The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36 km/s. The second star, Beta, has an orbital speed of 12 km/s. The orbital period is 137 d.
a) What is the mass of the star alpha?
b) What is the mass of the star beta?
c) One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole. The orbital period of A0620-0090 is 7.75 hours, the mass of V616 Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of the orbit of the orange star.
d) Find the radius of the orbit of the black hole.
e)Find the orbital speed of the orange star.
f)Find the orbital speed of the black hole.
m_a = 7.80 x 10^29
not sure about m_b
m_b is 2.34*10^30
radius of the orbit of the black hole
R = 3.40×108 m
orbital speed of the orange star
440 km/s
orbital speed of the black hole
77.0 km/s
I know I have to find angular speed in order to find the radius of orbit.
With the radius of orbit and velocity I can find centripetal acceleration. I then apply Newton's 2nd Law to solve for the mass.
I am stuck on finding angular speed...
Hey there Aliison,
For this problem, you don't actually need to use kepler's laws, all you have to do is 1) find both the radii of the of the stars to the center of the system (denoted by R_a and R_b using the period equation T=(2*pi*R)/v) then, since they are in stable orbits, you can relate the centripetal force (mV^2/R note that the radius here is either R_a or R_b) with he gravitational force between the two (G(m_a*m_b)/R^2 -- the R here is R_a+R_b) Hope this helps
To solve these problems, we can use Kepler's third law of planetary motion, which states that the square of the orbital period (T) of a planet is proportional to the cube of its semi-major axis (r). We can also use Newton's form of Kepler's third law, which relates the orbital period, orbital speed, and mass of the star.
Let's start with the first set of questions about the binary star system with stars Alpha and Beta.
a) To find the mass of star Alpha, we'll use Newton's form of Kepler's third law:
(T1 / T2) ^2 = (V2 / V1) ^3
Where T1 and T2 are the orbital periods of the two stars, and V1 and V2 are their respective orbital speeds. In this case, T1 = T2 = 137 days, V1 = 36 km/s, and V2 = 12 km/s.
Plugging in these values, we have:
(137 / 137) ^2 = (12 / 36) ^3
1 = (1/3) ^3
1 = 1/27
Hence, star Alpha has a mass of 27 times the mass of star Beta.
b) To find the mass of star Beta, we need to calculate the mass of star Alpha first. Using the earlier calculation, we found that star Alpha has a mass of 27 times that of Beta. So if we denote the mass of Beta as M, then the mass of Alpha would be 27M.
The total mass of the binary star system is the sum of the masses of Alpha and Beta:
Total mass = Mass of Alpha + Mass of Beta
Total mass = 27M + M
Total mass = 28M
Since we know the orbital speeds of both stars, we can use Newton's form of Kepler's third law again:
(T1 / T2) ^2 = (M1 / M2)
Plugging in the values, we have:
(137 / 137) ^2 = (28M / M)
1 = 28
So, M = 1/28 of the total mass of the binary star system. Therefore, the mass of star Beta is 1/28th of the total mass.
Now, let's move on to the second set of questions about the binary system A0620-0090.
c) To find the radius of the orbit of the orange star (V616 Monocerotis), we'll use Kepler's third law once again. Since we're given the orbital period (T) in hours, we need to convert it to days before using the equation.
r ^ 3 = (T / 365.25) ^ 2
We're also given that the mass of V616 Monocerotis is 0.67 times the mass of the Sun, which we'll denote as M_Orange. Since we're given the mass of the black hole (3.8 times the mass of the Sun), we'll denote it as M_BlackHole.
Using Newton's form of Kepler's third law:
(T_Orange / T_BlackHole) ^2 = (M_BlackHole / M_Orange)
Plugging in the values, we have:
(T_Orange / 7.75) ^ 2 = (3.8 / 0.67)
Simplifying and solving for T_Orange, we get:
T_Orange = 7.75 * sqrt(3.8 / 0.67)
Now we can substitute this value of T_Orange into the equation for the radius:
r ^ 3 = (T_Orange / 365.25) ^ 2
Solving for r, we get the radius of the orbit of the orange star.
d) To find the radius of the orbit of the black hole, we know that the radius of the orbit of the orange star will be the same as the radius of the orbit of the black hole since they're in a binary system. Therefore, the radius of the orbit of the black hole is the same as the radius we found in part c.
e) To find the orbital speed of the orange star, we'll use the equation for orbital speed:
V = (2πr) / T_Orange
Now we can substitute the radius (r) we found in part c and the orbital period (T_Orange) to calculate the orbital speed of the orange star.
f) To find the orbital speed of the black hole, we use the same equation as part e, but with the orbital period of the black hole (T_BlackHole) and the radius (r) we found in part d.
By following these steps and using the given values, you can find the mass of stars Alpha and Beta, as well as the relevant quantities for the A0620-0090 binary system.
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