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Question

Physics Question on Gravitation

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.

Answer

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
7 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=(4π2r3GM)12T = (\frac{4π^2 \,r^3 }{ GM})^{\frac{1}{2}}
=(4×(3.14)2×(4.73)3×10606.67×1011×5×1041)12=39.48×105.82×103033.35)12= (\frac{4 × (3.14)^2 × (4.73)^3 × 10^{60} }{6.67 ×10 ^{-11 }× 5 × 10^{41}}) ^{\frac{1}{2}} =\frac{ 39.48 × 105.82 ×10^{30} }{33.35 }) ^{\frac{1}{2}}

=(125.27×1030)12=1.12×1016=(125.27 × 10^{30} )^{\frac{1}{2}} = 1.12 × 10^{16} s
1 year = 365 x 324 x 60x60 s
1s=1365×324×60×601s = \frac{1}{365 ×324 × 60×60} year
1.12×1016s=1.12×1016365×324×60×60∴ 1.12 × 10^{16}s = \frac{1.12 × 10^{16} }{365 × 324 × 60×60}
=3.55×108= 3.55× 10 ^8 years