Talk:Kepler's laws of planetary motion/Archive 1

proofs

Hi,

I find this article, and particularly the "proofs" to be very misleading. The proofs assume that Newton's laws are correct, and then show that Kepler's laws are correct because a body moving according to Newton's formula do indeed move according to the laws of Kepler.

It should be the other way around.

The article uses the words "proof" and "prove" in a formal sense. You assume one thing, and you deduce another thing: This is a proof. The fact that Kepler's laws can be "proved" from Newton's in this sense shows that Newton's theory, in some sense, constitutes an "explanation" of Kepler's laws (Newton's laws are great becase they "explain" Kepler's laws and so many other observed phenomena at the same time). I think such proofs are worth having in this article. (Or, if they are just too long and technical, an external link might be okay. But the article should make clear that such proofs have been done!)
I don't think the article is misleading as it is. However, if these uses of "proof" and "prove" are confusing to some readers, maybe someone would like to replace these words with "deriviation" and "derive" (as in "we can derive Kepler's laws from Newton's laws). I don't like that change well enough to make it myself, but I would find it acceptible. --EH

Kepler and Brahe made observations about the position of planets around the sun, and plotted them on a map. Kepler found regularity in the planet's movements, which he expressed as his three laws. The three laws may be "proven" by making previsions about a planet's future position, and verify those predictions.

In contrast to the way the word "proof" is used in the article, the above paragraph uses "proven" in a way that really is not acceptible. The fact that the laws can be verified experimentally over and over again is fantastic evidence that the laws are true (at least to within some degree of approximation) and useful, but it is *not* proof. --EH

Later, Newton came with his law of gravitation, that he developed from Kepler's laws. Newton's law says "in order for planets to move according to Kepler's laws, they must be attracted to each other by a force ... etc ..." So, of course, Newton's equations do describe the Kepler's laws. They were made to do so !

--Stephan Leclercq 10:41, 28 Aug 2004 (UTC)

I agree that there is some room for improvement. For example, consideration of the first law can be made to be dependent only on the inverse-square nature of gravity--the rest of Newton's claims are completely irrelevant except only in so far as "inverse-square force" entails "inverse-square acceleration." It is simply not necessary to consider the larger Newtonian framework in order to "prove" Kepler's first law, or show how Newtonian gravity follows from Kepler's statements.
It is somewhat strange that the vector concept of angular momentum is brought in to do a job ordinary integration can deal with, particularly since the relevant equation was already derived, only to be abandoned completely... Conservation of angular momentum is already implicit in Kepler's first law, of course, but it would be better to show this explicitly rather than pulling it out of Newton's hat.
In component form we have:
$m(\ddot r - r\dot\theta^2) = f(r)$
$m(r\ddot\theta + 2\dot r\dot\theta) = 0$
Now consider the angular momentum:
$\mathbf{L} = \left|\mathbf{r} \times m\frac{d\mathbf{r}}{dt}\right| = \left|mr^2\dot\theta\right|$
So:
$r^2\dot\theta = \ell$
This was replaced with
In component form we have:
$\ddot r - r\dot\theta^2 = f(r)$
$r\ddot\theta + 2\dot r\dot\theta = 0$
Since $\dot r = dr/dt$ and $\ddot\theta={d\dot\theta}/{dt}$, the latter equation is equivalent to
$\frac{d\dot\theta}{\dot\theta} = -2\frac{dr}{r}$.
When integrated, this yields
$\log\dot\theta = -2\log r + \log\ell$,
$\ell = r^2\dot\theta$,
for some constant $\ell$, which can be shown to be the specific angular momentum.
The rest is straightforward, with $\ell^2$ replacing $m\ell^2$ everywhere it occurs, and
If the the central acceleration is inversely proportional to the square of the distance, as Newton's law of gravitation claims, we would have:
This is indeed the equation of a conic section with the origin at one focus. Thus, Newton's law of gravitation is a direct result of Kepler's first law.
--Stan Lioubomoudrov 17:05, 2 September 2005
--The one focus of the ellipse is at the Centre of Mass, is there any way of working out where the other is?

--JamesGallagher 10:41, 28 Aug 2004 (UTC) [date?]

Various ways, depending on what you already know about the ellipse. For instance, if you know the location of one focus and the center, the other is the same distance from the center on the opposite side.
—wwoods 18:50, 1 November 2006 (UTC)

This historical comment deserves to be in the article

The above comment by Stephan deserves to be in the article: how did Kepler, thanks to his observations, came to find his 3 laws, then how these laws where used by Newton to find the Law of Gravitation in 1/r^2 ... that is effectiveley made from the law of Kepler. Thanks --Nicop 19:25, 21 Oct 2004 (UTC)

Dependence on spatial dimensions

Assuming that $\mathbf{g}=-\nabla U$ and that $\mathbf{g}$ is divergence-free, i.e $\nabla^{2} U = 0$, in $N$ spatial dimension the potential satisfies $U(r)=C/r^{N-2}$. Hence in order to obtain circular and elliptic orbits, $N$ has to be $3$.

Third Law?

