Hunting a black hole. Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has orbital speed v = 270 km/s, orbital period T = 1.6 days, and approximate mass m1 = 6.0Ms, where Ms is the Sun's mass, 1.99 x 1030 kg. Assume that the visible star and its companion star, which is dark and unseen, are both in circular orbits. Find the ratio of the approximate mass m2 of the dark star to Ms.
Please help.
To find the ratio of the approximate mass m2 of the dark star to the Sun's mass (Ms), we can use Kepler's Third Law and the formula for orbital velocity.
Kepler's Third Law states that the square of the orbital period T is proportional to the cube of the semi-major axis length a:
T^2 = (4π^2 / G) * (a^3 / (m1 + m2))
where G is the gravitational constant.
From the given information, we know the orbital period T = 1.6 days and the mass of the visible star m1 = 6.0 Ms.
We also know the orbital speed v = 270 km/s. For a circular orbit, the orbital speed is related to the gravitational force as:
v = √(GM / r)
where G is the gravitational constant, M is the total mass (m1 + m2), and r is the distance between the two stars.
Since the orbit is circular, the centripetal force is provided by the gravitational force:
GM / r^2 = m1 * v^2 / r
By substituting GM / r^2 with m1 * v^2 / r, we simplify the equation to:
GM = m1 * v^2
By rearranging the equation, we get:
m2 * G = (v^2 * T^2) / (4π^2) - m1 * G
Now, let's solve for the ratio of m2 to Ms:
m2 / Ms = ((v^2 * T^2) / (4π^2 * G)) - m1 / Ms
Substituting the values given into the equation:
m2 / Ms = ((270,000 m/s)^2 * (1.6 days)^2 / (4π^2 * (6.67430 × 10^-11 m^3/(kg * s^2)))) - 6.0
Simplifying the equation above will give you the desired ratio of m2 to Ms.
To find the ratio of the mass of the dark star (m2) to the mass of the Sun (Ms), we can use Kepler's Third Law, which states that the square of the orbital period of a planet (or star) is proportional to the cube of the semi-major axis of its orbit.
First, we need to find the semi-major axis of the orbit of the visible star. The semi-major axis (a) can be calculated using the formula:
a = (G * (m1 + m2) * T^2 / (4 * π^2))^(1/3)
Where:
G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2)
m1 is the mass of the visible star (given as 6.0 Ms)
m2 is the mass of the dark star (unknown)
T is the orbital period of the visible star (given as 1.6 days)
Let's plug in the values and calculate the semi-major axis (a):
a = (6.67430 × 10^-11 m^3 kg^-1 s^-2 * (6.0 * 1.99 x 10^30 kg + m2 * 1.99 x 10^30 kg) * (1.6 * 24 * 60 * 60 s)^2 / (4 * π^2))^(1/3)
Now, we also know that the visible star has an orbital speed (v) of 270 km/s. The orbital speed can be related to the semi-major axis using the formula:
v = (2 * π * a) / T
Let's rearrange the equation to solve for the semi-major axis (a):
a = (v * T) / (2 * π)
Now, we can plug in the values for orbital speed (v) and orbital period (T) to calculate the semi-major axis (a):
a = (270 km/s * 1.6 days) / (2 * π)
Now, we have calculated the values of the semi-major axis (a) using two different formulas.
Equating the two equations for semi-major axis, we'll have:
(6.67430 × 10^-11 m^3 kg^-1 s^-2 * (6.0 * 1.99 x 10^30 kg + m2 * 1.99 x 10^30 kg) * (1.6 * 24 * 60 * 60 s)^2 / (4 * π^2))^(1/3) = (270 km/s * 1.6 days) / (2 * π)
Now, we can solve this equation to find the value of m2. Once we have that, we can calculate the ratio of m2 to Ms.