According to Kepler's third law, how many times larger is Neptune's average radius of orbit if it takes 164 Earth-years to orbit the Sun?

To understand how to determine how many times larger Neptune's average radius of orbit is compared to Earth’s, we can apply Kepler's third law of planetary motion, which states that the square of the period of orbit (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit. This relationship can be mathematically expressed as:

[ T^2 \propto a^3 ]

In a more practical context, if we take Earth’s orbital period to be 1 year and its average radius of orbit as 1 astronomical unit (AU), we can compare it with Neptune.

Given that Neptune takes 164 Earth-years to make one orbit around the Sun, you can set up the ratio using Kepler's third law:

1. Let ( T_{Neptune} = 164 ) years

2. The formula, when simplified, gives:

[ \left( \frac{T_{Neptune}}{T_{Earth}} \right)^2 = \left( \frac{a_{Neptune}}{a_{Earth}} \right)^3 ]

Substituting Earth’s period, we find:

[ \left( \frac{164}{1} \right)^