Is it just me, or is Kepler's third law not explained the way the other two are? Kaz

It's not just you. It needs to be explained better.--Bcrowell 5 July 2005 16:43 (UTC)
Fixed.--Bcrowell 5 July 2005 16:59 (UTC)
I found a great explanation of the Third law. Just a thought, if someone wanted to reword the article and reference: http://btc.montana.edu/ceres/html/58Orbits/58orbitsharm_law_act.htm.

I think the explanation is also innacurate. In the article it says, "The squares of the orbital periods of planets are directly proportional to the cubes of the semi-major axis of the orbits," but in all the other descriptions I found, including Wikipedia's article on Kepler, it says the squares of the orbital periods of planets are directly proportional to the cubes of the mean distance of the orbits from the sun. 74.210.24.47 22:27, 20 August 2007 (UTC)

It's the semimajor axis, which is sometimes sloppily referred to as the mean distance. Xihr 23:17, 20 August 2007 (UTC)

Plus Kepler himself thought it was the mean distance - he was proved wrong. It is the semi-major axis. 19:27, 30 May 2008 (UTC) —Preceding unsigned comment added by The Young Ones (talkcontribs)

relativistic case of Kepler's third law

I've added a section on the relativistic case of Kepler's third law. It seemed interesting enough to include, although its validity is much more limited than that of the corresponding Newtonian case. Fortunately, Mr. Crowell rearranged the sections so that I can include derivations without feeling that too much obfuscation is introduced. --Stan Lioubomoudrov 19:39, 2 September 2005

Your hard work is admirable, but I don't think this material is useful, or even correct as currently written. The end result of the derivation is a relationship between r and P, but these quantities only have meaning within the specific coordinate system employed. Coordinates in general relativity are arbitrary and meaningless in and of themselves.--Bcrowell 20:08, 2 September 2005 (UTC)
In what sense are they meaningless? The Schwarzschild t-coordinate is the one observed by an observer at infinity (in the distant limit, if you prefer)--this is physically meaningful. The r-coordinate having the property that a sphere with a radius r has area 4πr2. Admittedly, this is slightly less physically meaningful, but not by much--under normal conditions, the spatial geometry is so close to Euclidean that the Schwarzschild r-coordinate is exactly what is meant by radial distance in the Newtonian case. ----Stan Lioubomoudrov 20:40 (UTC) on Friday, September 2 2005
You've pointed out one way in which the r coordinate you used is analogous to the coordinates of Euclidean spacetime, but in other ways the analogy fails, e.g., the circumference of the orbit isn't equal to 2π times the distance from the planet to the sun. You could pick another coordinate system in which the circumference equaled 2π times the distance to the sun, and in that coordinate system, Kepler's third law would be false for circular orbits. What I'm saying is not at all controversial -- the meaninglessness of the coordinates is a basic, accepted feature of general relativity. You could pick coordinates in which the relationship between r and P would be that P7 was proportional to r11.
I've gone ahead and deleted the section. I hope you won't take this is a confrontational action, because I respect the work you've put into this article, but I don't think the section is correct or useful.--Bcrowell 22:50, 2 September 2005 (UTC)
Well, perhaps I can accept that it is not generally useful enough to be included, but I disagree with the rest of your judgement. The 'meaningless' of the coordinates in the sense that you provided is almost trivially true, and, as you say, very well-known, but let's be reasonable--the criticism of the possibility coordinates for which P7 ~ r11 holds just as true of Newtonian gravitation. Such a coordinate transformation presents no more conceptual, although much more practical, challenge than introducing rotation (and undoubtedly with more complicated epiphenomena than the centrifugal and Coriolis forces). The Schwarzschild metric, however, is fairly special in that the spatial curvature is Euclidean to great precision under normal (solar) conditions. This would not be true in P7 ~ r11 or rotating (in STR, at least, before one even considers gravity). In those cases, space is strongly curved, and nothing at all like Euclidean. The real point, then, is that the notion of space under the Schwarzschild coordinates matches the Kepler-Newtonian framework to a very significant precision. This fact is not something meaningless--on the contrary, it is what enables GPS corrections to work correctly despite an explicit substitition of Newton-Euclid radius for Schwarschild radius, as well as other uses. --- Stan Lioubomoudrov 00:43 (UTC) on Saturday, September 3 2005
The GPS example you gave is helpful, because it's coordinate-independent. It involves events (e.g., the GPS unit receiving a signal from the satellite) rather than just coordinates. Kepler's laws are statements about coordinates, not events. In any case, a derivation of Kepler's third law for circular orbits in general relativity is extremely misleading to the reader, since the third law fails spectacularly for non-circular orbits --- you can have an event horizon, so that some of the bound orbits aren't even periodic. Circular orbits don't even exist in general. The minimum radius (in the Schwarzchild coordinates) for a stable circular orbit is 6M, and the minimum radius for an unstable circular orbit is 3M.--Bcrowell 01:03, 3 September 2005 (UTC)
Hmm...I've done a little more studying, and I think your statement about the third law being true in Schwarzchild coordinates is just plain wrong. For instance, the unstable circular orbit of minimum radius 3M occurs for velocities approaching the speed of light. I have a hard time seeing how this could be reconciled with Kepler's third law. (Also, in general relativity, there can be "knife-edge" orbits that are arbitrarily close to circular for an arbitrarily large number of revolutions, after which the satellite flies back outward again, which means it would look to an outside observer like the satellite obeyed a certain relationship between P and r for a long time, and then decided to do something different :-)--Bcrowell 01:50, 3 September 2005 (UTC)
In the equatorial plane, dλ² = [1-2m/r]dt² - r²dφ², and since r²dφ² = (m/r)dt² by Kepler, one can substitute to get dλ² = (1-3m/r)dt², which becomes singular at r = 3m rather than the event horizon r = 2m. The non-existence of circular orbits below r = 3m is a direct consequence of Kepler's third law. I don't understand your criticism at all; the fact that Kepler's third law holds for circular orbits in general relativity has been known since 1916, when it was derived in Karl Schwarzschild's original paper, and the evidence you present in argument against it is not only compatible with it, but a direct consequence of it. Although I did not use Schwarzschild's original derivation, it's not as if I presented anything revolutionary. It was explicitly stated that the law was false in general, but true under more restrictive conditions. --Stan Lioubomoudrov 02:43 (UTC) on Saturday, September 3 2005
Hmm...sounds like you don't understand me and I don't understand you. But if we're agreed that the material isn't really relevant to the article anyway, maybe it's not worth trying to sort out.--Bcrowell 04:43, 3 September 2005 (UTC)
That's a significant if; you have not argued about its lack of utility (this would be strange, since you yourself cited very interesting results which follow from it), but only disputed its correctness. As far as I understand, your criticism is two-fold...
(1) You claimed that the proof of Kepler's third law for relativistic circular orbits is meaningless because the coordinates are meaningless, since there can be coordinates in which P7 ~ r11, or some other relation. My answer is that (a) such coordinates can be defined in Kepler-Newtonian mechanics just as well, so this reason alone is not strong enough. Of course, if one chooses different r-scale for Kepler's laws, one should transform the law to reflect the coordinate transformation, and use the transformed law instead. But that's precisely what one should do for the relativistic result as well. Furthermore, (b) in the context of Kepler's original laws (i.e., conditions typical of our solar system), Schwarzschid "r" is Kepler's "r" to a very significant degree, in fact about to 10-12 relative error even at Mercury's orbit (according to Schwarzschild's paper). This is, in fact, slightly greater precision than even the best modern orbital measurements allow (compare astronomical unit, standard gravitational parameter, Gaussian gravitational constant, etc.).
(2) You claimed that the fact that there are no circular orbits at r < 3m is incompatible with the relativistic Kepler's third law. My reply is that not only is it compatible, but that this non-existence is a direct result of Kepler's third law dt² = (r³/m)dφ². Take the Schwarzschild metric in the equatorial plane (θ = π/2, dθ = 0), assume a circular orbit (dr = 0), and the result is dλ² = [1-2m/r]dt² - r²dφ². Just by alrebraically rearranging Kepler's third law, r²dφ² = (m/r)dt², so direct substitution implies dλ² = (1-3m/r)dt². This is null at r = 3m (thus only light can orbit there) and spacelike at r < 3m (thus nothing can orbit there). The lack of such orbits is because Kepler's third law is true for circular relativistic orbits, not something contradictory to it.
If there is some other relevant argument against it, I cannot find it. You make other statements, yes, such as "some orbits are not even periodic", but note that the conditions for the validity of Kepler's third law under relativity were specified quite explicitly, and that such statements are equally true under Newtonian gravitation. --Stan Lioubomoudrov 23:53 (UTC) on Saturday, September 3 2005

