Mercury is an average distance of 5.79 x 107 km from the Sun. Estimate the length of the Mercurian year given that the Earth is 1.50 108 km from the Sun on the average.
An explanation of how to do this would be great, thanks.
Use Kepler's third law. There is a particuarly simple form of it
P^2 = a^3
if you measure the period (P) in years and the distance (a) in astronomical units (a.u.). The earth's distance from the sun is 1 au. Mercury's is 5.79*10^7/1.50*10^8 = 0.386 au
You will find the formula at
http://www.windows.ucar.edu/tour/link=/the_universe/uts/kepler3.html&edu=high
and probably also in your textbook.
To estimate the length of the Mercurian year, we can use the concept of orbital period. The orbital period is the time it takes for a planet to complete one revolution around the Sun.
The orbital period depends on the distance between the planet and the Sun. According to Kepler's Third Law of Planetary Motion, the square of the orbital period is directly proportional to the cube of the average distance between the planet and the Sun.
Mathematically, we can express this relationship as:
(T1^2 / T2^2) = (R1^3 / R2^3)
Where T1 and T2 are the orbital periods of two planets (in this case, Mercury and Earth), and R1 and R2 are the average distances between these planets and the Sun.
We can rearrange this equation to solve for the orbital period of Mercury (T1):
T1^2 = (T2^2 * R1^3) / R2^3
Given that the average distance between Earth and the Sun (R2) is 1.50 x 10^8 km and the average distance between Mercury and the Sun (R1) is 5.79 x 10^7 km, we can substitute these values into the equation:
T1^2 = (T2^2 * (5.79 x 10^7)^3) / (1.50 x 10^8)^3
Now, we can calculate the orbital period of Mercury:
T1 = sqrt((T2^2 * (5.79 x 10^7)^3) / (1.50 x 10^8)^3)
To find the value of T2 (orbital period of Earth), we can use the fact that the Earth's orbital period is approximately 365.25 days, or about 365.25 * 24 * 60 * 60 seconds. We can convert it to years by dividing by the number of seconds in a year.
Finally, we substitute the values into the equation and solve for the orbital period of Mercury:
T1 = sqrt((365.25^2 * (5.79 x 10^7)^3) / (1.50 x 10^8)^3)
Calculating the value, you should find the estimated length of the Mercurian year.
To estimate the length of the Mercurian year, we can use the fact that the orbital period of a planet is directly proportional to its average distance from the Sun.
The formula we can use is:
(orbital period of Mercury) / (orbital period of Earth) = √[(average distance of Mercury from the Sun) / (average distance of Earth from the Sun)]
Given that the average distance of Mercury from the Sun is 5.79 x 10^7 km and the average distance of Earth from the Sun is 1.50 x 10^8 km, we can substitute these values into the formula:
(orbital period of Mercury) / (orbital period of Earth) = √[(5.79 x 10^7) / (1.50 x 10^8)]
To solve for the orbital period of Mercury, we rearrange the equation:
(orbital period of Mercury) = (orbital period of Earth) * √[(5.79 x 10^7) / (1.50 x 10^8)]
Now we can substitute the known orbital period of Earth, which is approximately 365.25 days or 1 year, into the equation:
(orbital period of Mercury) = 1 year * √[(5.79 x 10^7) / (1.50 x 10^8)]
Calculating the square root and simplifying the expression:
(orbital period of Mercury) = 1 year * √(0.386)
(orbital period of Mercury) = 1 year * 0.6214
(orbital period of Mercury) ≈ 0.6214 years
Therefore, the estimated length of the Mercurian year is approximately 0.6214 years, or roughly 226.64 days.