Medium
Let us assume that our galaxy consists of $2.5 \times 10^{11}$ stars, each of one solar mass. How long will a star at a distance of $50,000 \; ly$ from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be $10^{5} \; ly$.
(N/A) Mass of our galaxy, $M = 2.5 \times 10^{11} \times (2.0 \times 10^{30} \; kg) = 5 \times 10^{41} \; kg$.
Radius of the orbit, $r = 50,000 \; ly = 5 \times 10^{4} \times 9.46 \times 10^{15} \; m = 4.73 \times 10^{20} \; m$.
Using Kepler's Third Law for the time period $T$ of a star revolving around the galactic centre:
$T = \sqrt{\frac{4 \pi^{2} r^{3}}{G M}}$.
Substituting the values:
$T = \sqrt{\frac{4 \times (3.14)^{2} \times (4.73 \times 10^{20})^{3}}{6.67 \times 10^{-11} \times 5 \times 10^{41}}}$.
$T = \sqrt{\frac{39.48 \times 105.82 \times 10^{60}}{33.35 \times 10^{30}}} = \sqrt{125.27 \times 10^{30}} \approx 1.12 \times 10^{16} \; s$.
Converting seconds to years:
$T = \frac{1.12 \times 10^{16}}{365.25 \times 24 \times 3600} \approx 3.55 \times 10^{8} \; \text{years}$.
Radius of the orbit, $r = 50,000 \; ly = 5 \times 10^{4} \times 9.46 \times 10^{15} \; m = 4.73 \times 10^{20} \; m$.
Using Kepler's Third Law for the time period $T$ of a star revolving around the galactic centre:
$T = \sqrt{\frac{4 \pi^{2} r^{3}}{G M}}$.
Substituting the values:
$T = \sqrt{\frac{4 \times (3.14)^{2} \times (4.73 \times 10^{20})^{3}}{6.67 \times 10^{-11} \times 5 \times 10^{41}}}$.
$T = \sqrt{\frac{39.48 \times 105.82 \times 10^{60}}{33.35 \times 10^{30}}} = \sqrt{125.27 \times 10^{30}} \approx 1.12 \times 10^{16} \; s$.
Converting seconds to years:
$T = \frac{1.12 \times 10^{16}}{365.25 \times 24 \times 3600} \approx 3.55 \times 10^{8} \; \text{years}$.
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