<p>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).</p>

<p>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.</p>

CC BY-SA 3.0
2016-10-13T13:41:03.490Z
2016-10-13T13:41:03.490Z
ProfRob
<p>One must <a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law" rel="nofollow">consider the mass</a> 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?</p>

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2016-10-13T04:56:12.783Z
2016-10-13T05:01:51.243Z
2016-10-13T05:01:51.243Z
LocalFluff
<p>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.</p>

<p>M1 + M2 = A3 / P2</p>

<p>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.</p>

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

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

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

CC BY-SA 3.0
2016-10-13T14:55:26.043Z
The correctness of Kepler's third law does not depend on what units are used.
ProfRob
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2016-10-13T14:49:57.250Z
2016-10-13T18:16:38.783Z
2016-10-13T18:16:38.783Z
ribarcheto94
<p>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. </p>

<p>My approach for moon Io:</p>

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

<p>$2.82\times10^{-3} AU$</p>

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

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

<p>Thus</p>

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

CC BY-SA 3.0
2016-10-12T23:30:09.433Z
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.
chirlu
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2016-10-13T03:44:19.693Z
I guess I don't really understand what they mean by "the same scaling". Can you please expand on that?
ribarcheto94
CC BY-SA 3.0
2016-10-13T05:13:21.230Z
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.
LaserYeti
CC BY-SA 3.0
2016-10-13T13:42:37.943Z
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?
Py-ser
CC BY-SA 3.0
2016-10-13T14:35:23.353Z
@LaserYeti What about Jupiter's 4th moon though? It's not included in that 4-2-1 relation/ratio.
ribarcheto94
CC BY-SA 3.0
2016-10-14T03:53:24.453Z
@ribarcheto94 I understand that I only listed three within the 4-2-1, the reason is that you will not usually find other satellites listed within the Jovian set that have a "nice" resonance pattern... the reality is that the 4-2-1 is a farce, and just for the intro astronomy students, so they can wrap their heads around the idea of orbital resonance. However if you wanted to continue the pattern, it would be something like **9-4-2-1**, none of these are exact, but they are fairly close so it should still work.
LaserYeti
<jupiter><kepler>
According to my calculations Jupiter's moons do not follow Kepler's 3rd law - Why is that?
ribarcheto94
# According to my calculations Jupiter's moons do not follow Kepler's 3rd law - Why is that?

(Asked by: username_0 on 2016-10-12)

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, $P^2(\mathrm{Earth\ years}) = a(AU)^3$

Thus

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

**Tags:** jupiter, kepler
> **Comments:**
> **username_1**: 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.  
> **username_0**: I guess I don't really understand what they mean by "the same scaling". Can you please expand on that?  
> **username_2**: 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.  
> **username_3**: 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?  
> **username_0**: @LaserYeti What about Jupiter's 4th moon though? It's not included in that 4-2-1 relation/ratio.  
> **username_2**: @ribarcheto94 I understand that I only listed three within the 4-2-1, the reason is that you will not usually find other satellites listed within the Jovian set that have a "nice" resonance pattern... the reality is that the 4-2-1 is a farce, and just for the intro astronomy students, so they can wrap their heads around the idea of orbital resonance. However if you wanted to continue the pattern, it would be something like **9-4-2-1**, none of these are exact, but they are fairly close so it should still work.  

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## 3 answers

### Answer by username_4 on 2016-10-13. Score: 10

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.


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### Answer by username_5 on 2016-10-13. Score: 2

One must [consider the mass](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law) 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?


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### Answer by username_0 on 2016-10-13. Score: 1

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.

> **Comments:**
> **username_4**: The correctness of Kepler's third law does not depend on what units are used.  

---

