If a hypothetical planet in our Solar System had a sidereal period of 11.1 Earth years and a circular orbit, how far from the Sun in Astronomical Units would it be? Enter your answer to the nearest 0.01 AU.
To determine the distance of the hypothetical planet from the Sun in Astronomical Units (AU), we can use Kepler's third law of planetary motion. This law states that the square of a planet's sidereal period (T) is proportional to the cube of its average distance from the Sun (r) raised to a constant power.
The formula for Kepler's third law is:
T^2 = k * r^3
Where T is the sidereal period of the planet, r is the distance from the Sun, and k is a constant.
First, let's convert the sidereal period of the planet from Earth years to Earth days since the standard unit for time used in Kepler's third law is days.
11.1 years * 365.25 days/year (taking into account leap years) = 4050.075 days.
Now, we can rearrange Kepler's third law to solve for r:
r = (T^2 / k)^(1/3)
To find the value of k, we can use the known values for Earth:
T = 365.25 days (Earth's sidereal period)
r = 1 AU (Earth's average distance from the Sun)
Plugging these values into the rearranged formula, we get:
1^2 = k * 1^3
k = 1
Now we can substitute the values for the hypothetical planet's sidereal period (T) and solve for r:
r = (4050.075^2 / 1)^(1/3)
r ≈ 34.49 AU
Therefore, if a hypothetical planet with a sidereal period of 11.1 years and a circular orbit existed in our Solar System, it would be approximately 34.49 AU away from the Sun.