a certain binary star system consists of two stars of masses m and 3m respectively separated by a distance of 1.5 x 10^12 m, measured between their centres. Theres stars revolve in sync with each other around their common centre of mass like a giant dumbbell. If the smaller star has a mass of 1.0 x 10^30 kg, what is their common period of revolution?
In principle, it is easy.
calculate the distance each star is from the center of mass, that is the radius of rotation for each star.
Hint: the mass m is 3 parts of separation distance, whilst the 3m star is one part away. So divide the distance by 4, and you know the two radius of roataions. You need this in the centripetal force calculation.
Forcegravity=centripetal force
for the small star...
G3m^2/distance^2=m w^2 (3/4 distance)
solve for w
but shouldn't the w for the large star be the same? Lets check..
G3m^2/distance^2=3m w^2 (1/4 distance)
hmmm. it is the same angular velocity.
Oh, period. Convert w (knowing w=2PI/period) to period.
To find the common period of revolution for the two stars in the binary system, we can use Kepler's Third Law of Planetary Motion, which can also be applied to binary star systems. According to Kepler's Third Law, the square of the period of revolution is proportional to the cube of the average distance between the objects.
The formula for Kepler's Third Law is:
T^2 = (4π^2 / G) * r^3
where:
T = Common period of revolution
G = Gravitational constant (6.674 × 10^-11 m^3⋅kg^-1⋅s^-2)
r = Average distance between the stars' centres
In this case, the average distance between the stars' centres is given as 1.5 × 10^12 m. However, we need to calculate the average distance from the centre of mass.
To do this, we can use the concept of the reduced mass (μ), which represents an adjustment of the masses to account for the rotation about the centre of mass. The formula for reduced mass is:
μ = (m1 * m2) / (m1 + m2)
where:
m1 = Mass of the smaller star
m2 = Mass of the larger star
Given that the mass of the smaller star (m1) is 1.0 × 10^30 kg, and the mass of the larger star (m2) is 3 times m1, we have:
m1 = 1.0 × 10^30 kg
m2 = 3m1 = 3.0 × 10^30 kg
Therefore, the reduced mass (μ) is calculated as:
μ = (m1 * m2) / (m1 + m2) = (1.0 × 10^30 kg * 3.0 × 10^30 kg) / (1.0 × 10^30 kg + 3.0 × 10^30 kg) = 0.75 × 10^30 kg
Now, using the reduced mass (μ) and the given average distance (r), we can substitute these values into Kepler's Third Law equation to find the square of the period of revolution. Once we find the square of the period, we can then take the square root to get the actual period (T).
T^2 = (4π^2 / G) * r^3
T^2 = (4 * π^2 / (6.674 × 10^-11 m^3⋅kg^-1⋅s^-2)) * (1.5 × 10^12 m)^3
T^2 = (5.94 × 10^22) * (3.375 × 10^36 m^3)
T^2 ≈ 2.00 × 10^59 m^3
Taking the square root of both sides to solve for T:
T ≈ √(2.00 × 10^59 m^3)
T ≈ 4.47 × 10^29 s
Therefore, the common period of revolution for the stars in this binary system is approximately 4.47 × 10^29 seconds.
To find the common period of revolution for the binary star system, we can use Kepler's Third Law of Planetary Motion. This law states that the square of the period of revolution, T, is directly proportional to the cube of the average distance, r, between the two masses (stars in this case). The equation can be written as:
T^2 = (4π^2 / G) * r^3
Where:
T = Period of revolution
G = Gravitational constant (approximately 6.67 x 10^-11 m^3 kg^-1 s^-2)
r = Average distance between the two masses (stars)
Given that the two stars have masses m and 3m respectively, and the average distance between their centers is 1.5 x 10^12 m, we can calculate r using the formula:
r = (1/2) * d
Where d is the distance between the two stars' centers.
Let's calculate the value of r:
d = 1.5 x 10^12 m
r = (1/2) * (1.5 x 10^12 m)
r = 0.75 x 10^12 m
r = 7.5 x 10^11 m
Now, let's substitute the values into the equation:
T^2 = (4π^2 / G) * (7.5 x 10^11 m)^3
Simplifying the equation:
T^2 = [(4π^2) / G] * (7.5 x 10^11 m)^3
T^2 = [(4 * 3.14^2) / (6.67 x 10^-11 m^3 kg^-1 s^-2)] * (7.5 x 10^11 m)^3
T^2 = (12.56 / 6.67) * (7.5 x 10^11 m)^3
T^2 = 1.879 x (7.5 x 10^11 m)^3
Calculating:
T^2 = 1.879 x (7.5 x 10^11)^3
T^2 = 1.879 x (5.0625 x 10^33)
T^2 = 9.5758125 x 10^33
Taking the square root of both sides to find T:
T = √(9.5758125 x 10^33)
T ≈ 9.786 x 10^16 seconds
Therefore, the common period of revolution for the binary star system is approximately 9.786 x 10^16 seconds.