We can consider the Sun to be a satellite of the Milky Way. The Sun revolves around the center of the Milky Way with a radius of 8 kpc. The period of revolution is 230 million years. What is the mass of the Milky Way?
The gravitational force between the center of the Milky Way and the Sun is in the radial direction and its magnitude is given by the Newton’s equation
F = GMm/r2 (1)
where G is the gravitational constant, M and m are the masses of the Milky Way and the Sun respectively and r is the radius of the orbit.
In case of the circular motion, the net force equals mass times centripetal acceleration, where the centripetal acceleration can be calculated by ω2r, where ω is the angular rate of rotation also known as angular velocity.
F = mac = mω2r. (2)
The angular velocity is given by
ω = 2π/T, (3)
where T is the period of revolution.
We substitute (3) into the equation (2) and we get
F = m4π2r/T2. (4)
We can equate equations (4) and (1) to obtain
m4π2r/T2 = GMm/r2.
We divide the last equation by m and we get
4π2r/T2 = GM/r2. (5)
We can rewrite the equation (5) as
M = 4π2r3/(GT2) (6)
1 kpc (kiloparsec) equals 3.086×1019 m. So, 8 kiloparsecs equal to 2.469×1020 m.
1 year equals 3.154×107 s and 230 million years equal to 7.253×1015 s.
We substitute the values into the equation (6) and solve
M = 4π2×(2.469×1020)3/(6.67408×10−11×(7.253×1015)2) ≈ 1.69×1041 kg.
Image Credit: NASA/Adler/U. Chicago/Wesleyan/JPL-Caltech