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?

###### Solution.

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/r^{2} (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 ω^{2}r, where ω is the angular rate of rotation also known as angular velocity.

Thereby,

F = ma_{c} = mω^{2}r. (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 = m4π^{2}r/T^{2}. (4)

We can equate equations (4) and (1) to obtain

m4π^{2}r/T^{2} = GMm/r^{2}.

We divide the last equation by m and we get

4π^{2}r/T^{2} = GM/r^{2}. (5)

We can rewrite the equation (5) as

M = 4π^{2}r^{3}/(GT^{2}) (6)

1 kpc (kiloparsec) equals 3.086×10^{19} m. So, 8 kiloparsecs equal to 2.469×10^{20} m.

1 year equals 3.154×10^{7} s and 230 million years equal to 7.253×10^{15} s.

We substitute the values into the equation (6) and solve

M = 4π^{2}×(2.469×10^{20})^{3}/(6.67408×10^{−11}×(7.253×10^{15})^{2}) ≈ 1.69×10^{41} kg.

Image Credit: NASA/Adler/U. Chicago/Wesleyan/JPL-Caltech