Answer: 88 Earth days
Explanation:
According to the Kepler Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
In other words, this law states a relation between the orbital period [tex]T[/tex] of a body (moon, planet, satellite) orbiting a greater body in space with the size [tex]a[/tex] of its orbit:
[tex]T^{2}=a^{3}[/tex] (1)
If we assume the orbit is circular and apply Newton's law of motion and the Universal Law of Gravity we have:
[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (2)
Where [tex]M[/tex] is the mass of the massive object and [tex]G[/tex] is the universal gravitation constant. If we assume [tex]M[/tex] constant and larger enough to consider [tex]G[/tex] really small, we can write a general form of this law:
[tex]MT^{2}=a^{3}[/tex] (3)
Where [tex]T[/tex] is in units of Earth years, [tex]a[/tex] is in AU (1 Astronomical Unit is the average distane between the Earth and the Sun) and [tex]M[/tex] is the mass of the central object in units of the mass of the Sun.
This means when we are making calculations with planets in our solar system [tex]M=1[/tex].
Hnece, in the case of Mercury:
[tex](1)T^{2}=(0.39 AU)^{3}[/tex] (4)
Isolating [tex]T[/tex]:
[tex]T=\sqrt{(0.39 AU)^{3}}[/tex] (5)
[tex]T=0.243 Earth-years \frac{365 days}{1 Earth-year}=88.6 days \approx 88 days[/tex] (6)
This means the period of Mercury is 88 days.
