Answers: 3
Mass of the Planet; 
Radius of the exoplanet's orbit; 
![r = \sqrt[3]{\frac{T^2GM}{4\pi ^2} }](/tpl/images/0001/4737/54e5c.png)
![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](/tpl/images/0001/4737/7ef83.png)
of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size 
of its orbit:
(1) Â 
is the period of the orbit of the exoplanet (considering 
)
is the Gravitational Constant and its value is 
 
is the mass of the star
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}}](/tpl/images/0001/4737/23008.png)
(2) Â ![a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}}](/tpl/images/0001/4737/b232c.png)
(3) Â 
This is the radius of the exoplanet's orbit
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Astronomers discover an exoplanet (a planet of a star other than the sun) that has an orbital period...
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