oAstronomers discover an exoplanet, a planet orbiting a star other than the Sun, that has an orbital period of 1.50 Earth years circular orbit around its star, which has a measured mass of $3.30 \times 10^{30} \mathrm{~kg}$. Determine the radius $r$ of the exoplanet's orbit. \[ r= \]

Step 1 ：Astronomers discover an exoplanet, a planet orbiting a star other than the Sun, that has an orbital period of 1.50 Earth years circular orbit around its star, which has a measured mass of \(3.30 \times 10^{30} \mathrm{~kg}\). We are asked to determine the radius \(r\) of the exoplanet's orbit.

Step 2 ：We can use Kepler's third law of planetary motion, which states that the square of the period of a planet's orbit is proportional to the cube of the semi-major axis of its orbit. In the case of a circular orbit, the semi-major axis is equal to the radius of the orbit. The formula for Kepler's third law is: \[T^2 = \frac{4\pi^2}{G M} r^3\] where: \(T\) is the period of the orbit, \(G\) is the gravitational constant, \(M\) is the mass of the central body (in this case, the star), and \(r\) is the radius of the orbit.

Step 3 ：We can rearrange this formula to solve for \(r\): \[r = \left( \frac{T^2 G M}{4\pi^2} \right)^{1/3}\]

Step 4 ：Given that \(T = 1.50\) Earth years, \(G = 6.674 \times 10^{-11} \mathrm{~m^3~kg^{-1}~s^{-2}}\), and \(M = 3.30 \times 10^{30} \mathrm{~kg}\), we can substitute these values into the formula to find \(r\).

Step 5 ：The calculated radius of the exoplanet's orbit is approximately \(2.32 \times 10^{11}\) meters. This is a reasonable result, as it is within the range of distances that planets in our own solar system orbit the Sun. For example, the Earth orbits the Sun at an average distance of about \(1.50 \times 10^{11}\) meters.

Step 6 ：Final Answer: The radius of the exoplanet's orbit is approximately \(\boxed{2.32 \times 10^{11}}\) meters.