Question

The distance from the Sun to the nearest star is about 4 ✕ 1016 m. The Milky Way galaxy is roughly a disk of diameter ~ 1021 m and thickness ~ 1019 m. Find the order of magnitude of the number of stars in the Milky Way. Assume the distance between the Sun and our nearest neighbor is typical.

Answer 1

Answer:

$\boxed{n=2.92\times 10^{10} \ stars}$

Explanation:

Order-of-magnitude estimates are useful when we want to get a very crude estimate of a quantity. Sometimes, to get the exact calculation of a problem is very difficult, or impossible, so we use order of magnitud in order to get a rough approximation. In this exercise, we need to find the order of magnitude of the number of stars in the Milky Way. We know:

The distance from the Sun to the nearest star is:

$4 \times 10^{16}m$

The Milky Way galaxy is roughly a disk of diameter:

$10^{21}$

The Milky Way galaxy is roughly a disk of thickness:

$10^{19}$

So we can approximate volume of the Milky Way:

$V=\pi r^2 h \\ \\ r=\frac{10^{21}}{2}m=5 \times 10^{20}m \\ \\ h=10^{19}m \\ \\ \\ V=\pi(5 \times 10^{20})^2(10^{19}) \\ \\ V=7.85\times 10^{60}m^3$

Now, let's estimate a rough density for the Milky Way. It is well known that in a sphere with a radius of  $4\times 10^{16}m$ there is a star, which is the sun. So the density of the Milky way is:

$\rho =\frac{n}{V} \\ \\ \rho:Density \ of \ Milky \ way \\ \\ n: Number \ of \ stars \\ \\ V:Volume$

For one star $n=1$ so we know the data at the neighborhood around the Sun, so the volume is a sphere:

$V=V_{sphere}=\frac{4}{3}\pi r^3 \\ \\ V_{sphere}=\frac{4}{3}\pi (4\times 10^{16})^3 \\ \\ V_{sphere}=2.68\times 10^{50}m^3 \\ \\ So: \\ \\ \rho=\frac{1}{2.68\times 10^{50}}=3.73\times 10^{-51}stars/m^3$

Finally, the numbers of stars can be found as:

$n=\rho V \\ \\ V:Volume \ of \ the \ Milky \ Way \\ \\ p:Density \ of \ the \ Milky \ Way \\ \\ n:Number \ of \ stars \\ \\ n=(3.73\times 10^{-51})(7.85\times 10^{60}) \\ \\ \boxed{n=2.92\times 10^{10} \ stars}$