Talk:Orbital period

Orbital Revolution
"one 360° revolution of the most massive body around the lesser massed body"? Or the lesser around the most massive??

It is actually orbiting around the point of the balance of the masses, however the point is valid. I therefore simplified this text even though it is not as specific. Note: I'd suggest you become a registered User. Also please sign post with four consecutive '~' so you can be identified. Thanks. Arianewiki1 (talk) 23:46, 22 January 2017 (UTC)

Spelling
Changed 'a' back to kilometers (see below). Checked this against: Wertz and Larson, Space Mission Analysis and Design (SMAD) 3rd ed. Table 6-2, pp 137.

R.A Barnes Undergraduate Satellite Engineer

None of this is valid, as the difference is American spelling versus the more universal spelling. Making changes to enforce a way of spelling is not acceptable even with an irrelevant reference. Arianewiki1 (talk) 23:53, 22 January 2017 (UTC)

- i made a change below to meters instead of km in semi major axis because the formula does not give the correct answer if you input "a" in kms.


 * Small body orbiting a central body
 * In astrodynamics the orbital period (in seconds) of a small body orbiting a central body in a circular or elliptical orbit is:
 * and
 * (standard gravitational parameter)
 * where:
 * is length of orbit's semi-major axis (km),
 * is length of orbit's semi-major axis (km),

David C.S Ong —Preceding unsigned comment added by 134.115.68.21 (talk) 09:58, 28 September 2007 (UTC) -

Whispers from the land of ignorants ...

This is a great article (Orbital Period page). I don't quite understand one section, however. The point under the 2-bodies-orbiting section: "... the sum of the semi-major axes of the ellipses in which the centers of the bodies move, or equivalently, the semi-major axis of the ellipse in which one body moves, in the frame of reference with the other body at the origin (which is equal to their constant separation for circular orbits) ..."

To use this in the calculation for determining the orbital period -- would I just sum the semi-major axes for both ellipses? E.g. two stars orbiting each other: "Star A" has a semi-major axis of 100 AU, and a mass of 1.5 Solar Masses. "Star B" has a semi-major axis of 300 AU, and a mass of 0.25 Solar Masses. Would I use 400 AU in the formula?

Thanks, Tesseract501(August 27, 2005)


 * Mass and semi-major axis are inversely proportional, so the example does not seem possible, but yes, we just add the semi-major axes.--Patrick 00:03, 28 August 2005 (UTC)


 * You'd use the total distance, 100+300 AU, and the total mass, 1.50+0.25 MSun.
 * Urhixidur 15:35, 14 December 2005 (UTC)

Letter P and letter T are both used in this article to signify orbital period. Which one should be preferred ? Bo Jacoby 10:32, 14 December 2005 (UTC)

I would say T, like Frequency uses.--Patrick 14:05, 14 December 2005 (UTC)

Diagram
It would be nice to see some diagrams showing the difference between the orbital periods. —Preceding unsigned comment added by Reddwarf2956 (talk • contribs) 09:53, 23 September 2009

Going backwards
This whole effort seems to be going in the wrong direction. To take only one example, the reference to the tropical year (as of 01:49, 12 September 2008) was well written and coherent. The current version (as of 10:19 UTC 14 November 2009) of the same topic is illiterate goobledegook. I hesitate to make an edit since it appears to be a waste of time. —Preceding unsigned comment added by 64.180.221.68 (talk)

Small body orbiting a central body Error
The formula provided states that T is in seconds, but the units in the formula work out that the result is seconds squared, not seconds. It's fundamentally wrong.

Unfortunately, I haven't found another reference to replace it from yet... —Preceding unsigned comment added by 67.164.83.84 (talk) 01:37, 18 November 2009 (UTC)


 * The formula is correct. The units of the gravitational constant G are m$3$/(kg•s$2$). The units of a$3$ are canceled by m$3$, while the units of M$1$ + M$2$ are canceled by kg. This leaves s$2$, which the square root sign reduces to seconds. — Joe Kress (talk) 09:54, 18 November 2009 (UTC)


 * Ah, yes. I was checking with wolfram, and had parens wrong.  Thanks for the correction.  —Preceding unsigned comment added by 128.170.62.95 (talk) 17:33, 18 November 2009 (UTC)

