Question

Mercury has an average disease to the sun of 0.39 AU. In two or more complete sentences, explain how to calculate the orbital period of mercury and then calculate it.

Answer 1

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 $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit:

$T^{2}=a^{3}$ (1)

If we assume the orbit is circular and apply Newton's law of motion and the Universal Law of Gravity we have:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (2)

Where $M$ is the mass of the massive object and $G$ is the universal gravitation constant. If we assume $M$ constant and larger enough to consider $G$  really small, we can write a general form of this law:

$MT^{2}=a^{3}$ (3)

Where $T$ is in units of Earth years,  $a$ is in AU (1 Astronomical Unit is the average distane between the Earth and the Sun) and  $M$ 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  $M=1$.

Hnece, in the case of Mercury:

$(1)T^{2}=(0.39 AU)^{3}$ (4)

Isolating $T$:

$T=\sqrt{(0.39 AU)^{3}}$ (5)

$T=0.243 Earth-years \frac{365 days}{1 Earth-year}=88.6 days \approx 88 days$ (6)

This means the period of Mercury is 88 days.