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binary_stars_2

Assume the mass of star 2 is 4 times the mass of star 1. Compare the radius of the orbit of star 1 to that of star 2.

Possible answers:

R1:R2=1:4

R1:R2=1:2

R1:R2=2:1

R1:R2=4:1

R1:R2=16:1

What's the answer?

I guess I'd use the gravitational force formula:

$ F = G\cdot \frac {m1 \cdot m2}{r^{2}} $

But I don't see how the $r$ from the formula, which is the distance between the two stars can be related to what the question is essentially asking - radius of the orbit of the smaller star.

And when the question says: "Compare the radius of the orbit of star 1 to that of star 2"

Wouldn't it mean that the bigger star (2) will stay stationary, and the smaller star would orbit around it? Or will they both orbit each other? I guess the latter.

If so, wouldn't they orbit each other the same way? e.g. always remain the same distance from each other?

Qmechanic
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Jack
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2 Answers2

3

I will be assuming that the system has some angular momentum about the center of mass initially. If the system has no angular momentum then both the stars would accelerate towards each other and end up colliding.

The problem is a two body problem. In such cases, both the stars would revolve around the center of mass.

If the distance between two stars is $x$ then the center of mass will be at a distance $\frac{x}{5}$ from the heavier star and at a distance $\frac{4x}{5}$ from the lighter one. Hence, the ratio of the distances $R_1:R_2$ will be $4 : 1$.

For cases in which one of the bodies is much heavier than other, the center of mass is much nearer to the heavier body and hence can be considered as motion of lighter body around heavier one. The earth-sun system is one such example.

I have a feeling that you are not aware of what two-body problem is. So I would suggest you to read first three chapters of the book Classical Mechanics by Herbert Goldstein.

user40330
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2

For two objects to remain in a stable circular orbit, the force acting on them must be equal to the centripetal force corresponding to their rotation. $$F=\frac{mv^2}{r}$$

or in terms of angular velocity $$F=m\omega^2r$$

where $r$ is the radius of orbit in this case.

As the gravitational force acting on the two stars is the same. $F$ is equal in both cases. Additionally the angular velocity must be equal in both cases. Otherwise the force will not keep acting along the centre of orbit which is obviously unphysical. Therefore we get:

$$m_1\omega^2r_1=m_2\omega^2r_2$$ $$4mr_1=mr_2$$ $$4r_1=r_2$$

The gravitational force is of limited use for your problem. If you knew the masses and wanted to work out the radial distances then you could use the gravitational force.

Floris
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nivag
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