Calculation of the Year of Mercury

Main Radius of Earth's Orbit: 1.5 x 10^11 km
Main Radius of Mercury's Orbit: 6 x 10^10 km

To calculate the year of Mercury, we can use Kepler's third law of planetary motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit.

Step 1: Convert the radii into meters:
Main radius of Earth's orbit = 1.5 x 10^11 km = 1.5 x 10^14 m
Main radius of Mercury's orbit = 6 x 10^10 km = 6 x 10^13 m

Step 2: Calculate the ratio of the semi-major axes:
Ratio of semi-major axes = (Main radius of Mercury's orbit) / (Main radius of Earth's orbit)
= (6 x 10^13 m) / (1.5 x 10^14 m)
= 0.4

Step 3: Calculate the ratio of the orbital periods:
According to Kepler's third law, the ratio of the orbital periods is equal to the square root of the ratio of semi-major axes raised to the power of 3/2.
Ratio of orbital periods = sqrt((Ratio of semi-major axes)^(3/2))
= sqrt(0.4^(3/2))
= sqrt(0.4^1.5)
≈ sqrt(0.212)

Step 4: Calculate the year of Mercury:
The orbital period of Earth is approximately 365.25 days.
Year of Mercury = (Orbital period of Earth) / (Ratio of orbital periods)
= 365.25 / sqrt(0.212)
≈ 87.97 days

Therefore, the year of Mercury is approximately 87.97 days.

Explanation:
- Kepler's third law of planetary motion relates the orbital period and the semi-major axis of a planet's orbit.
- By comparing the radii of Earth and Mercury's orbits, we can determine the ratio of their semi-major axes.
- Using this ratio, we can calculate the ratio of their orbital periods.
- Finally, by dividing the orbital period of Earth by this ratio, we can find the year of Mercury in days.
- The year of Mercury is approximately 87.97 days, which means it takes about 88 Earth days for Mercury to complete one orbit around the Sun.