Consider a double star system under the influence of the gravitational force between the stars. Star 1 has mass m1 = 1.72 × 1031 kg and Star 2 has mass m2 = 1.60 × 1031 kg. Assume that each star undergoes uniform circular motion about the center of mass of the system (cm). In the figure below r1 is the distance between Star 1 and cm, and r2 is the distance between Star 2 and cm.If the stars are always a fixed distance s=r1+r2 = 3.05 × 1018 m apart, what is the period of the orbit (in s)?
To find the period of the orbit, we need to use Kepler's Third Law and the concept of the reduced mass.
1. Find the reduced mass (µ):
The reduced mass is given by the formula:
µ = (m1 * m2) / (m1 + m2)
Substituting the given values:
µ = (1.72 × 10^31 kg * 1.60 × 10^31 kg) / (1.72 × 10^31 kg + 1.60 × 10^31 kg)
µ = 2.75 × 10^61 kg² / 3.32 × 10^31 kg
µ ≈ 8.28 × 10^29 kg
2. Use Kepler's Third Law to find the period (T):
Kepler's Third Law states that the square of the period of an orbit (T) is proportional to the cube of the semi-major axis (a) of the orbit.
In this case, we have the distance between the stars (s), which is equal to the sum of the semi-major axis of each star's orbit.
Therefore, s = r1 + r2 = 3.05 × 10^18 m
Since we have the distances from center of mass to each star (r1 and r2), we can write:
s = r1 + r2
Rearranging the equation:
r1 = s * m2 / (m1 + m2)
r2 = s * m1 / (m1 + m2)
Now, we have the semi-major axes (r1 and r2) for each star's orbit.
3. Calculate the period (T) using the semi-major axes:
T1 = 2π * √(r1³ / (G * (m1 + m2)))
T2 = 2π * √(r2³ / (G * (m1 + m2)))
G is the gravitational constant, which is approximately 6.67430 × 10^(-11) N(m/kg)².
Substituting the given values and simplifying the equations:
T1 = 2π * √((s * m2)^3 / (G * (m1 + m2)^2 * (m1 + m2)))
T2 = 2π * √((s * m1)^3 / (G * (m1 + m2)^2 * (m1 + m2)))
As both stars orbit around the center of mass, their periods are the same, so we can simply call it T.
T = T1 = T2
Simplifying the equation further:
T = 2π * √((s^3 * m1 * m2) / (G * (m1 + m2)^2 * (m1 + m2)))
4. Substitute the given values and calculate the period (T):
T = 2π * √((3.05 × 10^18 m)^3 * (1.72 × 10^31 kg) * (1.60 × 10^31 kg) / (6.67430 × 10^(-11) N(m/kg)² * (1.72 × 10^31 kg + 1.60 × 10^31 kg)^2 * (1.72 × 10^31 kg + 1.60 × 10^31 kg)))
Calculate the expression using a calculator to get the final answer.
Note: The answer to this calculation may seem very large, as it involves astronomical values.
To find the period of the orbit in this double star system, we can use Kepler's Third Law, which relates the period of an orbit to the semi-major axis of the orbit.
Kepler's Third Law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a):
T^2 = k * a^3
where k is a constant value.
In this case, the semi-major axis is equal to the sum of the distances from each star to the center of mass: a = r1 + r2.
Since the stars are always a fixed distance s = r1 + r2 apart, we can substitute s for a in the equation:
T^2 = k * s^3
We can solve for T by taking the square root of both sides:
T = √(k * s^3)
To find the value of the constant k, we can use the fact that the sum of the masses of the two stars is equal to the product of the constant and the square of the period:
m1 + m2 = k * T^2
Substituting the masses and the known value for the separation s:
1.72 x 10^31 kg + 1.60 x 10^31 kg = k * (3.05 x 10^18 m)^2
Simplifying and solving for k:
k = (1.72 x 10^31 kg + 1.60 x 10^31 kg) / (3.05 x 10^18 m)^2
Now we can substitute the value of k into the equation for the period:
T = √[(1.72 x 10^31 kg + 1.60 x 10^31 kg) / (3.05 x 10^18 m)^2] * s
Calculating this gives the period of the orbit.