"If a small planet were located 8 times as far as the Earth's distance from the Sun, what would be its period in Earth years?" (22.5 earth years)
As a note, we are given a table with all of the planets' masses and radii.
http://hyperphysics.phy-astr.gsu.edu/HBASE/kepler.html
see the law of periods
To figure out the period of a planet in Earth years, we need to use Kepler's third law of planetary motion, which states that the square of a planet's orbital period is proportional to the cube of its average distance from the Sun.
In this case, we know that the planet is located 8 times as far from the Sun as the Earth's distance. Let's call the Earth's distance from the Sun "r". So, the planet's distance would be 8r.
To determine the planet's period, we need to use the ratio of the distances cubed. Since we have 8r for the planet and r for the Earth, we can set up the following equation:
(Period of planet) ^ 2 / (Period of Earth) ^ 2 = (Distance of planet) ^ 3 / (Distance of Earth) ^ 3
Let's substitute the values:
(Period of planet) ^ 2 / (1 year) ^ 2 = (8r) ^ 3 / r ^ 3
Simplifying the equation further:
(Period of planet) ^ 2 / 1 = (8 ^ 3 * r ^ 3) / r ^ 3
(Period of planet) ^ 2 = 8 ^ 3
(Period of planet) ^ 2 = 512
Period of planet = √512
Calculating the square root of 512 using a calculator gives us:
Period of planet = 22.63 years (rounded to two decimal places)
So, the period of the small planet located 8 times as far as the Earth's distance from the Sun is approximately 22.63 Earth years.