Question

If Mercury's average orbital distance from the sun is 0.312 AU and k for our sun is 1.00 AU3/yr2, then what is Mercury's orbital period?

Answer 1

Take into account that by the Kepler's Third Law, you can use the following formula:

`k^{}=(a^3)/(T^2)`

where,

T: period of Mercury

a: average distance from Mercuty to Sun = 0.312AU

k: Kepler's constant = 1.00 AU^3/yr^2

Then, by replacing the previous values of the parameters, you obtain for T:

`T=\sqrt[]{(a^3)/(k)}=\sqrt[]{((0.312AU)^3)/(1.00(AU^3)/(yr^2))}\approx0.56yr`

Hence, the Mercury's orbital period is approximately 0.56 yr