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T 2 = R 3
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.
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).
Mercury’s orbital period would then be (88/365.25) or .241 Earth years.
Since T2 = R3, then (.241)2 = R3
.058081 = R3
Therefore R = the cube root of .058081 or .3873 astronomical units or 35,980,000 miles or 57,890,000 kilometers.
You can use this example for practice with this calculator.
A more complex problem is at the end of this page.
This calculator has been rewritten so that units can be in seconds, hours, days, kilometers, miles, astronomical units or light years.
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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