The whole discussion is not even on-topic for the article. Furthermore, Kepler's laws are violated spectacularly in general relativity: the orbits are not conic sections, and the r coordinate isn't necessarily oscillatory for bound orbits, so that in many cases there's nothing like a period that can even be defined. It's pointless and misleading to quote an example that makes it sound as though Kepler's laws are valid relativistically. The space you devoted to the long derivation was completely out of proportion to the importance of this piece of trivia relative to the rest of the article.--Bcrowell 01:26, 4 September 2005 (UTC)

Well, if you feel that its significance is disproportionate to its treatment, I can live with that. Every other reason you give, however, is an irrelevancy--I've explicitly stated the limited conditions under which it is true, and that it is false outside those conditions; it is only misleading to those that do not read. That relativistic orbits are not, in general, conic sections is completely irrelevant to my statements, as Kepler's third law is not a statement about shapes of spatial trajectories. That orbital period is not concretely definable in general is also irrelevant, as there are cases, many of which are stable, in which it is, under which my statements fell. I can accept that that the derivation is too long to be included, since the relativistic Kepler's third law is true only under conditions that are very restrictive, which is why I won't be reposting it. What I won't accept are your given reasons against the validity of what was stated. If your only substantive problem with it is that the length of exposition it received is disproportionate to its importance, you should have simply stated so, rather than made this into a discussion of whether or not the statements are true. --Stan Lioubomoudrov 02:09 (UTC) on Sunday, September 4 2005

Math versus concepts; relationship to Johannes Kepler article

This article is highly mathematical, and I suspect that 99% of the people who hit it are not going to have a clue what it's about. As several people have pointed out on this talk page, it also doesn't place Kepler's laws in the right and intellectual and historical context. I came here after hitting the Johannes Kepler article, which had sort of the opposite problem, being 95% about his astrology and mystical contemplation of the platonic solids, and only 5% about what would now be considered his valid scientific work. I did a little work on the Kepler article to add a nonmathematical treatment of Kepler's laws that tries to put them in the right intellectual context, and I guess what I'll do for now on this article is to put in some pointers to the Kepler article for people who want the basic conceptual framework rather than the derivations.--Bcrowell 5 July 2005 16:43 (UTC)

other heavenly bodies

Do these laws also work for Sedna? How about the asteroids in the asteroid belt?

