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Orbital period - Where does M1+M2 come from?

by cfsenel
Tags: orbital, period
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cfsenel
#1
Dec8-12, 09:58 AM
P: 4
I must say that I have not studied celestial mechanics other than the crumbs I learnt at high school. Now, what discomforts me is the orbital period formula I saw on Wikipedia:
[tex]T=2\pi\sqrt{\frac{a^3}{G(M_1+M_2)}}[/tex]
I do not understand where does this M1+M2 can possibly come from. My thinking is as follows: This formula must also be valid for circular orbits, so for simplicity I am considering a as the radius, instead of semi-major axes. Let M1=M be our sun, and M2=m is the orbiting planet. Equating the accelerations found from Newton's law of universal gravitation and from circular motion:
[tex]m\left(\frac{2\pi}{T}\right)^2r=G\frac{Mm}{r^2}[/tex]
Therefore,
[tex]T=2\pi\sqrt{\frac{a^3}{GM}}[/tex]
Where does m come into the picture, doesn't it just cancel? For M>>m, the given formula simplifies into what I found alright, but why, in the general case, it is true? The first thing that came into my mind was that I ignored general relativity by using Newton's law of gravity, but I think the effect of general relativity must be far smaller than the difference M+m makes. Another possibility that occurred to me is that in my derivation I assumed M stationary, which could be wrong since the planet also pulls the sun, but I do not know how to put that into the picture. Is that the reason of that M+m, or am I missing something else?
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mfb
#2
Dec8-12, 11:23 AM
Mentor
P: 11,601
a is the radius of the orbit, not the separation of the objects - the more massive object performs a small orbit, too.
They orbit their common center of mass, and their separation is ##a\frac{M+m}{M}## where a is the orbital radius of the mass m.

By the way: Your derived formula is not symmetric in the two objects - it cannot be right. Try to use the orbit of sun in your derivation ;).
cfsenel
#3
Dec8-12, 12:13 PM
P: 4
Alright, they orbit their common center of mass, makes sense, but I didn't quite follow: if they are both orbiting, a is the radius of which orbit? Since you gave their separation as ##a\frac{M+m}{M}## (by the way, how did you find this formula?) which is greater than a, it cannot be the radius of the larger orbit, so am I right to assume that you mean a is the radius of the orbit of the more massive object? Also, since they are both orbiting, what do you mean by seperation? Do you mean the distance between their orbits, so the radius of the less massive object is ##a+a\frac{M+m}{M}=a\frac{2M+m}{M}##?

At any rate, can I solve this following the simple method above (using circular motion formulae), or will I need to solve differential equations? It appears I cannot use that method because the distance between the objects doesn't seem to be constant, as they are both orbiting.

mfb
#4
Dec8-12, 12:17 PM
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P: 11,601
Orbital period - Where does M1+M2 come from?

I used the same a as in your post: The orbital radius of the object with mass m.

(by the way, how did you find this formula?)
m has a distance of a to the center of mass, so M has a distance of ##\frac{m}{M}a##. Add both and you get the separation of the objects.

Also, since they are both orbiting, what do you mean by seperation?
Distance between the objects at any point in time (assuming circular orbits, it is constant).

can I solve this following the simple method above
You can, just replace a by ##a\frac{M+m}{M}## in the formula for the gravitational force.
cfsenel
#5
Dec9-12, 05:15 AM
P: 4
From the way you add the radii of M and m to get their separation, I conclude that you portray a motion where M and m has the same period, and are always on the opposite sides of their orbits. I just realized this makes perfect sense, given that we must keep the center of mass at a constant position, it seems like the only (circular) solution. One thing I must add, for any future reader of this topic, is that in the first formula which I copied from Wikipedia, ##a## was the separation (semi-major axis for non circular orbits), not the radius of the less massive object (I was unable to make that distinction at the beginning). So, the derivation should have been (thanks to mfb):
[tex]m\left(\frac{2\pi}{T}\right)^2r=G\frac{Mm}{a^2}[/tex]
[tex]a=r\frac{M+m}{M} \Rightarrow m\left(\frac{2\pi}{T}\right)^2a\frac{M}{M+m}=G \frac{Mm}{a^2}[/tex]
[tex]T=2\pi\sqrt{\frac{a^3}{G(M+m)}}[/tex]
TurtleMeister
#6
Dec9-12, 05:12 PM
TurtleMeister's Avatar
P: 744
I don't understand what these equations have to do with the title of the thread. I thought the only reason that GM is used instead of G(M+m) is when M >> m and m's effect on T is negligible? cfsenel, your equations are confusing to me. It is unclear to me which variables represent the semi major axis and which ones represent the distance between the two bodies.
TurtleMeister
#7
Dec9-12, 08:54 PM
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P: 744
By the way, I think this is what mfb meant by making your equations symmetrical. But I'm learning this stuff too, so please check my work.

[tex]M_1\left(\frac{2\pi}{T}\right)^2\;r\frac{M_2}{M_1+M_2}\;=\;G\frac{M_1M_ 2}{r^2}[/tex]

[tex]M_2\left(\frac{2\pi}{T}\right)^2\;r\frac{M_1}{M_1+M_2}\;=\;G\frac{M_1M_ 2}{r^2}[/tex]
cfsenel
#8
Dec16-12, 10:03 AM
P: 4
TurtleMeister, I messed up a little bit initially with my usage of terminology, as I am not even learning it; I just remember couple of things from high school and freshmen physics. In my last post, ##a## is the distance between the two bodies and ##r## is the radius of the less massive object. (For reference, this is where I saw the formula) Since I was only dealing with circular orbits, I did not really have to deal with semi-major axes, but it is my understanding that ##a## is also the sum of the semi-major axes in this circular orbit case.
You are correct that GM is used only when M >> m; however, I guess that was the only case I was familiar with; apparently because we studied only that case in high school. Sorry for any confusions that I may have arose, as I said, I am only amateurishly interested in the subject; it would be better if someone with deeper knowledge answer if you still have confusions.
TurtleMeister
#9
Dec16-12, 12:42 PM
TurtleMeister's Avatar
P: 744
Thanks cfsenel. I understand it now. No problem.


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