# Why does Kepler's Third Law exist?

Martyn Arthur
ah...the equations are beyond me (for now at least) but I'm giving it a go.
Sorry if I am off track, but I am endeavoring..
My question was "is there a physical reason why all bodies rotating under gravity enjoy the same relationship between T and r".
So every planet, or indeed anybody rotating [indeed moving] with constant acceleration has constant momentum.
The relationship between T and r is thus a determinant of the relationship between the gravitational constant, the body's momentum and the body's distance from the Sun.
This relationship between T and r is a constant value which is obeyed by all bodies rotating with constant acceleration.
Kepler's equations, the third in particular, are a mathematical calculation 'proving' that the relationship is a consequence of the foregoing.
So the relationship is a "universal" constant but one which, unlike the speed of life, has an underlying mathematical explanation.
Thank, hopefully Martyn

Homework Helper
Gold Member
2022 Award
ah...the equations are beyond me (for now at least) but I'm giving it a go.
Sorry if I am off track, but I am endeavoring..
My question was "is there a physical reason why all bodies rotating under gravity enjoy the same relationship between T and r".
So every planet, or indeed anybody rotating [indeed moving] with constant acceleration has constant momentum.
The relationship between T and r is thus a determinant of the relationship between the gravitational constant, the body's momentum and the body's distance from the Sun.
This relationship between T and r is a constant value which is obeyed by all bodies rotating with constant acceleration.
Kepler's equations, the third in particular, are a mathematical calculation 'proving' that the relationship is a consequence of the foregoing.
So the relationship is a "universal" constant but one which, unlike the speed of life, has an underlying mathematical explanation.
Thank, hopefully Martyn
The gravitational force is proportional to mass, as is acceleration, according to Newton's laws of gravity and motion respectively. The motion of bodies under the force of gravity, whether parabolic projectile motion or elliptical orbits, is therefore independent of the mass of body.

That's quite fundamental. And I'm surprised if it has not been emphasised in the texts from which you are studying.

vanhees71 and topsquark
ah...the equations are beyond me (for now at least) but I'm giving it a go.
Sorry if I am off track, but I am endeavoring..
My question was "is there a physical reason why all bodies rotating under gravity enjoy the same relationship between T and r".
Did you even try to read the little bit of Landau & Lifschitz that I referenced? I guess not.

So every planet, or indeed anybody rotating [indeed moving] with constant acceleration has constant momentum.
False.

The relationship between T and r is thus a determinant of the relationship between the gravitational constant, the body's momentum and the body's distance from the Sun.
Not the body's momentum directly, but rather it's energy, i.e., ##m v^2 /2##.
Kepler's equations, the third in particular, are a mathematical calculation 'proving' that the relationship is a consequence of the foregoing.
So the relationship is a "universal" constant but one which, unlike the speed of life, has an underlying mathematical explanation.
Here's a cut down version of the scaling symmetry I tried to explain in my previous post.

Kinetic energy ##K## is a function of ##v^2##, which dimensions of ##L^2/T^2## (length squared over time squared). Gravitational potential energy ##V## is a function of ##1/R##, i.e., inverse length.

Now imagine that you rescale all lengths by a factor ##\alpha## and time by a factor ##\beta## (both are just +ve real numbers), i.e., $$r ~\to~ r' = \alpha r ~,~~~~ t ~\to~ t' = \beta t ~.$$ Under this rescaling, $$K ~\to~ K' = \alpha^2 K/\beta^2 ~,~~~~ \text{and}~~~ V ~\to~ V' = V/\alpha ~.$$ Question: what value must ##\beta## have such that ##K## and ##V## scale by the same overall factor. This is answered by putting $$\alpha^2 /\beta^2 = 1/\alpha ~,~~~~ \Rightarrow~~ \alpha^3 = \beta^2 ~.$$ That gives Kepler's 3rd law! Under these rescalings, $$T^2 ~\to~ \beta^2 T^2 ~,~~~~ R^3 ~\to~ \alpha^3 R^3 ~.$$ Multiplying the Lagrangian ##L = K-V## by a constant factor does not change the equations of motion. I.e., any spacetime rescaling such that ##\alpha^3 = \beta^2## must result in a new solution of the equations of motion, a different (but valid) orbit.

