A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of the motion, T=29.6 days, and the orbital speed v= 220 km/s of the stars can be measured from telescopic observations. What is the mass (kg) of each star?
To find the mass of each star in the binary star system, we can use Kepler's Third Law and the formula for orbital speed.
Step 1: Convert the period T from days to seconds.
Given: T = 29.6 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
T (in seconds) = 29.6 days * 24 hours * 60 minutes * 60 seconds
T = 2,550,720 seconds
Step 2: Convert the orbital speed v from km/s to m/s.
Given: v = 220 km/s
v (in meters/second) = 220 km/s * 1000 meters/second
v = 220,000 meters/second
Step 3: Use Kepler's Third Law to relate the mass (M) of each star and the period (T) of their motion.
Kepler's Third Law states: T^2 = (4π^2 / GM) * R^3
Where:
- T is the period of the motion
- G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the total mass of the binary star system
- R is the orbital radius (distance between the two stars)
Step 4: Rearrange the equation to solve for M.
M = (4π^2 / G) * (R^3 / T^2)
Step 5: Calculate the orbital radius R.
The orbital radius (R) can be calculated using the speed (v) and the period (T).
R = v * T / (2π)
Step 6: Substitute the values into the equation and calculate M.
M = (4π^2 / G) * (R^3 / T^2)
= (4 * π^2 / G) * ( (v * T / (2π))^3 / T^2)
Step 7: Calculate the mass (M) of each star by dividing the total mass (M) by 2.
Mass of each star = M / 2
Let's calculate the values step-by-step.
Gravitational constant (G) = 6.67430 × 10^-11 m^3 kg^-1 s^-2
π (pi) = 3.14159
T (in seconds) = 2,550,720 seconds
v (in meters/second) = 220,000 meters/second
R = v * T / (2π)
= 220,000 * 2,550,720 / (2 * 3.14159)
= 11,354,737,326.5 meters
M = (4π^2 / G) * ( (v * T / (2π))^3 / T^2)
= (4 * 3.14159^2 / 6.67430 × 10^-11) * ( (220,000 * 2,550,720 / (2 * 3.14159))^3 / (2,550,720)^2)
= 1.0799112654 × 10^31 kg
Mass of each star = M / 2
= 1.0799112654 × 10^31 kg / 2
= 5.399556327 × 10^30 kg
Therefore, the mass of each star in the binary star system is approximately 5.399556327 × 10^30 kg.
To find the mass of each star in the binary system, we can use Kepler's Third Law, which relates the period (T) and the orbital radius (r) to the total mass (M) of the system:
T^2 = (4π^2/G) * (r^3/M)
Where:
T = period of motion
r = orbital radius
G = gravitational constant
Since we are given the period (T) and the orbital speed (v), we first need to find the orbital radius (r).
The orbital speed (v) is related to the mass (M) and orbital radius (r) by:
v = √(GM/r)
Where:
M = total mass of the system
Rearranging the equation, we can solve for r:
r = GM/v^2
Given that the period (T) is the time taken to complete one revolution, and the orbital radius (r) is related to the distance traveled by the star, we can express the orbital radius (r) in terms of the orbital speed (v) and the period (T):
r = v * T
Substituting this expression for r in the previous equation:
v * T = GM/v^2
Simplifying further:
v^3 * T = GM
Now, we can isolate M:
M = (v^3 * T) / G
Finally, since we are looking for the mass of each star and the question states that both stars have equal masses, we divide the total mass by 2:
Mass of each star = [(v^3 * T) / G] / 2
Now we can plug in the given values:
T = 29.6 days = 2,553,600 seconds (since 1 day = 86,400 seconds)
v = 220 km/s = 220,000 m/s (since 1 km = 1000 m)
G = 6.67430 × 10^-11 m^3 kg^-1 s^-2
Substituting into the equation and calculating:
Mass of each star = [(220,000^3 * 2,553,600) / (6.67430 × 10^-11)] / 2
Therefore, the mass of each star in the binary star system is the calculated value.
Each has a force holding it to the other.
F=GM^2/d^2
this is equal to centripetal acceleration.
GM^2/d^2=v^2/(d/2)^2
check my thinking. solve for M
Using Newton's law and the law of universal graviation:
F=ma
Acceleration is centripital acceleration, so a=(v^2)/r
so G(m^2/r^2)=m(v^2/r)