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Question
Let us assume that our galaxy consists of 2.5 × 1011 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 105 ly
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Solution 1
Mass of our galaxy Milky Way, M = 2.5 × 1011 solar mass
Solar mass = Mass of Sun = 2.0 × 1036 kg
Mass of our galaxy, M = 2.5 × 1011 × 2 × 1036 = 5 ×1041 kg
Diameter of Milky Way, d = 105 ly
Radius of Milky Way, r = 5 × 104 ly
1 ly = 9.46 × 1015 m
∴r = 5 × 104 × 9.46 × 1015
= 4.73 ×1020 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 1015 m = 4.73 x 1020 m
M = 2.5 x 1011 solar mass = 2.5 x 1011 x (2 x 1030) kg = 5.0 x 1041kg
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
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