Question

Let us assume that our galaxy consists of $2.5 \times {10^{11}}$ stars each one of one solar mass. How long will a star at a distance of $50000ly$ from the galactic centre take to complete one revolution? Take the diameter of the milky way to be ${10^5}ly$.

Answer 1

To solve this problem, we need to use the formula of time period of rotation. But before applying the formula, we will convert each parameter in SI units. Finally when we get the final answer, first we will have the time in seconds and we will convert it into years.

Formula used:
$T = {\left( {\dfrac{{4{\pi ^2}{r^3}}}{{GM}}} \right)^{\dfrac{1}{2}}}$,
where, $T$ is the time period for one revolution, $r$ is the radius of the Milky Way, $G$ is the gravitational constant and $M$ is the mass of the galaxy

Complete step by step answer:
Our first step is to find the required quantities in SI units to be used in the formula.Now, we will find the radius of the Milky Way in the SI unit.We are given that the diameter of the Milky Way is ${10^5}ly$.
We know that $1ly = 9.46 \times {10^{15}}m$
Therefore the diameter of the Milky Way $= 9.46 \times {10^{15}} \times {10^5}m = 9.46 \times {10^{20}}m$
Radius of the Milky Way $r = \dfrac{{9.46 \times {{10}^{20}}}}{2}m = 4.73 \times {10^{20}}m$
Now, we will find the mass of the galaxy.It is given that our galaxy consists of $2.5 \times {10^{11}}$stars each one of one solar mass.
We know that one solar mass $= 2 \times {10^{30}}kg$
Therefore, the mass of the galaxy $M = 2.5 \times {10^{11}} \times 2 \times {10^{30}} = 5 \times {10^{41}}kg$
Now, we will find the time taken by the star to complete one revolution by using these values.
T = {\left( {\dfrac{{4{\pi ^2}{r^3}}}{{GM}}} \right)^{\dfrac{1}{2}}} \\\ \Rightarrow T = {\left( {\dfrac{{4 \times {{3.14}^2} \times {{4.73}^3} \times {{10}^{60}}}}{{6.67 \times {{10}^{ - 11}} \times 5 \times {{10}^{41}}}}} \right)^{\dfrac{1}{2}}} \\\ \Rightarrow T = 1.12 \times {10^{16}}s \\\ \Rightarrow T = \dfrac{{1.12 \times {{10}^{16}}}}{{365 \times 24 \times 60 \times 60}}years\\\ \therefore T = 3.55 \times {10^8}years \\\
Hence, a star at a distance of $50000ly$ from the galactic centre takes $3.55 \times {10^8}$ years to complete one revolution.

Note: Here, we have seen that the distance of the star from the galactic centre does not affect the time taken by the star to complete one revolution. Another important thing to keep in mind while solving this type of problem is to convert the quantities in SI units. Finally, in this question, the time is obtained in seconds and then we have converted in years.