Thank you for visiting A star orbiting a supermassive black hole has a semimajor axis of 1 78 10 16 meters and a semiminor axis of 1 70 10. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
The orbital-period of the stars in years is given as [tex] 5.21 . 10^{11} [/tex]. The unit of the orbital-period is generally represented as earth-years.
The orbital period, which is also known as the revolution period, is the amount of time a given astronomical object takes to complete one orbit around another object. For example, Earth completes its orbital period around the Sun every 365 days.
From Kepler’s third law of Planetary motion, we know that the square of the orbital period [tex] T [/tex], is proportional to the cube of the semi-major axis [tex] a [/tex] of the orbit. For finding the orbital period in years, first convert the given units to standard units.
We can use the given conversions,
1 solar mass ([tex] M⊙ = 1.9885 . 10^{30} kg [/tex])
From the question, we have,
Semi-major axis, [tex] a = 1.78 . 10^{16} m [/tex]
Semi-minor axis, [tex] b = 1.70 . 10^{16} m [/tex]
Mass of supermassive black hole, [tex] M = 6.20 . 10^{6} M⊙ [/tex] (solar masses)
The total mass of the supermassive black hole in kilograms is obtained as,
Mass, [tex] m = M . M⊙ [/tex]
[tex] \Rightarrow m = (6.20 . 10^{6}) . (1.9885 . 10^{30}) [/tex]
[tex] \Rightarrow m = 1.23 . 10^{37} kg [/tex]
The orbital period using Kepler’s third law is obtained as,
[tex] T^{2} = \frac{4\Pi^{2}}{Gm} a^{3} [/tex]
Here, G is the Gravitational constant, which is equal to [tex] 6.67430 . 10^{-11} m^{3}kg^{-1}s^{-2} [/tex]
Substituting all the values in the equation of orbital period using Kepler’s third law,
[tex] T^{2} = \frac{4\Pi^{2}}{(6.67430 . 10^{-11}) . (1.23 . 10^{37})} (1.78 . 10^{16})^{3} [/tex]
[tex] \Rightarrow T^{2} = 2.71 . 10^{23} [/tex]
[tex] \Therefore T = \sqrt{2.71 . 10^{23}} earth-years [/tex]
[tex] \Rightarrow T = 5.21 . 10^{11} earth-years [/tex]
For more details on the orbital period:
brainly.com/question/31543880
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