# Why does Kepler's Third Law exist?

• I
• Martyn Arthur
In summary, Kepler's third law states that the planets' relative orbits are determined by the gravitational force between them and the Sun.
Martyn Arthur
Hi; I understand Kepler 3 in terms of the relationship that it demonstrates.
I have researched the internet but can't find a reason why this relationship exists.
Is it somehow a consequence of some type of gravitational balance, if not is there some other mechanical reason?
Thanks
Martyn

topsquark
Martyn Arthur said:
Hi; I understand Kepler 3 in terms of the relationship that it demonstrates.
I have researched the internet but can't find a reason why this relationship exists.
Is it somehow a consequence of some type of gravitational balance, if not is there some other mechanical reason?
Thanks
Martyn
https://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Third_law

topsquark
Martyn Arthur said:
I have researched the internet but can't find a reason why this relationship exists.
It's easy enough to prove for a circular orbit, which you just do by insisting that the gravitational force be exactly the necessary centripetal force. The mass of the secondary drops out, leaving you only the radius and period as variables for a given primary, so one dictates the other.

For non-cylindrical orbits it's a bit harder to prove. If you are comfortable with rotating frames it's easy enough to convince yourself that the eccentricity makes no difference to the period, I think. Otherwise I can offer a plausibility argument, which is basically the same as the last paragraph: the satellite mass must drop out and then you only have ##GM##, ##T##, ##r## and dimensionless constants to play with. Dimensional analysis will get you Kepler 3.

topsquark
Thank you, but if I understand correctly this is an analysis deriving, calculating and demonstrating the accuracy of Kepler 3. Please correct me if I am wrong as I am sure will be done.
However Is there any known reason specifically why the planets physically / actually have those relative orbits?
Martyn

Martyn Arthur said:
However Is there any known reason specifically why the planets physically / actually have those relative orbits?
Wikipedia said:
Using Newton's law of gravitation (published 1687), this relation can be found in the case of a circular orbit by setting the centripetal force equal to the gravitational force.

topsquark
Martyn Arthur said:
Thank you, but if I understand correctly this is an analysis deriving, calculating and demonstrating the accuracy of Kepler 3. Please correct me if I am wrong as I am sure will be done.
However Is there any known reason specifically why the planets physically / actually have those relative orbits?
Martyn
It should be clear why it's true for a circular orbit.

topsquark
Martyn Arthur said:
However Is there any known reason specifically why the planets physically / actually have those relative orbits?
That's exactly what a derivation tells you - derivations are the logical steps linking your assumptions to your result. Kepler 3 follows from the assumption that the Sun's gravity (with it's ##1/r^2## behaviour) is the only force on the planets. That's all the reason there is.

Kepler predates Newton, of course, so historically his laws were all observationally based with no justification beyond "they fit the data". Only after Newton could we see that they are all just aspects of Newtonian gravity.

vanhees71, dextercioby, russ_watters and 3 others
Martyn Arthur said:
However Is there any known reason specifically why the planets physically / actually have those relative orbits?
What exactly do you mean by "relative orbits".
Kepler arrived at his law just from observation and measurement. Newton's Laws of Gravity weren't actually around so Kepler was looking for some 'law' that God might have applied when fabricating the Solar System. That was the sort of approach that Scientists made in those days. He just juggled around with the figures for orbit times and periods of all the visible planets (Tycho Brahe) and looked for the simplest mathematical relationship. It's the sort of thing you'd do with a set of data when you have no idea about what the Physics is behind it. You try a straight line, you try a square law etc.etc. and go for the one with the shortest error bars.
The 'mechanics' behind Keppler's laws is basically Newtons laws of gravity but the actual numbers involve more than just the Sun and each planet on its own. The effect of the rest of the Solar System makes each planet depart from the 'simple' version of law that he came up with. The error is shown in the tables in that link.

vanhees71, Ibix and topsquark
Thanks again for your patience, sorry if I am being obscure. If I understand correctly Kepler found the mathematical relationship and that is what they equations laws demonstrate.
Does his law and equations demonstate why specifically there a physical reason why that relationship exists, rather than simply proving it?
Thanks again
Martyn

We seem to be talking past each other.

Perhaps you could say what kind of answer you want. It might be useful to explain why you are asking about Kepler's third law, but are apparently happy with his other two. If you took your original question and replaced 3 with 2, what answer would you give?

Maarten Havinga and topsquark
Ah! So with the second law, it defines the equality of the area swept out in different segments of the orbit and I can see that the gravity /angular momentum physically causes this, cause and effect.
With the third law, I understand the effect but I am seeking to understand the equivalent of the angular momentum, what is the cause of the positioning, eg for example is it the cumulative effect of gravity?
Thanks
Martyn

I'm not sure what you mean by "positioning". There's no position in Kepler's third law - it just says that if there's a body in orbit at radius ##r## from the Sun then its period, ##T##, will have the same value of ##T^2/r^3## as every other planet.

