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The above equation was formulated in 1619 by the German mathematician and
astronomer Johannes Kepler (1571-1630). It expresses the mathematical
relationship of all celestial orbits. Basically, it states that the square
of the time of one orbital period (T2) is equal to the cube of
its average orbital radius (R3).
Example 1) The planet Mercury orbits the Sun in 88 days. What is its average
distance from the Sun?
For problems involving orbits around the Sun, it is convenient to use Earth
as the standard, where Earth's year of 365.25 days =1 and Earth's average distance
from the Sun (92,900,000 miles) would also equal 1 (this distance is also known as
an astronomical unit).
A more complex problem is at the end of this page.
Example 2) Europa, a moon of Jupiter, orbits the planet in 3.5 days. What
is its radius of orbit?
In this case we cannot use Earth as the standard because the Sun is NOT at the
center of Europa's orbit. Therefore, we must choose another moon of Jupiter's
in order to have a standard. Io orbits Jupiter in 1.75 days with an
orbital radius of 421,800 kilometers. So, Europa takes twice as much time as Io
to orbit Jupiter, making Europa's period = 2. Europa's radius of orbit would
then be the cube root of of the time squared (22 = 4). The cube
root of 4 = 1.5874 but this is in terms of Io units. So, to convert this to
kilometers, we multiply by Io's radius (421,800) and get 670,000 kilometers.
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