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:

$k^{}=\frac{a^3}{T^2}$

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:

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

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