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topsquark and vanhees71
Staff Emeritus
ah...the equations are beyond me

"And he said, Go, and tell this people, Hear you indeed, but understand not; and see indeed, but perceive not."

You want an explanation of a mathematical fact without using mathematics. This is impossible. You don't need to be an expert in mathematics to understand youwon't get what you want.

topsquark and sophiecentaur
brainpushups
@Martyn Arthur, if you can find access to, or purchase for yourself, the book Physics, The Human Adventure by Horton and Brush you'll find a nice discussion of Newton's synthesis of Kepler's laws that uses only algebra (but only addresses circular orbits) in Chapter 11. Further, the discussion of the historical development is far more detailed than most texts and will offer some good insight into the nature of theory in physics which you may find beneficial. The text is written at the undergraduate level for non-science majors and is well worth a read.

vanhees71 and topsquark
You want an explanation of a mathematical fact without using mathematics. [...]
I think you might have jumped to an overly harsh assessment.This all depends on what level of math the OP can handle. After all, he marked the thread as I-level. We'll have to wait for him to give us a clearer indication of what level of math he can handle.

sophiecentaur and vanhees71
Staff Emeritus
Well, he's been rejectiong all the derivations and pointers to books.

topsquark, sophiecentaur and vanhees71
Gold Member
Well, he's been rejectiong all the derivations and pointers to books.
Yes; only the very occasional genius is in a position to define just how they're going to learn a subject. Entering the Tour de France requires the possession of a bicycle.

topsquark
Gold Member
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
Yes.

Gold Member
You want an explanation of a mathematical fact without using mathematics. This is impossible.
With respect, it's a fact of nature - it's worked since before math was invented.

Nature is the territory; math is only the map. I think the OP gropes for an explanation in terms of the territory, not the map.

For example, this is a "explanation" of orbital velocity that involves no math:

topsquark
Martyn Arthur
Thank you for the refernce to the book,

### Physics, the Human Adventure: From Copernicus to Einstein and Beyond Hardcover – 1 Mar. 2001​

I am studying with the OU and it doesn't have any copies, I would like to read it but its too costly on amazon (I am a pensioner :-( )
But this dialogue has hugely helped me to start venturing into a domain about which I knew nothing.
You guys have been so patient and helpful that I say thank you again.
Assume I have no maths, not too far from the truth, pending progression on my course.
Please just look at it at a basic level absolutely, and forget please all equations.
Gravity, it is either a consequence of the interaction between masses or the curvature of space.
Is it fair to say that this is still in the melting pot?
Kepler's law has identified a universal relationship of planets, with other planets and the Sun.
Is the reason by Kepler 3 for this relationship either;
a. demonstrated to be a relationship by reference to know facts of gravity etc
b. an undefined reason;
For example, without known data, the apparent 'coincidental' distance between the Moon and the Sun without physical explanation that facilitates the eclipses (that have enabled us to learn so very much).

PeroK
Gold Member
Please just look at it at a basic level absolutely, and forget please all equations.
If you stipulate that no equations can be used, all you can say is that two masses attract and that the attractive force relates to the separation (further apart = less). You cannot demand an 'explanation' with more depth. Your "basic level" needs to take you back to Science as it was at least a thousand years ago.
Kepler uses Maths so you can't appreciate those laws without maths.

The coincidental similarity (not equality) between the angles subtended by Sun and Moon is not 'significant'. There are many other planets and their moons which do not demonstrate such perfect eclipses. Of all the Jovian moons, can you think of one that would exhibit the same behaviour? If so, which one and what about the others?

vanhees71 and topsquark
Gold Member
MHB
Assume I have no maths, not too far from the truth, pending progression on my course.
Please just look at it at a basic level absolutely, and forget please all equations.
I think it's great that you are trying to stretch yourself and I sympathize that your level of Mathematics isn't all that advanced. But you need to understand that without Mathematics we can only talk about generalities. There is no way we can describe Kepler's third in any quantitative way because what it talks about depends on the Mathematics; Kepler's third law is an equation, after all.

