# Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass.

Question:

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.

Solution:

Mass of our galaxy Milky Way, $M=2.5 \times 10^{11}$ solar mass

Solar mass $=$ Mass of Sun $=2.0 \times 10^{36} \mathrm{~kg}$

Mass of our galaxy, $M=2.5 \times 10^{11} \times 2 \times 10^{36}=5 \times 10^{41} \mathrm{~kg}$

Diameter of Milky Way, $d=10^{5}$ ly

Radius of Milky Way, $r=5 \times 10^{4}$ ly

$1 \mid y=9.46 \times 10^{15} \mathrm{~m}$

$\therefore r=5 \times 10^{4} \times 9.46 \times 10^{15}$

$=4.73 \times 10^{20} \mathrm{~m}$

Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:

$T=\left(\frac{4 \pi^{2} r^{3}}{\mathrm{GM}}\right)^{\frac{1}{2}}$

$=\left(\frac{4 \times(3.14)^{2} \times(4.73)^{3} \times 10^{60}}{6.67 \times 10^{-11} \times 5 \times 10^{41}}\right)^{\frac{1}{2}}=\left(\frac{39.48 \times 105.82 \times 10^{30}}{33.35}\right)^{\frac{1}{2}}$

$=\left(125.27 \times 10^{30}\right)^{\frac{1}{2}}=1.12 \times 10^{16} \mathrm{~s}$

1 year $=365 \times 324 \times 60 \times 60 \mathrm{~s}$

$\mathrm{Is}=\frac{1}{365 \times 24 \times 60 \times 60}$ years

$\therefore 1.12 \times 10^{16} \mathrm{~s}=\frac{1.12 \times 10^{16}}{365 \times 24 \times 60 \times 60}$

$=3.55 \times 10^{8}$ years