
Ryan G. answered 02/12/20
Math, physics and astronomy tutor
Hi Phebe,
So this is a classic Kepler's Third Law problem, which relates the orbital period of a body to it's distance from the force center. Mathematically,
T2 = [4*pi2/(GM)] * a3
where T is the period, M is the mass of the sun, a is the distance from the sun to the the planet and G is the universal gravitational constant.
Let's let the subscripts e stand for Earth and z stand for Zoltron. In addition, let's let the quantity in brackets be K and note that it is the SAME for all planets! Kepler's law for both planets:
Te2 = K*ae3 for Earth
and
Tz2 = K*az3 for Zoltron
Dividing the equation for Zoltron by the equation for the Earth gives,
Tz2/Te2 = az3/ae3 (the constant K cancels)
We need to solve for Tz. Doing so gives,
Tz2 = Te2 * (az3/ae3)
We know that az = 16 AU but we also know that ae = 1 AU so that az = 16*ae . Also, Te = 1 (year)
Therefore, Tz2 = 12 * 163. Taking the square root of both sides gives, Tz = 64 years (Earth years).