-Dan

vanhees71
Staff Emeritus
Kepler's Third Law is an equation.

vanhees71, strangerep and topsquark
Gold Member
Kepler's Third Law is an equation.
The inverse square law of light is an equation too, but it can be intuited with a model and some geometry. The variables in the equations (equatia? equatiae?) represent real world phenomena.

vanhees71 and topsquark
Gold Member
Keep being curious. Keep asking questions.
That's easy to say but we all know that learning by just asking questions seldom gets us very far. That is not much further than pop Science on the TV which may be 'enough' to enjoy the edited bits of Science but it never allows one to make valid predictions or conclusions.

To know the appropriate questions to ask, you must first have learned a significant amount. Q and A is very high risk.

topsquark
Gold Member
MHB
That's easy to say but we all know that learning by just asking questions seldom gets us very far. That is not much further than pop Science on the TV which may be 'enough' to enjoy the edited bits of Science but it never allows one to make valid predictions or conclusions.

To know the appropriate questions to ask, you must first have learned a significant amount. Q and A is very high risk.
I agree. It sounds to me like he's at the beginning of all of this. A teacher in an organized course is the best thing for him. I guess I'm assuming he already knows that and would be taking a course at some point so I didn't bring that up.

-Dan

sophiecentaur
Staff Emeritus
The inverse square law of light is an equation too, but it can be intuited with a model and some geometry.
You mean $A = 4\pi r^2$? Looks like an equation to me!

sophiecentaur, Motore and strangerep
Gold Member
You mean $A = 4\pi r^2$? Looks like an equation to me!
It is.
But the phenomena itself can be intuited with the use of a visual aid, like so:

Sure there's technically a bit of simple math there, but 99% of the meaning of the equation is embodied in the geometry of the diagram.

sophiecentaur
Homework Helper
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2022 Award
I am studying with the OU ...
I strongly suspect the OU physics course will include some equations.

Gold Member
A teacher in an organized course is the best thing for him.
I think you could be being a bit more polite than I. I would suggest that "best' should be replaced with "essential" for whoever wants PF advice. It is mostly hard stuff and very little of it avoids Maths.

Even @DaveC426913 's diagram is 'maths' - just dressed up in a more friendly way.

topsquark
Gold Member
Even @DaveC426913 's diagram is 'maths' - just dressed up in a more friendly way.
Agree. I just think it's the level of maths the OP is looking for.

topsquark, sophiecentaur and PeroK
Martyn Arthur
I continue to be profoundly grateful for your patience of you all, the recent posting reassured me that I am not being regarded as a 'nuisance'.
May I please approach this from a different perceptive, forgetting Kepler and equations completely?
Separately [as a comparison] the relationship between the distances between Earth, the Sun, and the Moon are, without any 'apparent' reason such that eclipses are caused to occur.
It could then follow that the explanation for the two of the two minor bodies relative to the sun is a simple consequence of the compound gravitational forces acting in accord with the gravitational constant.
There is a gravitational constant, it is what it is, and we do not seek further.

Likewise is it the case that the defined relative orbiting situations, in a particular mode/sequence, are a physical result of the gravitational constant et al.

From my perspective, objectively / simply of trying to understand.
I am trying to pitch this in the most basic format. is the specific relationship between orbital relationships defined by science [then as defined by Kepler, not forgetting ever Newton, or is it just the way it is?
Martyn Arthur...thanks ongoing for the patience and understanding of you guys,
Martyn Arthur [agan]

Gold Member
It could then follow that the explanation for the two of the two minor bodies relative to the sun is a simple consequence of the compound gravitational forces acting in accord with the gravitational constant.
There is a gravitational constant, it is what it is, and we do not seek further.
That our Moon often eclipses the Sun is a matter of coincidence - there's no deeper factors*. There's lots of Moons, and they inhabit a wide range of distance from their primaries.

