Kepler's 3rd law: ratios don't fit data

I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.

Kepler's third law gives the equation $$P^2 = a^3$$ where $$P$$ is the period of orbit and $$a$$ the distance.

I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.

By solving the equation for $$a$$, I get $$a = (P^2)^{1/3}$$.

When I plug in the numbers, they don't correspond.

So my questions are:

1. Are there unit standards I need for both $$P$$ and $$a$$? Currently $$P$$ is in minutes, $$a$$ in kilometres.
2. Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.
• hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6 – Kyle Kanos Apr 22 at 18:45
• The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units. – jacob1729 Apr 22 at 21:06
• BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth? – G. Smith Apr 23 at 0:49
• No satellite orbits earth at an altitude of 705 of any commonly used unit--that's either so far out it's lost or it's deep under the surface. Kepler's third law is talking about the distance between the centers, not the distance between the surfaces. – Loren Pechtel Apr 23 at 4:55
• Aqua has a 95 minute orbit revolution and is at 705 km altitude above earth’s surface. I have taken into account earth’s radius... – triple7 Apr 23 at 5:11

that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.

Instead, I'd be checking whether $$T^{2}/a^{3}$$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)

• Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period? – triple7 Apr 22 at 18:46
• @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body. – Jerry Schirmer Apr 22 at 18:59
• Yep, finally found it. Thanks – triple7 Apr 22 at 19:24
• @triple7 Would you be willing to make a post on Physics Meta describing which screen-reader you use and how it interacts with MathJax? We frequently encourage people to use MathJax exactly to improve accessibility, and if it's having the opposite effect, we'd like to improve that. – rob Apr 23 at 12:56
• @triple7 Go to this link for making new Meta questions and construct a question the same way you did here. Let us know which screenreader software you use (name, version, vendor) and other information about your setup (operating system, unusual hardware) that might be relevant. If you can tell whether the issue is any MathJax versus MathJax above a certain complexity, then good and bad examples would be helpful --- though I can imagine that would be hard to debug without help. Good tags would be "bug," "support," "mathjax." Hope to help. – rob Apr 24 at 2:00

The general form of Kepler's period law is $$T^2 = \frac{4\pi^2}{G(M+m)}a^3$$. Often, we make the simplifying assumption that $$M\gg m$$, so that $$M+m \approx M$$.

Kepler's period law only takes the form $$T^2 = a^3$$ (forgetting about the units) when you use certain quantities- in this case, $$M$$ being solar mass, $$T$$ being an Earth year, and $$a$$ being an astronomical unit.

Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!

• Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/ – triple7 Apr 23 at 4:29
• I don't have the rep to make a small edit, but \gg will render as $\gg$ while >> renders as the IMO much uglier $>>$. – a CVn Apr 23 at 12:03

Kepler's third law claims that $$p^2 \propto a^3$$. The equality sign you use is incorrect.

As others have pointed out, the equality sign is wrong. I wanted to write an answer to say why it has to be.

Suppose the equation $$P^2=a^3$$ (ie with the equality) did hold. I measure the orbital period of Jupiter (11.86 Earth Years) and the semi-major axis (5.204 AU), square one and cube the other and hey: it seems to work.

But then suppose that actually, I don't want to work in AU, so I measure the distance in km instead. I still measure the period in years because that's convenient enough. All of a sudden, the numerical value of the semi-major axis is much much larger because kilometres are much smaller than AU. So now the equality is broken.

The upshot is that you can't write equations that equate a left hand side measured in units of [Time]^2 with a right hand side measured in units of [Length]^3. Or in fact, any equation with different units on both sides - because if it does hold true for one example, then simply changing you length scale will make it stop being true.

PS: For planets, Kepler's 3rd law will hold with equality if you measure in AU and Earth years, which are reasonable sized quantities. For satellites you may find their orbital radius to be an unhelpfully small number if you measure them all in AU though. The foolproof option is to just use the full constant of proportionality $$4\pi^2 / (GM)$$, or you could pick one satellite at random and measure all other satellite parameters in units of that satellites "year" and orbital radius.

The equality sign is incorrect, in general. It's more generally a relationship of proportionality. Here's a good mnemonic for remembering that ratio.

$$m\frac{d^2\vec{r}}{dt^2}=\frac{-GMm}{r^3}\vec{r}$$ is Newton's law of gravitation.

Dividing by $$m$$ and moving everything to one side, we have:

$$\frac{d^2\vec{r}}{dt^2}+\frac{GM}{r^3}\vec{r}=0$$

Let's assume orbital motion is roughly simple harmonic. Then we have $$\omega=\frac{GM}{r_{avg}^3}$$ where $$\omega$$ is the angular velocity = $$2\pi/T$$ where $$T$$ is the orbital period.

Then we have:

$$\frac{4\pi^2}{T^2}=\frac{GM}{r_{avg}^3}$$

So:

$$\frac{r_{avg}^3}{T^2}=\frac{GM}{4\pi^2}$$

I stress that this is a ball park, albeit very accurate, mnemonic. The $$M$$ used is only approximately correct and is based on the assumption that the orbited body is much more massive than the object doing the orbiting.

It's much more accurate to say the Harmonic mean of the apses is equal to $$\frac{GMm^2}{L^2}$$ where $$G$$ is Newton's Gravitational Constant, $$M$$ is the mass of the orbited body, $$m$$ is the mass of the orbitting body and $$L$$ is the angular momentum. Again M is an approximation.