Units of measure
This article contains several mathematical equations relating various physical quantities. Only one of these equations specifies any units of measure. It is good style and certainly a favor to the reader if all the units of physical quantities are specified explicitly. When such a specification is absent the results are meaningless. For example, the equation:  T = 3.3 \sqrt{(a/R)^3} clearly results from some choice of units for T, a and R (and also implicitly for the constant value 3.3). We are told that T is in hours, but what are the units of a and R? Meters, miles, inches, astronomical units? I could do the calculations to answer that question myself but I feel that people should do their own homework, otherwise they won't learn. — Preceding unsigned comment added by 154.5.32.113 (talk) 09:45, 30 May 2011 (UTC)

This sentence is gibberish.
" It differs from the sidereal period because the node is a coinciding of planes rather than a linear coinciding, and the object's line of nodes typically precesses or recesses slowly in relation to orbital cycle. "

Two coincident planes really constitute a single plane. I think the author of this sentence means to say that the line of nodes is the line of intersection of the plane of the orbit and the plane of the ecliptic. (I don't know what he means by a "linear coinciding.") And it is true that as the plane of the ecliptic precesses this line of nodes rotates about the normal to the orbit. Therefore any period measured with respect to this direction will also change over time.154.5.45.119 (talk) 23:25, 30 August 2011 (UTC)

"Citation needed" for orbiting body of water
Under section 2.2 (Orbital period as a function of central body's density), the editor provided the equation for calculating the orbital period of a body around another given the density of the second body. He then proceeded to illustrate a possible use of the equation by showing the result of the calculation if the body orbiting the earth were made of water. For some reason, that example has been flagged, [ citation needed ]. Could someone elaborate on that? Julesmazur (talk) 19:28, 12 February 2013 (UTC)
 * I removed the citation needed tags on that and two other simple calculations, but I did leave the one on the "universal unit of time" bit, which, to my eye, seems like balderdash. I tried to find an outside reference page on http://exploration.grc.nasa.gov/education/rocket/shortr.html but it never gets to orbital periods (*tsk *tsk) - Featherwinglove (talk) 21:55, 9 September 2013 (UTC)
 * I agree with the removal of the "citation neeeded" for the simple calculations in "Orbital period as a function of central body's density" but someone put them back in. Does anyone support the "citation needed" for combining $$T = 2\pi\sqrt{a^3/\mu}$$, $$\mu = GM$$, $$M = \rho V$$, and $$V = \frac{4 \pi}{3} a^3$$ in a single equation $$T = 2\pi\sqrt{ \frac{a^3}{G \rho 4 \frac{1}{3} \pi a^3}} = \sqrt{ \frac{3 \pi}{G \rho}}$$ ? 209.131.62.116 (talk) 21:10, 12 November 2013 (UTC)

synodic period and gravity
It seems like there should be some comment about gravitation peaks and valleys in respect to the synodic period that elapses between two successive conjunctions with the Star–planet_1 line in the same linear order. Maybe even something with how this relates to orbital stability and orbital resonance. John W. Nicholson (talk) 11:36, 2 March 2013 (UTC)

Orbital period as a function of central body's density
Is this whole section rubbish? What I mean by this question is: did Galileo proved that gravity accelerates objects at the same rate no matter how much they weigh? Gravity is central reason for the orbital period of any body. So, why is this section here? Yes, the earth has a moon. No, it does not orbit in "T = 1.4 hours". So, something is wrong. John W. Nicholson (talk) 21:00, 19 September 2013 (UTC)
 * It is not completely rubbish, but it might be (fairly trivial) own research. The formula follows from the real formula above when M=4pia^3 rho/3 is substituted. The absurdly short orbit is just above the ground (compare to the 92.8 minute orbit of ISS); the moon has a far bigger a. The talk about universal units is unsourced. Overall it mostly confuses the issue, and I have not seen anybody actually use this approach to periods. Anders Sandberg (talk) 22:43, 19 September 2013 (UTC)


 * Even with that substitution of $$ M=(4/3)\pi a^3 \rho$$, there is a problem. Namely, because of the difference in meaning of a^3, a^3/a^3 do not cancel (one is the radius measured from center and above central body surface (R), second is radius of the central body (r) so R^3/r^3). Also, the units for $$\rho$$ has to include units for r^3 as to maintain the total mass. So, it is looking more like what you said "own research" and needs to be removed. It would seem to be useful if used in some way that varies in density (like with or without a ring system), but still needs a source. 1.4 hours is 84 minutes, that is close to the ISS orbit. I would expect the difference to be the ratio of altitude above the earth to earth's radius. John W. Nicholson (talk) 01:27, 20 September 2013 (UTC)


