A comet has an orbital period of 96 years. What is its average distance (in AU) from the Sun?

Answer 1

To find the answer, we look to Keplars Third Law of Planetary Motion.

If:

T = Orbital period in Earth years (96)

a = Average distance from the sun (semi-major axis)

The third law of planetary motion is described with:

T² = a³

^{In English, this states that the square of the orbital period (T) is proportional (==) cube of the average distance from the sun (a). We interpret this in astronomical units (AU).}

Now, if T² = a³ the opposite is naturally true as well ... a³ = T²

Now that things are defined, and the equation is laid out to give us the right answer, we can plug in variables, rearrange, and simplify.

Lets get that cube root on the (a) variable out of there by doing identical operations to both sides of the equation.

I don't know how to denote "cube roots" on here, so instead, I'll use ^ (1/3) to denote the cube root.

and with the exponent rule

a= 96^{3/2 (that's a two-thirds fraction: 96 to the power of two-thirds)}

^{Which comes out to approximately 20.97AU as}