Sorry, bad stuff on my part, please substitute "what is the physical cause, causing a body in orbit at radius r from the Sun such that its period, T, will have the same value of T2/r3 as every other planet".
Is it just coincidence, as for example the separation of the sun from the moon facilitates eclipses? (Part of my assignment is to explain Kepler's laws).
Thanks
Martyn

Let's stick to circular orbits for now. At a given radius, there is only one speed that leads to a circular orbit. If you travel faster than that your orbit rises. If you travel slower than that, your orbit falls. The critical speed depends on the radius because (a) the strength of gravity changes with radius, and (b) the tightness of the curve changes with radius. If you work out the maths on all that, Kepler's third law drops out.

So no, it's not a coincidence. It's just that there is only one orbital speed for a circular orbit of a given radius around a fixed mass.

The maths for elliptical orbits is messier, but the argument is the same - there is only one period you can have for a given major axis, and if you have a different speed you end up in a different orbit.

sophiecentaur and topsquark
Ah so the relationship is a determinant of gravity and law 3 calculates the relationship in the discussion. I think I am right in saying then that it is purely the nature of gravity, how it functions, that is the cause of the relationship.
If so thank you very much for your patient help in this.
Thanks
Martyn

Martyn Arthur said:
Does his law and equations demonstate why specifically there a physical reason why that relationship exists, rather than simply proving it?
It was not derived from first principles of Physics; it was just an equation that agrees with observation.
However, Newton's Laws of gravitation predict his results with good agreement. I don't think you could say that 'simply proves it' as it just follows the same principle that applies to all of Physics. You make a model, you take observations and, if they follow the model, you accept the model. If other results come along, the model can be modified.

Can you suggest any other physical models that work any other way?

Martyn Arthur said:
Thanks again for your patience, sorry if I am being obscure. If I understand correctly Kepler found the mathematical relationship and that is what they equations laws demonstrate.
Does his law and equations demonstate why specifically there a physical reason why that relationship exists, rather than simply proving it?
Thanks again
Martyn
It's a mathematical relationship. It must ultimately depend on the mathematics of gravity and geometry. In a way that is why Newton marks the beginning of modern physics: because the mathematics determines the outcome.

sophiecentaur, topsquark and Ibix
Martyn Arthur said:
I think I am right in saying then that it is purely the nature of gravity, how it functions, that is the cause of the relationship.
Yes. What the derivation on Wikipedia (linked in #3) does is say: firstly, for an object to move in a circle of radius ##r## with orbital period ##T## something must supply a force equal to ##mr(2\pi/T)^2## pointing at the center of the circle. That is a general truth about circular motion, and the force can be anything - it may be tension in a piece of string with a ball on it whirled around your head or whatever. Secondly, gravity supplies a force ##GMm/r^2##. Thirdly, if an object is to move in a circle under the force of gravity then the force gravity supplies must be the same as the force needed for the planet to be traveling in a circle (or else the planet wouldn't be traveling in a circle). That is, this equation must hold: ##mr\left(\frac{2\pi}T\right)^2=\frac{GMm}{r^2}##. That ##r^3\propto T^2## follows as a matter of algebra.

So the only information in there is how gravity behaves and how things moving in circles behave. The rest is just algebra. Again, elliptical orbits are mathematically messier, but the same general argument applies.

DrClaude, topsquark, berkeman and 1 other person
You have been so very patient.
So planet a is subject to gravity x and forms a particular orbit subject to gravitational forces acting on that mass
Planet b has a different mass and forms a different orbit subject to gravitational forces acting on that different mass.
Kepler 3 has demonstrated the relationship between those two masses as defined in his law, and consequent upon the different gravitational forces acting.
This law then defines how other bodies' orbits occur in relative proportions consequent upon their location and the relative gravitational forces acting on them.
Thank you so very much.

No.

Mass is irrelevant, and a quick look at Kepler's Third Law shows that mass does not even appear.

topsquark
how can mass be irrelevant to the operation of gravity?

m divides out:

$$F = ma$$
$$G\frac{Mm}{r^2} = ma$$
$$a = G\frac{M}{r^2}$$

topsquark and berkeman
Martyn Arthur said:
how can mass be irrelevant to the operation of gravity?
Mass of the orbiting body.

We are dealing with masses where the smaller one is much smaller than the larger one (like Earth and a brick).

By observation, a brick falls as fast as a cinder block. The mass of the falling object is irrelevant to its rate of falling (as well as its orbital speed).

If we used a rock the mass of the Moon, we would have a different scenario.

topsquark
DaveC426913 said:
Mass of the orbiting body.
It doesn't have to be orbiting. Being dropped off the Leaning Power of Pisa works just fine.

topsquark
It doesn't have to be orbiting. Being dropped off the Leaning Power of Pisa works just fine.
Yes, as I subsequently imply.

