The distance from the Sun to the nearest star is about 4 x 10^16 m. The Milky Way galaxy is roughly a disk of diameter ~10^21 m and thickness ~10^19 m. How do you find the order of magnitude of the number of stars in the Milky Way?

Oct 29, 2017

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:
$V = \pi \cdot {r}^{2} \cdot h$

Plugging in our numbers (and assuming that $\pi \approx 3$)
$V = \pi \cdot {\left({10}^{21} m\right)}^{2} \cdot \left({10}^{19} m\right)$
$V = 3 \times {10}^{61} {m}^{3}$
Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter ($\rho$) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of $4 \times {10}^{16}$m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

$\rho = \frac{n}{V}$

Using the volume of a sphere
$V = \frac{4}{3} \pi {r}^{3}$
$\rho = \frac{1}{\frac{4}{3} \pi {\left(4 \times {10}^{16} m\right)}^{3}}$
$\rho = \frac{1}{256} {10}^{- 48}$ stars / ${m}^{3}$

Going back to the density equation:
$\rho = \frac{n}{V}$
$n = \rho V$

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

$n = \left(\frac{1}{256} {10}^{- 48} {m}^{- 3}\right) \cdot \left(3 \times {10}^{61} {m}^{3}\right)$
$n = \frac{3}{256} {10}^{13}$
$n = 1 \times {10}^{11}$ stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.