Answer:
T = 7.61*10^6 s
Explanation:
In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:
[tex]T=\sqrt{\frac{4\pi^2r^3}{GM_s}}[/tex] (1)
T: period of Mercury
r: distance between Mercury and Sun
Ms: mass of the sun = 1.98*10^30 kg
G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2
You replace the values of all parameters in the equation (1):
[tex]T=\sqrt{\frac{4\pi^2(5.79*10^{10}m)^3}{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}}\\\\T=7.61*10^6s*\frac{1h}{3600s}*\frac{1d}{24h}=88.13\ days[/tex]
The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days