They work for anything orbiting the sun. They can also be applied any set of satellites orbiting any single body, e.g., the satellites orbiting the Earth.--Bcrowell 05:03, 20 July 2005 (UTC)

Third law

Wouldn't it be better to actually have a proof of the third law for an ellipse (assuming that we know the Law of universal gravitation, which is already assumed), rather than just saying "Newton also proved that in the case of an elliptical orbit, the semimajor axis could be substituted for the radius." --Borbrav 19:35, 23 October 2005 (UTC)

Proposals

Proof of relationship between tan(T/2) and tan(E/2) doesn't seem complicated to me. Should it be added here?

An interesting corollary from the 2nd law is the formula for the speed of the body, moving on elliptical orbit. Is it mentioned in other Wiki articles? If not, then should it be mentioned here? Unfortunately, I didn't find it's proof so far (probably because I have little time for this and astronomy is only my hobby). Cmapm 02:25, 5 November 2005 (UTC)

Kepler's Third Law regarding different masses

Posts moved to Reference desk. Zhatt 00:26, 30 November 2005 (UTC)

Kepler's law and dimensional analysis

To me, the easier connection between Kepler's and Newton's laws is an elementary dimensional argument. Consider a circular motion of a planet around a star. Let T be measured in years per radian rather than in years per turn. Then the acceleration a = RT−2 , and so R2a = R3T−2 . That the right-hand side is constant is Kepler's third law. That the left-hand side is constant is Newton's law of gravitation. So either law follows from the other. Bo Jacoby 14:31, 13 December 2005 (UTC)

Keplerian motion and Mercury

The paragraph discussing the precession the perihelion of Mercury is not really accurate. Kepler's laws don't predict precession at all. They do not incorporate any of the perturbing effects of other planets, and are never used for any kind of accurate ephemeris. It is more correct to say that the the precession of the perihelion of Mercury violates Newton's laws, and probably the paragraph just doesn't belong in this article. DonPMitchell 23:10, 15 December 2005 (UTC)

SO(4)

I'm having trouble quickly finding any article that discusses the SO(4) symmetry of planetary motion. Just in case you haven't heard of this, here's a quick review. The conservation of angular momentum implies, by Noether's theorem, that there is a symmetry group, the rotation group. However, in inverse-square force-law motion, there is an additional conserved quantity, the Runge-Lenz vector. Working backwards from this, it can be deduced that the fully symmetry group of planetary motion is not just SO(3), but the larger SO(4). I was hoping that there'd be an article spelling out te details, or at least mentionin it, but it seems not ?? linas 17:16, 13 June 2006 (UTC)

Never mind; I just edited the article on the Runge-Lenz vector to spell this out in slightly greater detail. Perhaps someone can amend this article to at least hint at this (important) result? linas 18:06, 13 June 2006 (UTC)

units

"So the expression T2a–3 has the same value for all planets in the solar system as it has for Earth.

That value is

Thus, not only does the length of the orbit increase with distance, the orbital speed decreases, so that the increase of the sidereal period is more than proportional."

The value of T2a-3 is not listed with any units. Could someone please put units on it?

Hi, I want the formulas to find the average distance of pluto from the sun using Kepler's laws

On the next line of the article it says, "(with T in seconds, a in meters)".Roachmeister 11:48, 18 May 2006 (UTC)

Using year and astronomical unit (AU), T2a−3=1 [year]2[AU]−3, because the period of the motion of earth around the sun is a (siderial) year, and the mean distance between earth and sun is the AU. So, observe the period of Pluto T [year] and compute the distance a=T2/3 [AU]. Bo Jacoby 14:00, 25 July 2006 (UTC)

proving kepler's third law

The third law is about two planets, but in the 'proof' only one planet is considered. It does not prove what it claims to prove. Can anybody follow the 'proof'? Bo Jacoby 16:30, 1 January 2007 (UTC).

The two-body problem can always be reduced to a one-body problem with a central force via a change of variables (to center-of-mass and relative coordinates); I think this is mentioned in the article, although I haven't read through it in detail. This fact is rather basic to the understanding of planetary motion and Kepler's laws; if you aren't familiar with it, you should probably read some introductory material on the topic before making substantial edits to the article. A classic introductory textbook that gives a clear derivation is An Introduction to Mechanics by Kleppner and Kolenkow (1973). —Steven G. Johnson 21:18, 3 January 2007 (UTC)

Thank you, I am aware of this. But Kepler's third law relate the motions of two very light planets around the same heavy sun, so it is a simple case of a three-body problem. It says that P2a–3 has the same value for two planets A and B :

PA2aA–3 =PB2aB–3 .

This cannot be proved by considering only one planet and one sun. Bo Jacoby 00:33, 4 January 2007 (UTC).

