The planet Mercury orbits the sun in 88 days what its average distance from the sun? Kepler’s Third Law: T 2= P3
So T^2 is PROPRTIONAL to P^3
T^2/R^3 = 4 pi^2 / G M
where G = graviational constant 6.67408 x 10^-11 m^3/kg-s^2 M= sun Mass = 1.989 x 10^30kg
(88 days x 24 hours x 3600 sec/hr)^2 / P^3 = 2.973949 x 10^-19
P = 5.7927 x 10^10 m
= 57,927,719 km from Sun