Question

Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.87 Earth years in its circular orbit around its sun, which is a star with a measured mass of 3.59 x 1030 kg. Find the radius of the exoplanet's orbit.

Answer 1

Answer: $4.487(10)^{11}m$

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 $T$ of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size $a$ of its orbit:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)  

Where:

$T=3.87Earth-years=122044320s$ is the period of the orbit of the exoplanet (considering $1Earth-year=365days$)

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$  

$M=3.59(10)^{30}kg$ is the mass of the star

$a$ 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:

$a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}}$ (2)  

$a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}}$ (3)  

Finally:

$a=4.487(10)^{11}m$ This is the radius of the exoplanet's orbit