Kepler's Third Law:
The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
* T² ∝ a³
Where:
* T is the orbital period (in years)
* a is the semi-major axis (in Astronomical Units, AU)
Calculations:
1. Given: The semi-major axis (a) = 4 AU
2. Find: The orbital period (T)
Using Kepler's Third Law:
* T² = a³
* T² = 4³
* T² = 64
* T = √64
* T = 8 years
Therefore, the period of revolution of the hypothetical planet would be 8 years.