Question

Question :

Let us assume that our galaxy consists of 2.5 x10^11 stars
each of one solar mass. How long will a star at a distance of 50000 light years from the galactic centre take to complete one revolution ? Take the diameter of the milky way to be 10 light years.

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Thanks in advance v_v​

Answer 1

Solution :

$\bullet \: \sf \: Numbers \: of \: stars \: in \: our \: galaxy \: (N) = 2.5 \: \times \: {10}^{11}$

$\bullet \: \sf \: Mass \: of \: each \: star \: (m) = One \: solar \: mass \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \qquad \: \: \: \: \: \: \: \: = \sf \: 2 \: \times \: 10^{30}$

$\bullet \: \sf \: Total \: mass \: of \: the \: stars \: in \: one \: galaxy = N \: \times \: m$

$: \implies \sf 2.5 \times {10}^{11} \times 2 \times {10}^{30}$

$: \implies\sf 5.0 \times {10}^{41} \: kg$

$\bullet \: \sf \: Radius \: of \: orbit \: of \: \: a \: star \: (r) = 50000 \: light \: years$

$\star \: \sf \: But \: 1 \: light \: year = 9.46 \times {10}^{5} \: m$

Therefore,

$\bigstar \: \: \sf r = 50000 \times 9.46 \times {10}^{15} \: m \: \: \bigstar$

$\bigstar \: \sf \: Diameter \: of \: milky \: way = {10}^{5} \: light \: years. \: \bigstar$

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The centripetal force required for orbital motion is obtained from the gravitational force.

Therefore,

$\dag \: \underline{ \boxed {\boxed{\bf \: Centripetal \: force = Gravitational \: force} }}\: \dag$

So,

$: \implies \sf \dfrac{m {v}^{2}}{r} = \dfrac{GMm}{ {r}^{2} }$

$: \implies \sf v^{2} = \dfrac{GM}{r}$

$: \implies \sf (r \omega)^{2} = \dfrac{GM}{r} \: \: \: \: \big \lgroup \bf \because \: v = r \omega \big \rgroup$

$: \implies \sf \omega^{2} = \dfrac{GM}{r^{3} }$

$: \implies \sf \bigg \{\dfrac{2\pi}{T} \bigg \} = \dfrac{GM}{r^{3} } \: \bigg \lgroup \bf {\because \: \omega = \dfrac{2\pi}{T} } \bigg \rgroup$

$: \implies \sf T^2 = \dfrac{ 4\pi^{2} r^{3}}{GM}$

$: \implies \sf T = \sqrt{ \dfrac{4 \times (3.14)^{2} \: \times \: (5 \times 9.46 \times 10^{15}) ^{3} }{6.67 \times 10^{ - 11} \times 5 \times {10}^{41} }}$

$: \implies \sf T = \sqrt{12527.5 \times {10}^{28} }$

$: \implies \sf T = 111.93 \times {10}^{14 } \: s$

$: \implies \sf T =\dfrac{ 111.93 \times {10}^{14 }}{365 \times 24 \times 3600} \: years$

$: \implies \underline{ \boxed{\sf T =3.55 \times {10}^{8} \: years}}$

Answer 2

Correct 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 50000 light years from the galactic center take to complete one revolution? Take the diameter of the milky way to be 10⁵ light

Given :-

Mass of Sombrero Galaxy = $\sf 2 \times 10^{11}$ solar mass

Diameter of Sombrero Galaxy = $\sf 5 \times 10^{5}\: ly$

To Find :-

Time consumed by the star at a distance of 50000 light years from the galactic center take to complete one revolution.

Solution :-

We know that,

• d = Diameter
• m = Mass
• r = Radius

Given that,

Mass of Sombrero Galaxy, (m) = $\sf 2 \times 10^{11}$ solar mass

Solar mass = Mass of Sun = $\sf 2.0 \times 10^{36}\: kg$

Mass of the galaxy, (m) = $\sf 2.5 \times 10^{11} \times 2 \times 10^{30} = 5 \times 10^{41}\: kg$

Diameter of Sombrero Galaxy = $\sf 5 \times 10^{5}\: ly$

According to the question,

1 light year = $\sf 9.46 \times 10^{15}\: m$

Therefore, r = $\sf 5 \times 10^{4} \times 9.46 \times 10^{15}$

Radius $\implies \sf 4.73 \times 10^{20}\: m$

As this star revolves around the massive black hole in center of the Sombrero galaxy, its time period can be found with the relation:

$\sf T= \bigg[\dfrac{4 \pi ^{2} r^{3}}{GM} \bigg]^{\dfrac{1}{2}}$

$\sf \sqrt{\dfrac{4 \times 3.14^{2} \times 4.73^{3} \times 10^{60}}{6.67 \times 10^{-11} \times 5 \times 10^{41}} }$

$\implies \sf 1.12 \times 10^{16}\ s$

Now we know, 1 year = 365 × 24 × 60 × 60 s

$\sf 1 \: s=\dfrac{1}{365 \times 24 \times 60 \times 60}$

Therefore,

$\sf 1.12 \times 10^{16}s = \dfrac{1.12 \times 10^{16}}{365\times 24\times 60\times 60} \ years$

$\implies \sf 3.55 \times 10^{8}\ years$

Therefore, it will take 3.55 × 10^8 years by the star at a distance of 50000 light years from the galactic center take to complete one revolution.