Question

let us assume that our galaxy consist of 2.5*10^11 stars each of one solar mass . how long will a star at a distance of 50000 light year from a the galactic center take to complete one revolution take at diameter of the milky way to be 10^5
light years

Answer 1

Explanation:

$\Large{\red{\underline{\underline{\sf{\blue{Given:}}}}}}$

$\sf{Number\:of\:stars\:in\:our\:galaxy\:=\:2.5\times 10^{11}}$

$\sf{Mass\:of\:each\:star\:=\:1\:solar\,mass\:=\:2.5\times 10^{30}kg}$

$\sf Mass\:of\:our\:galaxy\:=\:2.5\times 10^{11}\times 2\times 10^{36}kg\:=\:5\times 10^{41}kg$

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$\Large{\red{\underline{\underline{\sf{\green{To\:Find:}}}}}}$

$\sf Time\:taken\:for\:a\:star\:to\:complete\:one\:revolution$

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$\Large{\red{\underline{\underline{\sf{\pink{Solution:}}}}}}$

$\sf Diameter\:of\:milky\:way\:=\:10^5ly$

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$\sf \quad d\:=\:10^5\times 3\times 10^8{ms}^{-1}\times 1$

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$\sf The\:distance\:of\:the\:star\:=\:radius\:of\:it's\:orbit\:=\:5\times 10^4ly$

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$\textsf{therefore, the\:time\:period\:of\:the\:revolving\:is\:given\:by,}$

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$\sf \quad T^2\:=\: \dfrac{4\pi^2R_E^3}{GM_E}$

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$\sf \quad =\:\dfrac{4\pi^2}{GM_E}\times \dfrac{d^3}{8}$

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$\sf T^2\:=\: \dfrac{\pi^2\times (3\times 10^{13})^3\times 365\times 24\times 60\times 60}{2\times 6.67\times 10^{-11}\times 5\times 10^{41}}yr^2$

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$\sf T^2\:=\: \dfrac{27\pi^2\times 10^{39}\times 365\times 24\times 3600}{6.67\times 10^{31}}yr^2$

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$\sf T^2\:=\:1.259\times 10^{17}yr^2$

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$\boxed{\large{\sf\orange{T\:=\:3.549\times 10^8yrs}}}$

Answer 2

Answer:

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

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

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

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

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

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

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

Therefore,

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

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

___________________.....

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,

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

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

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

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

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

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

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

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

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

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

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