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. question 8.5 gravitation

Answer 1

Number of stars in our Galaxy (N) = 2.5 × 10¹¹
Mass of each stars = 2 × 10^30kg
So, mass of the stars of the galaxy(M) = 2.5× 10¹¹× 2× 10^30
= 5 × 10⁴¹ Kg
Radius of orbit of a star (r) = 50000 light-years

We know,
1 light years = 9.46 × 10^15 m
So, r = 50000×9.46×10^15 m
= 5 × 9.46 × 10^19 m

Centripital force = Gravitational force
mv²/r = GMm/r²
v² = GM/r
(2πr/T)² = GM/r [ v = 2πr/T
4π²r²/T² = GM/r
T² = 4π²r³/GM
Put the values of r , G and M
T = √{ 4×(3.14)²× (5×9.46×10^19)³/6.67×10^-11×5×10⁴¹}
= 111.93 × 10¹⁴ sec
= 111.93 × 10¹⁴/(365×24×3600) yr
= 3.55 × 10^8 yr

Answer 2

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

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.