Binary stars
Imagine two stars A and B forming a binary star system, where they revolve around a common point in space under their mutual gravitational attraction alone. If A is heavier than B and they both revolve in circular orbits, then which star will have a smaller orbital radius?
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4 solutions
Why would both stars have the same angular velocity ω ?
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Both stars make one revolution in the same period of time (because otherwise the center of mass of the system would clearly not be stationary (or at constant velocity)).
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Thanks, I see now. There is no external force on the system, so the center of mass of the system has constant velocity.
What about the Sun, is it at rest or moving at a constant velocity and the whole solar system moves with it?
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@Rohit Gupta – The center of the Sun isn't the center of mass of the system consisting of the Sun and any planet; that is to say, the Sun slightly 'wobbles' because any planet also exerts an equal gravitational pull on the Sun. (It turns out that the Sun is so large and massive that for any planet besides Jupiter, the center of mass of the Sun-planet system is inside the Sun, though not at its center.) So for any planet, that center of mass is stationary or at constant velocity, were it not for external forces (which includes the other planets); multiple-body systems are very difficult to study. And this is not to mention that we are neglecting the effects of any object outside the solar system.
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@Anton Wu – I meant to ask if the center of mass of the solar system is at rest or moving at constant velocity in space?
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@Rohit Gupta – Yes, the center of mass of the solar system should be at constant velocity since it is of a closed system (this is of course assuming that the effect of any external object, e.g., the black hole at the center of the Milky Way, is negligible, which is not the case in reality).
Won't the gravitational force F g also contain the radius r in its formula?
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The ' r ' in the formula for F g (call it d for clarity) is the distance between the stars, not the orbital radius, the point being that the d is the same for both stars.
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Ohh yeah right. Can we say that the orbital radius is the distance of the star to the center of mass of the system?
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@Rohit Gupta – Yes, the center of mass of the system is the center of both stars' orbits.
what does it mean by centre of mass of the system?
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The center of mass of a system is the unique point with the property that, in studying external forces, it is equivalent to suppose that all of the system's mass is concentrated at that point. In this case, since there are no external forces, the center of mass should be stationary.
The calculation of the center of mass of a system is essentially taking a weighted average of the positions of the system's objects. So in the case of binary stars, if star A has mass m A and position x A and star B has mass m B and position x B , then the position of the center of mass is:
m A + m B m A ⋅ x A + m A + m B m B ⋅ x B .
Just see the figure and tick it if don't know physics much...
Centre of mass of the system is closer to the heavier star.
How do we know the point around which the stars revolve is the center of mass of the system?
In the absence of an external force, the center of mass of the system is stationary. The center of mass must be closer to the heavier star, so it has the smaller orbit. By the way, the orbits are circular only in a frame of reference where the center of mass is stationary and the angular momentum vector is constant. The problem should really state the frame of reference, but that would be a strong hint.
For further clarifications,
1) In the absence of external force, the center of mass need not be stationary rather it can move at a constant velocity.
2) If we have a binary star whose center of mass is stationary, the stars can still move in elliptical orbits about the center of mass.
Radius of a body revolving around a circular path is inversely proportional to its mass
Can you elaborate how is this true?
From the formula for centripetal acceleration ( a c = ω 2 r ), we can calculate the orbital radius r :
r = ω 2 a c = ω 2 F g ⋅ m 1
where m is the star's mass, ω is its angular velocity, F g is the gravitational force that it experiences.
Since the gravitational force on each of the stars is the same (by Newton's third law), the star with the larger mass will have a smaller orbital radius. Thus, the answer is A .