Using the Solar System reference data, if a small planet orbited the Sun at 5 times Jupiter’s distance, what would its orbital period in Earth years?

Reference Data: The Solar System

Reference Data:
The Solar System

Using the Solar System reference data, if a small planet orbited the Sun at 5 times Jupiter’s distan...

Answer

Using Kepler’s third law with distances in AU and periods in Earth years, $P^2=a^3$. Jupiter is about $5.2\,\text{AU}$ from the Sun, so $a=5\times 5.2=26\,\text{AU}$ and $P=\sqrt{26^3}=\sqrt{17576}\approx 132.6$ years. So the planet’s orbital period would be about $133$ Earth years.

Explanation

What the problem is asking

You are scaling an orbit outward from Jupiter and then finding how much longer the planet takes to go around the Sun. The key idea is that orbital period grows quickly with distance.

Use Kepler’s third law in Earth units

For objects orbiting the Sun, if you measure:

$a$ in astronomical units (AU)
$P$ in Earth years

then Kepler’s third law is: $$P^2=a^3$$

Convert “5 times Jupiter’s distance” into AU

Jupiter’s average distance from the Sun is about $5.2\,\text{AU}$.

So the new planet’s orbital radius is: $$a=5\times 5.2=26\,\text{AU}$$

Compute the orbital period

Plug into $P=\sqrt{a^3}$: $$P=\sqrt{26^3}=\sqrt{17576}\approx 132.6$$

So the period is about $133$ Earth years.

Quick reasonableness check

Neptune orbits at about $30\,\text{AU}$ and takes about $165$ years, so a $26\,\text{AU}$ orbit being around $133$ years makes sense.