Martyn Arthur said:
it is purely the nature of gravity, how it functions, that is the cause of the relationship.
Of course. I would say that is tautologous. "The nature" is a very old fashioned qualitative descriptions of things - like "nature abhors a vacuum".
From what you write, I conclude that you believe that somewhere there is ultimate truth and that we could actually achieve it. I just can't bring myself to take that view. Afaiac it's all a matter of continuous improvement of models and theories.

dextercioby and topsquark
Martyn Arthur said:
So planet a is subject to gravity x and forms a particular orbit subject to gravitational forces acting on that mass
Yes.
Martyn Arthur said:
Planet b has a different mass and forms a different orbit subject to gravitational forces acting on that different mass.
Well, the mass isn't really important here. The planets are where they are as an accident of history, not physical law. Kepler 3 says that any planet, transplanted to Jupiter's orbit with Jupiter's speed, would have the same orbital period Jupiter has, whether it's a gas giant or a dwarf planet, or a dinky little artificial satellite.
Martyn Arthur said:
how can mass be irrelevant to the operation of gravity?
Gravitational force is proportional to the mass of the planet, but it's the acceleration that dictates the path and acceleration is force divided by mass - so it's independent of the mass of the planet (still depends on the mass of the star, though). This is actually a key observation on the road to general relativity.

topsquark and PeroK
Ibix said:
The planets are where they are as an accident of history, not physical law.
That's probably a bit simplistic. The density and composition influences where they show up and also the existence of other planets around will affect their 'final' ( at least long term) orbit. The effect of Jupiter on the Asteroids is a more short term thing. But, in the end, everything affects everything else to some extent and the Solar System has settled down into the present arrangement after probably quite a bit of jostling around in the beginning. The forecast expansion of the Sun to include the orbits of planets out as far as Earth will certainly change things. The much reduced density of the Solar material will, I imagine, allow the rocky planets to exist intact for some while, even when they've been swallowed up. Kepler is based on point masses so the orbits will be then somewhat different and we'll probably be more like stars in a galaxy, where the central attractor is only part of the gravitational environment (all the other stars). No more simple elliptical orbits, for a start.

topsquark
sophiecentaur said:
That's probably a bit simplistic.
Probably. But the relevant point here is that Kepler's laws do nothing to predict the arrangement of the planets.

sophiecentaur and topsquark
That would come centuries later - requiring computers.

Ibix and topsquark
Ibix said:
Probably. But the relevant point here is that Kepler's laws do nothing to predict the arrangement of the planets.
This is a bit late to the party dude but I think I've sussed out the OP's real question. The way that the tables are presented of planets, their orbits and 'that coefficient' may look as if the relative positions of the planets is somehow related to Kepler's law. But this is putting the cart before the horse.
Fact is that you could take our Sun and a different set of planets and they would also fit on that straight(ish) line. So why isn't there a planet half way between Mars and Earth, sitting on that line? Such a body would / could destabilise the whole set up because of its interaction with, particularly, the nearest planetary orbits. The planets were, of course, formed from bands of dust and rock and gas, which separated out into massive spheres. The presence of massive Jupiter (so the theory goes) prevented all the rock in the Asteroid belt from ever merging together so an extra planet wouldn't have formed where the Asteroid belt sits.
It's all due to gravitational effects but it's a many-body problem and Kepler was only looking at two bodies at a time and making many assumptions.

Ibix and topsquark
@sophiecentaur I was about to make a similar post, but more simply: to the OP, do you understand that K3 applies to any other solar system as well?

K3 describes the relationship between period and radius but it does not determine either.

dextercioby, Ibix, topsquark and 2 others
Are we discussing Kepler's Law or Bode's Law?

Ibix and topsquark
Martyn Arthur said:
I have researched the internet but can't find a reason why [Kepler 3] exists.
Is it somehow a consequence of some type of gravitational balance, if not is there some other mechanical reason?
Since no one else has mentioned this...

A short answer is that K3L follows from a property of the Lagrangian known as Mechanical Similarity. The examples section in that Wiki page mentions Kepler 3, but not by that name. Landau & Lifschitz vol1 p22 has more detail.

In essence, if some of the variables in a Lagrangian are homogeneous (meaning that, e.g., for some constant (where might be different for different variables), there are cases where the homogeneities in the different variables can combine to result in merely multiplying the Lagrangian by a constant factor. This doesn't change the equations of motion, but does correspond to the existence of similarly shaped orbits of different size and energy. This symmetry doesn't commute with the Hamiltonian, hence is not associated with a conserved quantity.

One can therefore regard Kepler 3 as arising because of the ways that kinetic energy and gravitational potential energy scale under independent rescalings of space and time.

There are other cases where mechanical similarity is useful. L&L give examples.

Kepler 3 is interesting because it does not derive from an underlying conservation law, but rather this less well-known feature of scaling similarity.

sophiecentaur, vanhees71, DrClaude and 3 others

• Introductory Physics Homework Help
Replies
2
Views
467
• Introductory Physics Homework Help
Replies
7
Views
177
• Introductory Physics Homework Help
Replies
8
Views
1K
• Mechanics
Replies
4
Views
692
• Mechanics
Replies
117
Views
6K
• Astronomy and Astrophysics
Replies
1
Views
1K
• Mechanics
Replies
2
Views
1K
• Thermodynamics
Replies
2
Views
1K
• Mechanics
Replies
2
Views
853
• Astronomy and Astrophysics
Replies
3
Views
1K