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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 - Physics

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

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 8 Gravitation
Q 5 | Page 201
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