Kepler's third law conceptually requires only a single orbiting body, in which case it relates the orbital period and radius (semimajor axis) for different initial conditions. In the case of relating the orbits of multiple planets orbiting the same star simultaneously, it (like all of Kepler's laws) applies only in the limit where the gravitational interactions between the planets can be neglected. It is no secret that Kepler's laws involve numerous (very good) approximations when applied to a real solar system. —Steven G. Johnson 03:38, 4 January 2007 (UTC)

The first and second Kepler laws guarantee that that the orbital elements P and a are individually conserved. One planet does not have different orbits or different initial conditions. So, if there is only one planet, then is trivially true, and void of information, to claim that P2a–3 has the same value for all planets. Apart from this sophistry, the subsection on proving Kepler's third law needs improvement. Bo Jacoby 01:52, 5 January 2007 (UTC).

Units of "p" in derivation of Kepler's first law

Based on the definition of

\ell = r^2 \dot{\theta}

in the derivation of Kepler's second law, I believe that the units of \ell are m^2 rad sec^{-1}. If this is true, then $p$ as defined does not have units of meters, but rather meters times square radians, in which case a normalization of 1/(2\pi)^2 would seem to be required. Do you agree?

Overall this is a very nice Wikipedia entry. Thanks for taking the time to write and maintain it. —The preceding unsigned comment was added by Jbunniii (talkcontribs) 18:09, 22 April 2007 (UTC).

Thank you for the nice words. The length $p=\ell ^2 G^{-1}M^{-1}$ has the geometrical interpretation of semi latus rectum. If you divide by (2π)2 the result ceases to have a simple geometrical interpretation. The unit radian is often implicite because meter per meter is dimensionless. Bo Jacoby 22:40, 23 April 2007 (UTC).

Clarification of notation

Edited page to reflect the "one dot, two dots" shorthand for first and second derivative w.r.t. time. This has been a bug-a-boo of mine whilst teaching, as my beginning physics students have been known to write/type faster than they think. Not that I have ever been guily of this...well, hardly ever..{laughter}. Drieux 21:57, 15 July 2007 (UTC)

Kepler's First Law

From the article: "In the case where one particle is not much lighter than the other, it turns out that each particle moves around their common centre of mass, so that the general two body problem is reduced to the special case where one particle is much lighter than the other."

This is incorrect. The two bodies always move around their common center of mass.

When one body is large, the center of mass may be inside that body, but they both orbit that common center (in separate ellipses, unless the masses are equal, in which case they're antipodal on a circular orbit). In fact, this is how astronomers use the wiggle of a star to find a large planet orbiting it: the common center is within the star, but far enough away from the center of the star (due to the size of the planet) so that the star's movement around the center of mass shows up as a "wobble" which is detectable even at large distances.

I've always used the image of a pipe between the two bodies in question, with the ability of the mass to flow through the piple from one body to the other. The center of mass will travel along this pipe until the situation reverses (the center of mass now within the second body), thus showing the symmetry of the situation.

I think the reason for the confusion is that textbooks start with the reduction of the two body problem to the single body under a central force, but never translate the results back at the end of the derivation.

Rwilsker 23:06, 15 July 2007 (UTC)

Yes, "The two bodies always move around their common center of mass" even "In the case where one particle is not much lighter than the other". So the quote from the article is quite correct. However, Kepler's laws are about the motion of light planets around the heavy sun, and the treatment of the general two-body problem, including the motion of binary stars, belongs in some other article. Bo Jacoby 10:23, 17 July 2007 (UTC).
I've clarified the article and added an animation. -- Beland (talk) 15:15, 11 August 2008 (UTC)

The first section of the article includes no reference as to what the algebra means! Very bad maths! The variables should always be stated or it is meaningless! —The preceding unsigned comment was added by 80.1.72.245 (talk) 18:13, 11 January 2007 (UTC).

I don't follow. Every variable is explained, and there are links to further explanation. Bo Jacoby 16:53, 12 January 2007 (UTC).

Foundation left by Copernicus?

Could someone explain this sentence: "His third law is based on the foundation left by Copernicus, because he uses a mathematical expression to show the correlation between T (time for one revolution) and D (distance from the sun)."? I don't understand how this is based on a foundation left by Copernicus. If this is true, it's interesting and probably bears a little further exposition. I also have to ask, is there a source for this, or is it synthesis? I'll delete the passage in a week if I don't hear anything. --Ben Kovitz 15:04, 11 October 2007 (UTC)

Don't delete it unless you verify that Copernicus missed this. Copernicus could find the ratios of the distances of planets from the sun, and their orbit period. It is hard to imagine that he didn't notice some approximate form of Kepler's third law. Kepler's law is not an approximate relation between orbit size and period, it's exact in the Newtonian two body problem--- the orbit period is proportional to the 3/2 power of the semi-major axis, not just any old "radius".Likebox (talk) 20:32, 4 February 2008 (UTC)

Units?

What units for the distances involved? I'm struggling to see any mentioned at all - JVG (talk) 15:39, 28 March 2008 (UTC)

That's because the laws do not depend on the units used. Bo Jacoby (talk) 07:15, 29 March 2008 (UTC).

