Let us assume that our galaxy consists of 2.5 × 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 1

Mass of our galaxy Milky Way, *M* = 2.5 × 10^{11} solar mass

Solar mass = Mass of Sun = 2.0 × 10^{36} kg

Mass of our galaxy, *M* = 2.5 × 10^{11} × 2 × 10^{36} = 5 ×10^{41} kg

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

Radius of Milky Way, *r* = 5 × 10^{4} ly

1 ly = 9.46 × 10^{15} m

∴*r* = 5 × 10^{4} × 9.46 × 10^{15}

= 4.73 ×10^{20} m

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

`T = ((4pi^2r^3)/(GM))^(1/2)`

= `((4xx(3.14)^2xx (4.73)^2 xx 10^(60))/(6.67xx10^(-11)xx5xx10^(41)))^(1/2) = ((39.48xx105.82xx10^(30))/33.35)^"1/2"`

= `(125.27 xx 10^(30))^(1/2) = 1.12 xx 10^(16) s`

1 year = `365 xx 324 xx 60 xx 60 s`

1s = `1/(365xx24xx60xx60)` year

`:.1.12xx10^(16) s = (1.12 xx 10^(16))/(365xx24xx60xx60)`

= `3.55 xx 10^8` year

#### Solution 2

Here, r = 50000 ly = 50000 x 9.46 x 10^{15} m = 4.73 x 10^{20} m

M = 2.5 x 10^{11} solar mass = 2.5 x 10^{11} x (2 x 10^{30}) kg = 5.0 x 10^{41}kg

We know that

`M =(4pi^2r^3)/(GT^2)`

or` T = ((4pi^2r^3)/"GM")^(1/2) = [(4xx(22/7)^2xx(4.73xx10^(20))^3)/(6.67xx10^(-11)xx(5.0xx10^(41)))]^(1/2)`

= `1.12 xx 10^(16)` s