Answer: [tex]T=7.59(10^{6})s[/tex]
According to the Third Kepler’s 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 (planet Mercury in this case) orbiting a greater body in space (the Sun) with the size [tex]r[/tex] of its orbit.
This Law is originally expressed as follows:
[tex]T^{2}=\frac{4\pi^{2}}{GM}r^{3}[/tex] (1)
Where;
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674x10^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M=1.99×10^{30}kg[/tex] is the mass of the Sun
[tex]r=5.79×10^{10}m is the distance from Mercury to the Sun (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
[tex]T=\sqrt{\frac{4\pi^{2}}{GM}r^{3}}[/tex] (2)
[tex]T=2\pi\sqrt{\frac{r^{3}}{GM} [/tex] (3)
[tex]T=2\pi\sqrt{\frac{(5.79(10^{10}m))^{3}}{(6.674x10^{-11}\frac{m^{3}}{kgs^{2}})(1.99×10^{30}kg)}[/tex] (4)
Solving and taking into account that [tex]1N=1kg\frac{m}{s^{2}}[/tex]:
[tex]T=2\pi\sqrt{1.46(10^{11})s^{2}}[/tex] (5)
Finally:
[tex]T=7.59(10^{6})s[/tex]>>>>This is the period of revolution for the planet Mercury