The laws work for any units provided they form a consistent set. Like a lot of people I have a strong personal preference for SI, a consistent set, but it might be worth mentioning that many astrodynamics textbooks and software packages use "canonical" units normalized for the sun or a particular planet. For earth orbits, the canonical length unit is the earth radius (er), the canonical mass unit is the earth mass (em), the canonical velocity is the velocity of an "earth skimming" satellite (in a circular orbit with a semi-major axis of 1 er), and the canonical time unit is the time taken to move one earth radius at one canonical velocity unit. The math is simplified because constants like GM (μ, the earth's gravitational constant) become unity, but it's possible to forget them or make other mistakes when converting to non-canonical units such as SI, or to orbits around another central body. Karn (talk) 04:10, 7 June 2008 (UTC)

true anomaly

The true anomaly is called both θ and ν in the article. It should be unified. Which letter should be preferred? Bo Jacoby (talk) 08:15, 7 June 2008 (UTC).

My impression from the literature is that ν (nu) is universally (or at least near-universally) used for true anomaly. However, there may be derivations based on formulas from analytic geometry that traditionally use other symbols, such as θ (theta). To follow both conventions, then, means at some point you have to change symbols and that could be confusing. I'm not sure what to do. Suggestions? Karn (talk) 00:55, 11 June 2008 (UTC)

Pick one and be consistent. They're symbols; if there's a universal one to use, then use it. If not, then use something sensible. Xihr (talk) 07:25, 12 June 2008 (UTC)
I made it ν (nu) throughout. Bo Jacoby (talk) 06:54, 26 July 2008 (UTC).

Incorrect conclusion

"Thus, Kepler's first law is a direct result of Newton's law of gravitation and Newton's second law of motion." This is not correct. Kepler's first law says that the orbit is an ellipse, while Newton's law also admits eccentricities greater than one, in which case the orbit is not an ellipse. Rather than the difficult integration of Newtons laws into Kepler's first law, one could differentiate Kepler's first and second laws to show that the Kepler motion satisfies Newton's laws. Bo Jacoby 08:50, 1 January 2007 (UTC).

Apparently since changed. -- Beland (talk) 15:15, 11 August 2008 (UTC)

1. The symbol for the polar coordinate angle, the socalled true anomaly, is usually θ, but here it is T. I would have changed it to θ except that the nice drawing calls it T and I do not know how to edit the drawing.
2. The intermediate steps from the full-angle formula
$\tan T=\frac{\sqrt{1-\varepsilon^2}\cdot\sin E}{\cos E-\varepsilon}$
to the half-angle formula
$\tan\frac T2=\sqrt\frac{1+\varepsilon}{1-\varepsilon}\cdot\tan\frac E2$
is not shown, and it is not obvious to me.
3. The proof of the third law needs simplification.

Thank you. Bo Jacoby 21:40, 2 January 2007 (UTC).

Step 2 above has now been made. Steps 1 and 3 still remains. Bo Jacoby 21:51, 2 February 2007 (UTC).

Step 1 is obsolete; it's now nu everywhere. -- Beland (talk) 15:15, 11 August 2008 (UTC)

"The general equation, which was derived from Newton's law of gravity,"

$\left({\frac{P}{2\pi}}\right)^2 = {a^3 \over G (M+m)},$

Can someone explain why this formula doesn't work out if you rearrange to solve for $\ M$ using data for the Earth and the Sun??

(((149597876.6^3)/ (365.26/2*PI())^2) /0.0000000000667248)-5.98E+24)=1.52415096012299E+29
(7.66% of the sun) GabrielVelasquez (talk) 01:26, 26 June 2008 (UTC)

Wrong units, for starters. You need all figures you use to compute the value to be in a consistent unit system; you're mixing up kilometers for distance units with SI (where it should be meters). Xihr (talk) 01:55, 26 June 2008 (UTC)

(((149597876600^3)/(365.26/2*PI())^2)/0.0000000000667248)-5E+24 = 1.52421076012294E+38 (7,663,201,408.36%)
Oh yeah, much better. I don't think you are bothering to check this. If I rework the formula for time or distance I get exactly 365.36 days or 1.4959e8 km if I multiply M by 1.188014124, but I don't know why I should have to, that's the question here. The only way that formula comes any were near being equal is if I use Kilometers. GabrielVelasquez (talk) 18:56, 2 July 2008 (UTC)
By the way, what are you refering to when you make the statement "Consistant unit system," - do you know what that means? or does using fancy big phraseology make you feel important... because irrelevant statements don't impress all of us. GabrielVelasquez (talk) 18:58, 2 July 2008 (UTC)

Uh, I told you exactly what you were doing wrong. Feel free to rant if it makes you feel better, but it's not going to help you fix your problem. Xihr (talk) 19:49, 2 July 2008 (UTC)
What you said doesn't work, doesn't add up, there is no rant in that fact, in other words you act like you are helping but your so called help has become part of the problem. You don't know what you are talking about. I was asking for a real answer, not your useless opinion. What you call a rant is my explaining what is the problem, which your ignorance has not helped at all. GabrielVelasquez (talk) 04:34, 5 July 2008 (UTC)

The way to find an error in a calculation is to present the calculation slowly and carefully, omitting no step and no assumption. When this is done you will most likely find the error yourself. What are your values for G and P and m and a? What is the formula for M? What are the units used? What are the intermediate results? Have fun. Bo Jacoby (talk) 07:19, 5 July 2008 (UTC).

