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.
Answers
Answer:
Orbital period of I0 , TI0 = 1.769 days = 1.769 × 24 × 60 × 60 s
Orbital radius of I0 , RI0 = 4.22 × 108 m
Satellite I0 is revolving around the Jupiter
Mass of the latter is given by the relation:
MJ = 4π2RI03 / GTI02 .....(i)
Where,
MJ = Mass of Jupiter
G = Universal gravitational constant
Orbital period of the earth,
Te = 365.25 days = 365.25 × 24 × 60 × 60 s
Orbital radius of the Earth,
Re = 1 AU = 1.496 × 1011 m
Mass of sun is given as:
Ms = 4π2Re3 / GTe2 ......(ii)
∴ Ms / MJ = (4π2Re3 / GTe2) × (GTI02 / 4π2RI03) = (Re3 × TI02) / (RI03 × Te2)
Substituting the values, we get:
= (1.769 × 24 × 60 × 60 / 365.25 × 24 × 60 × 60)2 × (1.496 × 1011 / 4.22 × 108)3
= 1045.04
∴ Ms / MJ ~ 1000
Ms ~ 1000 × MJ
Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.
Answer:
3.55 × 108 years.
Explanation:
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 = ( 4π2r3 / GM)1/2
= [ (4 × 3.142 × 4.733 × 1060) / (6.67 × 10-11 × 5 × 1041) ]1/2
= (39.48 × 105.82 × 1030 / 33.35 )1/2
= 1.12 × 1016 s
1 year = 365 × 324 × 60 × 60 s
1s = 1 / (365 × 324 × 60 × 60) years
∴ 1.12 × 1016 s = 1.12 × 1016 / (365 × 24 × 60 × 60) = 3.55 × 108 years.
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