A planet whose distance from the sun is 3 au would have an orbital period of how many earth-years?

The orbital period will be 5.196 earth years. The period of an orbit is proportional to the square root of the cube of the semi-major axis. The semi-major axis of earth is 1 AU (actually 1.000001018 AU, but for this problem 1 is close enough). The hypothetical planet will have a semi-major axis of 3 AU. So making the ratios with earth's year being 1, earth's semi-major axis being 1, the planet's year being y, and the planet's semi-major axis being 3, we get the following ratio. 1 / 1^ (3/2) = y / 3^ (3/2) Raise the denominators to the 3/2 powers. 1/1 = y/5.196 Multiply both sides by 5.196 5.196 = y So the orbit period will be 5.196 earth years.