Xihr was correct in his statement. And the phrase "consistent units" means exactly that - it is not a fancy phrase - it means that all the values you put into the equation should have units that are consistent with each other. The value you put in for G is in SI units (means it can be expressed in the base units kg, m, s). The value you inserted for period is in days rather than seconds), and the value you put in for radial distance is in km (rather than m)- so the units are not consistent. Please refrain from rude statements to people who are trying to help you out. Hope this helps. PhySusie (talk) 11:24, 5 July 2008 (UTC)
Obviously what he said was not correct if you had to elaborate to clarify. Arrogance is what makes the both of you assume that if you give some general squawk everyone is supposed to know your specific meaning. God help your students if you are teachers. I'm still left to work it out because you don't seem to know, I'm the one who asked the question (and not for your worthless famous squawking ego catharsis) and I still don't have my answer:
• solar mass $M_{\odot}=1.98892\times10^{30}\hbox{ kg}$
• astronomical unit $AU_{\odot}=149,597,870.691 \hbox{ km}$
• Earth mass M = 5.9736 × 1024 kg
• Gravitational constant $G = \left(6.67428 \plusmn 0.00067 \right) \times 10^{-11} \ \mbox{m}^3 \ \mbox{kg}^{-1} \ \mbox{s}^{-2}.$
• Sidereal year $P_{\odot}=365.25636042 \hbox{ days}$

(((149597870691^3)/((365.26*23.933*60*60)/2*PI())^2)/0.0000000000667248)-5.9736E+24 = 2.05267087548276E+28
(1.03%) So much for meters and seconds. GabrielVelasquez (talk) 16:14, 5 July 2008 (UTC)

(((149597870691^3)*(2*PI())^2) /(((365.256*23.933*60*60)^2)*0.0000000000666748))-5.9736E+24 = 2.00160766004847E+30
...correct algebra but that ain't it either, as the article for solar mass says 1.98892e30 kg.
I'll just have to wait for someone who understands this for a definitive answer.
GabrielVelasquez (talk) 21:04, 5 July 2008 (UTC)

I'm sure the irony of you ranting about arrogance is not lost on anyone but yourself. Please behavior more appropriately; you are pushing close to violating WP:NPA. Not to mention that this is not a forum in the first place. Xihr (talk) 20:27, 5 July 2008 (UTC)

Sure, you can't answer the question I asked about the formula in the Article so I must be ranting, sure that makes perfect sense to anyone who reads this, good job. This is about the formula in the article in case you conveniently forgot, I'm obviously still working on it - but if this is not a forum then why are you still bothering; Your violation of WP:NPA is pretty obvious, good luck with that. by the way I'll be providing you with a long list of Wikipedia policies that you definitely need to be rereading in detail these particular seconds and meters. And of course that can't be just my opinion, since I linked it to a policy (WP:NPA) <-- that right there, it must be fact. Right.
GabrielVelasquez (talk) 21:04, 5 July 2008 (UTC)

It looks like you already have a correct enough answer above - considering you are using an approximation the two values are very close. The equation you start with is for an elliptical orbit with 'a' being the semi-major axis. The value you are plugging in is the average Sun-Earth distance - which is a very good approximation considering our orbit is not very elliptical. So it looks to me like you have solved your problem. Besides that, this is not really the place for a question like this - next time you might try posting on the Science Reference Desk and people can help you out there. This page is for discussion about the article itself. And again, please try to be polite to those people trying to help you. PhySusie (talk) 11:24, 6 July 2008 (UTC)
I don't agree with you on either point, and again it is another one of your assumptions that I am not refering to the article, if this formula is false (see above comments: 3rd Law)it should be stated as so in the article. I would not suggest the formula be removed because in the end it is mathematics history at the very least, but the answer is not good enough considering that it is the average distance. GabrielVelasquez (talk) 18:25, 7 July 2008 (UTC)
The equation is not false - it has held true over the centuries. I don't think this discussion is going to find something new that the scientists have missed. But thanks for checking it out anyway. It is a good way to catch a mistake in the formula in case it was entered into the article incorrectly. PhySusie (talk) 18:38, 7 July 2008 (UTC)
File:Missing Star masses by Kepler 3rd Law.jpg
I feel I owe you this explination of why I had to check this formula. These are the extra solar planets that are listed with mass, semi-major axis, period, and yet oddly with the star mass listed as unknown. where in RED, I added the figures.

