I'm asked to collect data on the orbital properties of the four Galilean satellites of Jupiter and show that they obey the same scaling as in Kepler's 3rd law.

My approach for moon Io:

Online, I found that the distance from Io to Jupiter is 422,000km ~= 0.00282089577 AU which is

$2.82\times10^{-3} AU$

Io's orbit around Jupiter is 1.77 Earth days; 1.77/365 ~= $ 4.85\times10^{-3}\, \mathrm{Earth\ years}$

By Kepler's 3rd law, $P^2(\mathrm{Earth\ years}) = a(AU)^3$


$(4.85\times10^{-3})^2 = (2.82\times10^{-3})^3$ which is obviously not true even just by looking at the powers of 10 after distributing the exponent on each side.

  • 7
    $\begingroup$ Reread the task: that they obey the same scaling as in Kepler’s third law. It's to be expected the moons don’t move at the same speed as the planets; that’s because they orbit a different body, with a different mass. Coincidentally, you can’t say anything about ratios by looking at only one moon. $\endgroup$
    – chirlu
    Oct 12, 2016 at 23:30
  • $\begingroup$ I guess I don't really understand what they mean by "the same scaling". Can you please expand on that? $\endgroup$ Oct 13, 2016 at 3:44
  • $\begingroup$ I also feel that because the Kepler's law that you're to be using is the "modified" or Newtonian version, $P^2=\frac{a^3}{(M+m)}$ or the full equation, $P^2=\frac{4\pi^2a^3}{G(M+m)}$. Either way I think you're simply looking for the 4-2-1 relation of the moon's orbits wrt the proportionate distances they are from Jupiter. $\endgroup$
    – LaserYeti
    Oct 13, 2016 at 5:13
  • $\begingroup$ How can you mess up with the 3rd Kepler law in such a way? You never thought to check the correctness of the equation before to make a post here? $\endgroup$
    – Py-ser
    Oct 13, 2016 at 13:42
  • $\begingroup$ @LaserYeti What about Jupiter's 4th moon though? It's not included in that 4-2-1 relation/ratio. $\endgroup$ Oct 13, 2016 at 14:35

3 Answers 3


Kepler's third law is that $R^3/P^2$ is a constant. However it is not a universal constant; it depends on the mass of the body that is being orbited. $$\frac{R^3}{P^2} \simeq \frac{GM}{4\pi^2},$$ where $M$ is the mass of the orbiting body (assuming that $M\gg$ the mass of the moons).

Depending on your level of sophistication, you could try plotting $R^3$ against $P^2$, or better, $\log R$ against $\log P$, to demonstrate consistency with Kepler's third law.


One must consider the mass of the object being orbited. Kepler's third law is true for all planets orbiting the Sun, and for all moons orbiting Jupiter, but not across different gravity wells. This was not understood until Newton, and must've posed an interesting problem (I've never thought of before) already in Kepler's time since the orbital periods and relative distances of the four Galilean moons were known then. And of course the orbit and relative distance of Earth's own moon. Maybe it was even a key inspiration for Newton's thinking about gravity?


This is what they are looking for. I tried it with numbers and it worked. Thank you for your input, you put me on the right track.

M1 + M2 = A3 / P2

Consistent units must be used to make this equation work. If the data are not given in a consistent system of units, they must be converted.

The masses must be measured in solar masses, where one solar mass is 1.99 X 1033 grams, or 1.99 X 1030 kilograms.

The semi-major axis must be measured in Astronomical Units, where 1 AU is 149,600,000 kilometers, or 93,000,000 miles.

The orbital period must be measured in years, where 1 year is 365.25 days.

  • 2
    $\begingroup$ The correctness of Kepler's third law does not depend on what units are used. $\endgroup$
    – ProfRob
    Oct 13, 2016 at 14:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .