# 15.6 Galaxy Mass

Is it possible to weigh a galaxy? In fact, it is possible to measure the mass of a galaxy using the motion of one star within it! Kepler’s law can be used to measure the masses of stars in binary orbits. With mass in solar units, a in AU, and P in years, the equation is:

P^{2} = a^{3} / (M_{A} + M_{B})

Now take M_{A} to be the mass of the galaxy within the Sun’s orbit, and M_{B} to be the mass of the Sun. M_{A} is much, much greater than M_{B}, therefore we can replace the Sun by the single number M_{G} (mass of the galaxy). Consider *a*, the semi-major axis, to be *r*, the radius of the Sun's orbit around the galactic center. (Newton showed that the Sun only responds to mass inside its orbit, so this calculation cannot give us the mass outside the Sun's orbit.) Finally, inserting numerical constants into the equation, we have:

P^{2} = 4π^{2}r^{3} / G M_{G}

In a circular orbit, the period is given by the circumference of the orbit divided by the orbital velocity, or P = 2πr/v. Substituting this expression into the equation for Kepler's third law, and rearranging, gives the new result:

M_{G} = rv^{2}/G

G is the gravitational constant, with a value of 6.67 × 10^{-11} in units of Newton × m^{2}/kg^{2}. Now we can insert the known values of r and v, being careful to use units of meters for r and km/s for v. The orbital velocity of the Sun is v = 225 km/s, or 2.25 × 10^{5} m/s. The distance to the galactic center is r = 8500 pc, or 8500 × 3 × 10^{16} m. The result is M_{G} = 2.55 × 10^{20} x (2.25 × 10^{5})^{2} / 6.67 × 10^{-11} = 1.9 × 10^{41} kg. One solar mass is 2 × 10^{30} kg, so the total mass interior to the Sun's orbit is 1.9 × 10^{41} / 2 × 10^{30} ≈ 10^{11} solar masses. This enormous number reflects only the material between our orbit and the galactic center so it is obviously a lower bound on the total mass of our galaxy.

Now we can also see why the flat rotation curve of the Milky Way implies a steadily increasing mass in the outer regions of the galaxy. The equation written above can be used to calculate the mass within any circular orbit in the disk. A flat rotation curve means that the velocity does not increase or decrease with radius. In effect, v is a constant. G is also a constant, so we deduce that:

The Milky Way. Click here for original source URL.

Schematic rotation curve of a galaxy. Click here for original source URL.

M_{G} ∝ r

This result is remarkable. In the case of the solar system, the Keplerian orbits of the planets are responding only to the Sun. When the mass is all contained in the center of the system, the equation above shows that v ∝ 1/πr. The orbital speeds of the planets reduce with distance from the Sun — a declining rotation curve. In our galaxy, the rotation curve is flat out to a large radius. So the mass of the galaxy continues to increase out to the largest radii that can be measured.