I feel I owe you this explination of why I had to check this formula. These are the extra solar planets that are listed with mass, semi-major axis, period, and yet oddly with the star mass listed as unknown. where in grey, I added the figures.[1]
GabrielVelasquez (talk) 22:37, 9 July 2008 (UTC)

Also what you think or don't think is going to happen, what is new or not, so far has not proven to mean much compared to the fact that one of these here is proving to be false: The 14 unknown mass stars that are not unknown, or Keplers formula, or these professional scientists data on extra solar planets. GabrielVelasquez (talk) 22:47, 9 July 2008 (UTC)
You are performing the computations incorrectly because you are using incompatible units, as was pointed out from the beginning. Regardless of how important you think this is, it is original research and inappropriate for Wikipedia -- and it is incorrect original research, at that. Remember that Wikipedia is not a forum, please move on before you engage in more personal attacks. Xihr (talk) 06:34, 10 July 2008 (UTC)
If you (GV) really want to know: the problem is that you incorrectly used sideral days. A year is 24*365.25 hours, not 23.933*365.25 hours. If you had asked in a way that I consider civil I would have answered right away. Han-Kwang (t) 11:32, 17 July 2008 (UTC)
I find 23.933 more accurate, with the right algebra. About my question about he formula in the article, I don't think you are correct and I am not interested in your opinion either, the question has already been addressed with regard to the article. You write as though your answering "right away" is somehow important to anyone, which it is not to me; June 26th was when the question was first posed and there was nothing wrong with it as you again incorrectly insinuate. I don't believe you are civil <-- see it must be true because I put a link there. The units have been confirmed and your late and useless input is no longer needed. Thanks GabrielVelasquez (talk) 18:17, 18 July 2008 (UTC)
Thank you for your amusing contributions. I enjoyed reading it. Han-Kwang (t) 22:22, 18 July 2008 (UTC)

(PS:) None of these people read physics text books, where this formula listed as having a "k" which is what I was hoping someone would mention, correct or not.
For the record the correct formula, which none of these geniuses figured or thought to add to the article is:

• M=((((((2*PI())^2)*((d*149597876600)^3))/(((p*23.933*60*60)^2)*0.0000000000667259))-(m*317.877*5.98E+24)))/1.98891E+30

GabrielVelasquez (talk) 23:40, 29 August 2009 (UTC)

A question

In the second paragraph, does the term "legendarily precise" have to be there? Zain Ebrahim (talk) 18:03, 16 July 2008 (UTC)

It seems that it is changed now. Han-Kwang (t) 11:32, 17 July 2008 (UTC)
He was right; I went ahead and fixed it.  Xihr  19:28, 17 July 2008 (UTC)

Ellipses

The main article merely shows that Kepler's laws agree with Newton's laws. It does not show how Kepler came across ellipses in the first place. See http://www.wlym.com/~animations/part4/47/index.html —Preceding unsigned comment added by 86.167.246.75 (talk) 09:59, 1 August 2008 (UTC)

The first reference in the main article is ungrammatical. The grammar has been improved slightly.

Book 5

Only Book 5 is re-printed by Hawking, as the first four books put Kepler in a poor light. —Preceding unsigned comment added by 92.12.146.16 (talk) 12:19, 4 August 2008 (UTC)

edit on the 9 February 2007 by user 193.80.109.10

Quote: "The general equation, which Kepler did not know, is be derived by equating Newton's Law of Gravity with Uniform Circular Motion ($a=(4\pi^2r)/t^2$), which is valid for (near) circular orbits".

It is a good point, but note that in the rest of the article a is semi major axis and not acceleration, and that the period is called P rather than t. Please improve. Bo Jacoby 20:16, 20 February 2007 (UTC).

In order to improve the article I sugest to use

$m\ddot\mathbf{r} = GMm \frac {r^3}(- {\mathbf{r}})$

or the classical

$m\ddot\mathbf{r} = {G Mm}\frac {r^2}(-\hat{\mathbf{r}})$

$m\ddot\mathbf{r} = \frac{Mm}{r^2}(-\hat{\mathbf{r}})G$

BTW m is not important here, so it can be nice if you remark the fact that two particles with different mass has the same period in the same orbit.

2 July 2007 Drostie

Drostie's edit assumes knowledge about vector cross product, and makes the next section incomprehensible. The connection to conservation of angular momentum is important, but could be explained in the article on angular momentum, which also has the link to Noether's theorem. Please explain the idea behind this edit. Bo Jacoby 18:04, 8 July 2007 (UTC).

FWIW, I find the cross-product explanation clearer than the description in terms of angular/radial coordinates. I think someone who made it that far down into the article can be assumed to have basic knowledge of both differential and vector calculus. If you aim for laymen with a high-school physics background, then I regret to say that you can safely assume that everything after the article lead section is fully incomprehensible. Actually, I wrote the current lead section after getting a complaint from someone outside the Wikipedia community that he had no idea what Kepler's laws were after looking up this article. Han-Kwang 21:27, 8 July 2007 (UTC)
The article by Hyman (cited at footnote #1) presents an extremely understandable presentation and derivation of Kepler's laws, without using any trigonometry, or even any vectors. Perhaps we might consider using this approach in the article.Ferrylodge 22:04, 8 July 2007 (UTC)

The notation needed for understanding the derivation of the first law was removed by Drostie. If you prefer vectors rather than polar coordinates, please use it throughout the whole article. Bo Jacoby 15:17, 9 July 2007 (UTC).

quote: "Kepler's laws are formulated below, and are derived from Newton's laws, using heliocentric polar coordinates. However, Kepler's laws can alternatively be formulated and derived using Cartesian coordinates.[1]". This statement has to be changed if Drostie's edit is going to prevail. Bo Jacoby 08:57, 11 July 2007 (UTC).

Since reverted. -- Beland (talk) 15:23, 11 August 2008 (UTC)