According to Kepler\'s 3rd Law the square of the orbital period of a planet is p
ID: 2265524 • Letter: A
Question
According to Kepler's 3rd Law the square of the orbital period of a planet is proportional to the
cube of the semi-major axis. Verify this from the data in the
Planet Orbital Period (julian years) Semi-Major Axis (AU)
Mercury 0:2408467 0:38709893
Venus 0:61519726 0:72333199
Earth 1:0000174 1:00000011
Mars 1:8808476 1:52366231
Jupiter 11:862615 5:20336301
Saturn 29:447298 9:53707032
Uranus 84:016846 19:19126393
Neptune 164:79132 30:06896248
Here a Julian year is 365:25 days and an AU - Astronomical Unit, is 1:496 1011 m. Try the calculation first using the units of years and AU, and then change to meters and seconds.
Explanation / Answer
Where the period (T) is in years and
(R) is the object's mean distance from
the sun in this case measured in
astronomical (earth=1au) units, kepler's
law reduces to:
T^2/R^3 = 1
T = sqrt((5.07au)^3) = 11.416 Yr
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