# Two equal-mass stars orbit each other

1. Mar 24, 2015

### Calpalned

1. The problem statement, all variables and given/known data
Two equal-mass stars maintain a constant distance apart of $8.0 *(10^11)$ m it is eight times ten to the eleventh, not eight times ten to the first power times one. I don't know why Latex forbids exponents of two digits or more and rotate about a point midway between them at a rate of one revolution every $12.6$ years. What is the mass of each star?

2. Relevant equations
(1) $a_R=\frac{v^2}{R}$
(2) Law of universal gravitation $\frac{GM}{R^2}$

3. The attempt at a solution
My solutions guide manual says "set the gravity force on one of the stars equal to the centripetal force..." $F_G = G\frac{M^2}{d^2} = F_R = M\frac{v^2}{d/2}$ then appears in the guide. Here $d= 2R =$diameter. I am confused because I don't see how $d$ can be used in place of $R$ in equation for universal gravity. Additionally, where did $M^2$ come from? Is it the combined mass of the two stars? Finally, in the last part $= M\frac{v^2}{d/2}$ what is M? The two stars are orbiting each other, so there is nothing in the center. Why isn't $M = 0$kg?

PS: The question states "two equal mass stars..." Correct me if I'm wrong but I am under the assumption that if two objects orbit each other (or a cen
tral point between them) their masses MUST be equal or nearly equal... Is this always true?

Thank you.

2. Mar 24, 2015

### Staff: Mentor

Put things that you want Latex to treat as a single group within curly brackets. Thus: 10^{11}.
Newton's law of universal gravitation expresses the force due to gravity between two masses. You've only shown one, and you haven't shown an equals sign to indicate what the expression yields.

Thus: $F = G\frac{M_1 M_2}{R^2}$

The "R" in Newton's law of gravitation is the distance between the centers of the two interacting masses. Here that distance happens to be d.
There are two masses involved in Newton's law. They just so happen to be identical in this problem.
You pick either of the stars and determine the required centripetal force. By symmetry the centripetal force on the other star will be the same.

That there is no body at the center of rotation of the system doesn't matter. It's still the center of mass of the system, and the two stars are orbiting it.
No, having unequal masses simply moves the location of the center of mass around which the bodies orbit. It becomes a matter of convention to declare the more massive object the "primary" and consider the smaller one to be the satellite, particularly if the center of gravity happens to be located within the body of said primary.

3. Mar 24, 2015

### Dick

In $v^2/R$ the $R$ is the distance from each star to the center of rotation. In the law of gravitation $R$ is the distance between the two stars. They are different. The gravitation law says $F=G M_1 M_2/R^2$ where $M_1$ and $M_2$ are the masses of the two bodies. Doesn't that tell you where the $M^2$ comes from? And yes, if two objects orbit each other around a center midway between at a constant distance they have equal mass. A lot of these things you can learn by understanding the equations and using them instead of asking questions and being confused about EVERYTHING.

4. Mar 24, 2015

### SammyS

Staff Emeritus
You need to place braces around the 11 .

$\text{10^{11} gives } 10^{11}\ .$
There is nothing special about using $r$ for a variable in $\displaystyle\ F_\text{gravity}=G\frac{m_1\cdot m_2}{r^2} \$ .

$r$ is simply the distance between the two point masses.

In your case the distance is d.

Regarding the question in your PS: If two isolated objects orbit each other , both objects orbit about the point which is the "center of mass" of the two objects, no matter what there relative masses may be.

5. Mar 25, 2015

### Calpalned

When I solve physics problems, a lot of the time I feel that I did every math calculation correctly, but I end up with the wrong answer. Would any of you be able to point out my mistake?

$F_G = mg$
$\frac{mv^2}{R} = \frac{MMG}{R^2}$ where $d = 2R$
$v^2 = \frac{mg}{(d/2)^2}$ I solved for velocity, which is the same for the two stars because they have equal masses.
$(d\pi)^2 = mg(\frac{2}{d})^2$
$d^2\pi^2 = \frac{4mg}{d^2}$
$m = \frac{d^4\pi^2}{4G}$ here, $d= 8.0 * 10^{11}$ meters
$m = 1.515 * 10^58$ My answer
$m = 9.6 * 10^29$ Correct answer

6. Mar 25, 2015

### Dick

Check units on your expressions. $\frac{d^4\pi^2}{4G}$ does not have the units of a mass. Go back and try to figure out where you started to go wrong. And how did you manage to get an expression for velocity without using that the period is 12.6 years?

7. Mar 25, 2015

### Calpalned

My solutions guide shows that the orbit of the two stars is an elongated ellipse (see picture). It states that $F_R = M \frac{v^2}{d/2} = M \frac{2(\pi d/T)^2}{d}$
It is clear that the writer used the identity $vT$ = displacement = $2\pi R$ Is $2 \pi R$ valid for ellipses?

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8. Mar 25, 2015

### Staff: Mentor

The problem statement says that: "The two stars don't crash into each other because of their circular motion. " So I would consider the image to depict a view of the orbit that is more edge-on than overhead.

9. Mar 25, 2015

### Calpalned

Thank you, I see it now. However, if the orbit were the shape of an ellipse, the circumference will not be $2\pi R$ right? Thanks Gniell

10. Mar 25, 2015

### Phynos

Double check that substitution you did for the velocity.

11. Mar 25, 2015

### Staff: Mentor

Right. If the orbit were elliptical the circumference would not be $2 \pi R$, and furthermore, the velocity would not be constant around the orbit.

12. Mar 25, 2015

### Calpalned

I redid the problem and I think I understand it. I just have one last question. One of the first steps to solving this problem is to set the universal law of gravitation equal to centripetal acceleration.
$\frac {MMG}{d^2} = \frac {Mv^2}{d/2}$
I just want to make sure I have understand this correctly: The law of universal gravitation uses the distance between the two masses while centripetal acceleration uses the distance between one of the masses and the central point of rotation. If not, how can I figure out what to use for r in the two equations?

Last edited: Mar 25, 2015
13. Mar 25, 2015

### Phynos

The gravitational force on one star comes from the opposing star, which is a distance d away.

They are both in orbit about the center of mass of that system, which is halfway between them, so d/2.

14. Mar 25, 2015

### Calpalned

I see that my mistake occurred in the first step.
$\frac {v^2}{r} = \frac {MG}{r^2}$ is not correct.
Rather, it should be $\frac {v^2}{(d/2)} = \frac {MG}{d^2}$

Thanks to your help, I caught another mistake. $v = \frac {d \pi}{T}$ I forgot to divide by T