__Question 8.5__:
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__:*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 = ( 4π

^{2}r

^{3}/ GM)

^{1/2}

= [ (4 X 3.14

^{2}X 4.73

^{3}X 10

^{60}) / (6.67 X 10

^{-11}X 5 X 10

^{41}) ]

^{1/2}

= (39.48 X 105.82 X 10

^{30}/ 33.35 )

^{1/2}

= 1.12 X 10

^{16}s

1 year = 365 X 324 X 60 X 60 s

1s = 1 / (365 X 324 X 60 X 60) years

∴ 1.12 X 10

^{16}s = 1.12 X 10

^{16}/ (365 X 24 X 60 X 60) = 3.55 X 10

^{8}years

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