## A closer look at Kepler’s third law

Posted: March 25, 2016 by oldbrew in Celestial Mechanics
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Kepler’s laws [credit: thesimplephysicist.com]

The Physics Classroom website says:
‘Kepler’s third law provides an accurate description of the period and distance for a planet’s orbits about the sun. Additionally, the same law that describes the T²/R³ ratio for the planets’ orbits about the sun also accurately describes the T²/R³ ratio for any satellite (whether a moon or a man-made satellite) about any planet. There is something much deeper to be found in this T²/R³ ratio – something that must relate to basic fundamental principles of motion.’

But is it really quite simple?

Take a few basic formulae for spheres:
Circumference = 2pi r
Area (flat surface of half sphere) = pi r²
Volume = (4pi r³) / 3

If we’re talking about planetary bodies:
Radius = distance from the Sun (aka ‘semi-major axis’)
Circumference = orbital distance (i.e. one ‘lap’ of the Sun)
Area = the area swept by one orbit (see graphic)
Volume = the volume of space needed to contain that orbit

The only variable is the radius. Pi and the numbers are constants so can be discarded for this exercise.
Therefore circumference varies as r, area varies as r², and volume varies as r³.
Variation = the difference between any two planetary bodies A and B.
So: volume variation / (circumference x area) variation = a constant (because r³ = r² x r).

For example, compare Venus and Earth:
Radius or SMA ratio = 0.723:1
Circumference ratio = 0.723²:1² = 0.522729:1
Volume ratio = 0.723³:1³ = 0.377933:1

Obviously: 0.377933 = 0.723 x 0.522729
Planetary data: http://nssdc.gsfc.nasa.gov/planetary/factsheet/

It also follows that:
(Circumference/area) ratio x orbit speed ratio = (orbit period ratio)²
[i.e. the ratio of any two planetary bodies]

So is it true that ‘There is something much deeper to be found in this T²/R³ ratio’?
Or is it quite straightforward?
= = =
Update, October 2020 : Miles Mathis theorises on the third law here.

Mathis: But to really understand the Third Law simply and intuitively, I find it best to ditch all that entirely and start over from scratch. We then get the cube straight from the volume equation. As you know, volume increases as the cube of the radius, so, conversely, the charge emitted by the Sun will decrease in power by the cube. You will say the orbit is 2D, but the interaction of the charge field with the planet is 3D. Both the Sun and the planet are 3D, and so is any path made by the planet in orbit, so the 3D volume equation must be used.

1. Curious George says:

Your argument is purely geometric. No physics involved at all. You even managed to leave the orbital period out completely. Or, you could make the same argument for squares and cubes instead of circles and spheres.

I can’t blame you; the General Relativity also reduced physics to geometry – but on deeper grounds indeed.

Kepler’s laws can be derived from Newton’s gravity law, see http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/KeplersLaws.htm

2. oldbrew says:

I did point out that:
(Circumference/area) ratio x orbit speed ratio = (orbit period ratio)²

Orbit speed falls off as the inverse square of the semi-major axis (= radius).

3. oldbrew says:

Kepler is in the news: ‘Astronomers see supernova shockwaves for the first time’

‘Scientists watched the earliest moments of two old stars exploding using the Kepler Space Telescope, named after Johannes Kepler, who discovered the last observed supernova in the Milky Way in 1604.’
http://www.telegraph.co.uk/news/science/space/12200630/Astronomers-see-supernova-shockwaves-for-the-first-time.html

4. Raghu Singh says:

Yes, Kepler’s third law of orbital motions apply, for instances, to the planets orbiting the sun and to the electron ”orbiting” the nucleus. Well, here is one more for physics.

Take an ideal one-dimensional rod of mass M constituted of atoms of mass δM (<< M) and charge E spaced equally by d. Such an atom, under the electrostatic forces of its neighboring atoms, oscillates between them with frequency v and displacement e (< d). Guess what! I get this relationship: ν**2 is inversely proportional to d**3. (I am sure you know that frequency times time-period = 1.)

5. Berényi Péter says:

The constancy of the T²/R³ ratio in the solar system is related to the inverse-square law of gravitation. Its value is 4π²/(G×M☉), where G is the gravitational constant (6.67408×10⁻¹¹ m³ kg⁻¹ s⁻²) and M☉ is the solar mass (1.98855×10³⁰ kg). In other systems M☉ is to be substituted by the mass of the dominant central body, whatever it happens to be.

However, if we dig deeper, the inverse-square law itself depends on several factors.

First of all space is three dimensional and we have no idea why is that so. If string theory or one of its variants is to be believed, in fact there are many more dimensions, just most of them got curled up in a complicated topology to a tiny itsy-bitsy size, some 10²⁰ times smaller than a proton in the first instant before inflation, while three of them expanded to enormous (cosmological) distances for no good reason, so they are directly observable. The topology of curled-up dimensions is unknown and it is unspecified by theory, but there is a vast number of possibilities, each one resulting in different physics.

Then space must be utterly flat for the inverse-square law to work. That’s not entirely true, of course, but around small bodies like the Sun it is a pretty good approximation, because gravity is extremely weak, so curvature of spacetime is negligible (surface of a sphere is proportional to the square of its radius).

Finally the gravitational force should not decay exponentially with distance as others do (e.g. the nuclear and weak forces). That’s because it is supposed to be mediated by a mass zero (spin 2) boson, called graviton, which can’t decay into anything else (although I do not see what prohibits graviton decay into a photon pair). However, it is only a theoretical construct, was never actually observed and probably never will be. A Jupiter mass detector close to a neutron star is expected to register a single event in a decade, but must be shielded against neutrinos and the size of shield required ensures the entire thing would collapse into a black hole fast. If gravitons do exist, we still do not have an idea why they are not coupled to the Higgs field, which is supposed to lend rest mass to others.

These are the known unknowns, but there may be unknown unknowns as well.

6. oldbrew says:

BP says: ‘Then space must be utterly flat for the inverse-square law to work.’

Not sure about that but solar systems and galaxies do seem to be based on a fairly flat disc.

[Andromeda galaxy]

WHY IS THE SOLAR SYSTEM FLAT?
http://www.universetoday.com/108355/why-is-the-solar-system-flat/

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