Kepler’s 3 Law applies to any system of small ‘planets’ orbiting a massive ob je
ID: 2060972 • Letter: K
Question
Kepler’s 3 Law applies to any system of small ‘planets’ orbiting a massive ob ject. For example, it applies
to the moons of Jupiter, for which the law states
R3 = GMJupiter / 42 T^2
In this equation, T is the orbital period of a given moon and R is the mean distance of the moon from the
center of Jupiter. For moons executing elliptical orbits, it can be shown that the mean distance equals the
semi-major axis of the ellipse.
The Wikipedia website http://en.wikipedia.org/wiki/Moons of Jupiter tabulates all of the physical data for
the moons of Jupiter, including the orbital period in (earth) days and the semi-ma jor axis in km. "For each
of the six moons numbered 5 through 10 (Io through Leda), calculate the ratio R3/T 2 in m3/s2 . Then, from the average value of these six ratios, calculate the mass of Jupiter." How does your value compare with the value given in Appendix F of the textbook (p. A-8)?
Explanation / Answer
Io Semi Major Axis = 421,700 km and Period is 1.769137786 days
Europa Semi Major Axis = 671,034 km and Period is 3.551181041 days
Ganymede Semi major axis = 1,070,412 km and Period is 7.15455296 days
Callisto Semi major axis = 1,882,709 km and Period is 16.6890184 days
Themsito Semi major axis = 7,393,216 km and Period is 129.87 days
Leda Semi major axis = 11,187,781 km and Period is 241.75 days
Then we need to covert all Axis to meters and all days to seconds to put into R3/T2
For Io = (421700000)3/(152853.5047)2 = 3.210 X 1015 m3/s2
For Eurpoa = (671034000)3/(306822.0419)2 = 3.210 X 1015 m3/s2
For Ganymede = (1070412000)3/(618153.3757)2 = 3.210 X 1015 m3/s2
For Callisto = (1882709000)3/(1441931.19)2 = 3.210 X 1015 m3/s2
For Themisto = (7393216000)3/(11220768)2 = 3.210 X 1015 m3/s2
For Leda = (11187781000)3/(20887200)2 = 3.210 X 1015 m3/s2
Then, R3/T2= GMJ/ 42
(3.210 X 1015 m3/s2) = (6.67 X 10-11)(MJ)/(42)
Mass of Jupiter = 1.90 X 1027 kg
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