*(I mean, except for the usual limits on multiple body systems: a moon can't get too far from its primary or it won't have a stable orbit, and it can't get too close or it will disintegrate.)

topsquark
Gold Member
Separately [as a comparison] the relationship between the distances between Earth, the Sun, and the Moon are, without any 'apparent' reason such that eclipses are caused to occur.
You are seeing something significant here but is it really? Should there be an "apparent reason" why two sixes turn up together? There is a perfectly reasonable explanation of the way the Solar System was formed in a rough disc shape (do some Googling about it). That means that most of the objects orbit round the Sun near to its equatorial plane. Also, most of the planets rotate (and their moons) with axes much the same as the axis of those orbital planes. To get an eclipse, all that's necessary is for a large (or just large enough) object to pass between the Sun and another object. That will cause the object (or part of it) further away to be briefly in shadow. Solar Eclipses are very impressive because the Sun is a major part of our visual experience and the eclipse is total (but not always). Every so often, other objects go across the Sun (from our POV). Transits of Venus are quite common.

topsquark
Gold Member
is the specific relationship between orbital relationships defined by science [then as defined by Kepler, not forgetting ever Newton, or is it just the way it is?
Everything in our lives is "the way it is". Science attempts to apply models to help us predict what will happen or explain what has happened in terms of those models. That's all. You will appreciate that most of those models involve Maths and Maths is a lot more reliable than arm waving.
Hopefully you will get more familiar and competent with Maths or your personal picture of Science will be limited to stuff that Science Journalists provide.

topsquark
Martyn Arthur
erm... reference "the gravitational force is proportional to mass, as is acceleration, according to Newton's laws of gravity and motion respectively. The motion of bodies under the force of gravity, whether parabolic projectile motion or elliptical orbits, is therefore independent of the mass of the body."
Given the standard equation F = G(m1m2)/R^2 how can the mass of a body under the force of gravity be irrelevant?
An increase in the mass of a body orbiting the sun would change the relationship.

Gold Member
erm... reference "the gravitational force is proportional to mass, as is acceleration, according to Newton's laws of gravity and motion respectively. The motion of bodies under the force of gravity, whether parabolic projectile motion or elliptical orbits, is therefore independent of the mass of the body."
Given the standard equation F = G(m1m2)/R^2 how can the mass of a body under the force of gravity be irrelevant?
An increase in the mass of a body orbiting the sun would change the relationship.
You would normally read, along with this statement that it only applies for small bodies on much larger ones - as with artillery shells and the Earth or the Earth orbiting the Sun. It's valid for so many situations that people can be a bit sloppy about using the caveat.
Your version of the Force equation is more like the correct one. It's only correct when the bodies can be considered to be point masses.

Martyn Arthur
That's fair thanks, now reading well ahead into my course I read that the K3 relationship is defined as a consequence of the gravitational effect on orbiting bodies, which seems to tie in with my understanding that K3 demonstrates a factual situation, a relationship that is a consequence of the gravitational consequence of the masses of bodies orbiting, here the sun.
Please I just want to understand, is it the case that K3 demonstrates that the orbital periods of bodies orbiting the sun are a direct consequence of the relationship between the masses of those bodies and the mass of the sun.
Hence there is a rule that would extend to all bodies with an orbital relationship.
Hence if the mutual gravitational force is too weak the orbiting body exits the orbit, if too great it crashes into the central object, else it occupies an orbital period commensurate with the model prescribed by K3.

Gold Member
Sorry but you are demonstrating that , without the Maths as a model, one can drop into a black hole at any minute.
If you gave the Sun twice the mass, K3 would still apply - just with reduced orbital radii. The ‘equation’ is the only way to communicate something like this. You keep demonstrate this. Wait till you have the Maths and it will all be clear.

vanhees71 and topsquark
Martyn Arthur
Guys, I thought this forum was for anyone seeking to understand.
If you were giving an introductory talk to a class of 14-year-olds in a school how would you explain things, or having regard to a couple (just a couple of earlier comments) is this below this forum?
My undergraduate stating second-year course says the relationship is essentially due to the effect of gravitational forces why do you find it so impossible to quantify the core figures, or indeed are they not known?