Astronomy - Determine the sum of the stellar masses

[tosv]

Homework Statement

A double star located at a distance of 10 light years from us. The maximum angle between the stars, as seen from Earth, is 2 arcseconds. (1 arcsecond = 1 / 3600 degrees), we can assume that the stellar orbit is circular and that this angle gives us the real distance between the stars. We observe that the rotation period of the stars is 4 years. What is the sum of stellar masses?

Homework Equations

Newton's law of universal gravitation:
$$F_{G}=G\frac{m_{1}m_{2}}{r^{2}}$$
Centripetal force:
$$F_c=\frac{mv^2}{r}$$
Kepler's third law:
$$P^{2} \propto a^{3}$$

The Attempt at a Solution

The stellar orbit is circular, so the gravitational force should be equal to the centripetalforce, but how do I proceed?

 

[gneill]

You can work out the actual distance between the two bodies by simple geometry. You should also realize that the both bodies will orbit a common center of mass, so you can work out the individual radii of their orbits (symbolically) as fractions of that distance which depend upon their relative masses. Their individual velocities can similarly be determined (symbolically) from the resulting circumferences of their orbits and the given period of the orbit.

These, along with your gravitational force and centripetal force expressions, should allow you to find an expression for the sum of the two masses.

 

[tosv]

I rewrite the centripetal force as:
$$F_{c}=m\cdot \omega^{2}\cdot r$$

So now I can express the equality between the gravitational force and centripetal force for respective star to be
$$M\omega^{2}R=G\frac{Mm}{(R+r)^{2}}$$
$$m\omega^{2}r=G\frac{Mm}{(R+r)^{2}}$$

I simply the expression and then and I add the left-hand side of the equations together and the same procedure to the right-hand side and I find this expression for the sum of the masses:
$$M+m=\frac{\omega^{2}\cdot (R+r)^{3}}{G}$$

The angular velocity can be expressed as
$$\omega=\frac{2\pi}{T}$$

I finally found this expression for the sum of the masses:
$$M+m=\frac{(\frac{2\pi}{T})^{2}\cdot (R+r)^{3}}{G}$$
But I’m not sure how I’ll proceed and calculate the distance (R+r), should I relate the distance from Earth and the maximum angle between the stars?

 

[gneill]

But I’m not sure how I’ll proceed and calculate the distance (R+r), should I relate the distance from Earth and the maximum angle between the stars?

Yes. Basic trig.

 

[rwisz]

As a scientist, the first step in solving this problem would be to gather all the necessary information and understand the relationships between the different variables. From the given information, we know that the distance between the stars is 10 light years (which is equivalent to approximately 9.461e+16 meters) and the maximum angle between them is 2 arcseconds (which is equivalent to approximately 0.0005556 degrees). We also know that the rotation period of the stars is 4 years.

Next, we can use Kepler's third law to determine the sum of the stellar masses. This law states that the square of the orbital period (P) is proportional to the cube of the semi-major axis (a) of the orbit. In this case, since the orbit is circular, the semi-major axis is equal to the distance between the stars (10 light years). Therefore, we can write the equation as follows:

P^{2} \propto a^{3}
(4 years)^{2} \propto (10 light years)^{3}
16 \propto 1000
P^{2} = \frac{1000}{16}
P^{2} = 62.5

Now, we can use this value of P^{2} to solve for the sum of the stellar masses using Newton's law of universal gravitation. Rearranging the equation, we get:

m_{1}m_{2} = \frac{F_{G}r^{2}}{G}

Since the gravitational force is equal to the centripetal force, we can substitute the equation for the centripetal force (F_{c}=\frac{mv^2}{r}) into the equation for the gravitational force. This gives us:

m_{1}m_{2} = \frac{\frac{mv^{2}}{r} r^{2}}{G}
m_{1}m_{2} = \frac{mv^{2}}{G}

We know that v (the orbital speed) can be calculated using the formula for centripetal force, so we can substitute that into the equation:

m_{1}m_{2} = \frac{\frac{m\frac{2\pi r}{P}^{2}}{G}}
m_{1}m_{2} = \frac{4\pi^{2}m}{GP}

Now, we can substitute the value of P^{2