Orbital period of mercury can be calculated using the third law of Kepler which states that period squared equals to distance cubed, that is, (T1/T2)^2 = (a1/a2)^3 Where T1 is the orbital period of mercury, T2 is the earth orbital period and is equal to 1 year, that is, 365.25 days. a1 is the mercury axis and is equal to 0.39AU while a2 is the earth axis and is equal to 1AU. T^2 = (0.39)^3 = 0.059319 Therefore, T1 = 88 days.