law (for circular orbits) to the case where the bodies orbit their center of mass. Kepler's Third Law assumes the body around which we are orbiting is fixed in space.

I now realize that the problem is much more subtle than I previously thought.

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  • $\begingroup$ No, the question is much worse than you previously thought. I would argue that every choices except (E), $T=\frac\pi 4 \sqrt{\frac {d^3}{GM}}$ is a valid response. Option (E) can be ruled out. None of the others can. The wording of the question does not necessarily imply a circular orbit. This is a terrible test question! $\endgroup$ – David Hammen Jan 11 '15 at 8:52
  • $\begingroup$ just fyi it is always okay to accept your own answer if you feel it answers your question the best, which is often the case! $\endgroup$ – uhoh Feb 14 '19 at 6:21

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