Answer: [tex]4.487(10)^{11}m[/tex]
Explanation:
This problem can be solved using the Third Kepler’s Law of Planetary motion:
“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
This law states a relation between the orbital period [tex]T[/tex] of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size [tex]a[/tex] of its orbit:
[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (1)
Where:
[tex]T=3.87Earth-years=122044320s[/tex] is the period of the orbit of the exoplanet (considering [tex]1Earth-year=365days[/tex])
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M=3.59(10)^{30}kg[/tex] is the mass of the star
[tex]a[/tex] is orbital radius of the orbit the exoplanet describes around its star.
Now, if we want to find the radius, we have to rewrite (1) as:
[tex]a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}}[/tex] (2)
[tex]a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}}[/tex] (3)
Finally:
[tex]a=4.487(10)^{11}m[/tex] This is the radius of the exoplanet's orbit
Given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.
Given the data in the question;
To determine the radius of the exoplanet's orbit, we use the equation from Kepler's Third Law:
[tex]T^2 = \frac{4\pi^2 }{GM}r^3\\[/tex]
Where, T is the period of the orbit of the exoplanet, G is the Gravitational Constant, M is the mass of the star and r is orbital radius.
We make "r", the subject of the formula
[tex]r = \sqrt[3]{\frac{T^2GM}{4\pi ^2} }[/tex]
We substitute our given values into the equation
[tex]r = \sqrt[3]{\frac{(122044320s)^2*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} } \\\\r = \sqrt[3]{\frac{(1.48948*10^{16}s^2)*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} }\\\\r = \sqrt[3]{\frac{3.5689*10^{36}m^3}{4*\pi ^2} }\\\\r = \sqrt[3]{9.04*10^{34}m^3}\\\\r = 4.488*10^{11}m[/tex]
Therefore, given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.
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