 * The calculation is of a minimum period orbit, how long does it take to orbit a body at minimum altitude? Turns out it's dependent only on the density of the body. At minimum altitude the orbital radius is equal to the radius of the body, and so the 'a' is the same, and it cancels fine. And at that altitude for bodies with the same density as the Earth, it takes 1.4 hours. The ISS is pretty much at minimum altitude, because it's only just above the atmosphere, and the atmosphere is relatively thin compared to the Earth's radius, so the ISS's orbit indeed takes about 1.4 hours.Teapeat (talk) 02:14, 20 September 2013 (UTC)


 * It's quite a cute result really, it means that if you were just above a spherical rocky asteroid a hundred metres across, it would take the same amount of time to do one orbit as the ISS does. I've heard of this before; I first heard about this when there was the NEAR probe in orbit around Eros. In that case the period differed from orbit to orbit, because Eros is very lumpy and so the orbit was chaotic. But it should work fine for dwarf planets upwards.Teapeat (talk) 02:14, 20 September 2013 (UTC)


 * I did a search to see who added it: but it existed in the article before this. It's probably a standard result.Teapeat (talk) 02:14, 20 September 2013 (UTC)


 * Looking around, I see that it is a not uncommon physics exercise and indeed a cute result. I think the section could remain, but it is far too prominent right now: the two body orbits are important and ought to come above it. I suggest that the section is rewritten slightly so it does not take too much space, explains why it is a very special case, and is moved down a bit.Anders Sandberg (talk) 08:08, 20 September 2013 (UTC)


 * I vote to remove it completely. It's a fine physics exercise, but does nothing to explain orbital periods.  The period of an orbit is a function of the mass of the central object.  This whole density calculation only works with mean density anyway, which is essentially a stand-in for mass.  No planet has a uniform density.  Since it's completely theoretical (and so not a great example) and it does nothing to explain orbital period, I say take it out.  If no one objects, I will do so in a few days. Sanchazo (talk) 19:51, 22 November 2016 (UTC)

"citation needed" too much here?
Well I have to beg the question: aren't there too many requests for citation here. For example the line (under two bodies orbitting each other, slightly out of place there):

"In a parabolic or hyperbolic trajectory, the motion is not periodic, and the duration of the full trajectory is infinite.". Isn't this kind of "logical" in the sense that it is *obvious* that any hyperbole or parabole is non-periodic. If one knows what those shapes are it is, so maybe add a link to parabole/hyperbole descriptions (though even that is consdered common sense and doesn't have to be specified). Adding a citation here would literary be looking for a citation out-of-place, no paper will discuss this. — Preceding unsigned comment added by 82.139.125.134 (talk) 19:40, 4 June 2015 (UTC)


 * No, not too much. Vast sets of statements in this article are unsourced.  Material not sourced, is difficult to verify and may be challenged and removed.  In the case of this article, some of the claims have been challenged for a long period of time, and are therefore subject to removal from Wikipedia pretty much anytime an editor chooses to start the cleanup process.
 * Ditto with mathematical derivations. Unless supported by sources, they are original research, which tends not to stay in Wikipedia long term.  Cheers.  N2e (talk) 15:51, 8 March 2016 (UTC)


 * I agree with 82.139.125.134: Requesting a citation for the fact that a parabola is infinite is ridiculous. That's part of the definition, everybody who knows what a parabola is knows that is infinite. A link to Parabola or Parabolic_trajectory might be in order, though. Similar for some of the derivations. The claim that "the orbital period is independent of size" in the preceding paragraph might be surprising at first, but it isn't hard to verify: If you know that volume (and hence mass) scales at the third power of the radius (which you probably learned in primary school) and know basic algebra (probably 5th or 6th grade) you can verify that in your head in about 30 seconds. --Ligneus (talk) 20:55, 4 October 2016 (UTC)

Too technical
This whole article is pretty difficult to understand, even as someone with a non-physics scientific background. Could someone at least add what the Earth equivalents are of the different kinds of orbital period? If I understand rightly, I think the sidereal period of Earth is 1 year, yes? — OwenBlacker (Talk) 11:05, 8 March 2016 (UTC)


 * Yes, approximately. I believe it is defined in some other articles, for example, Geosynchronous orbit, and you might be able to pick up a source and maybe citation from there or some other article with it defined, in order to improve this article.  Go for it.  N2e (talk) 15:37, 8 March 2016 (UTC)


 * Agreed, so I vastly improved the introduction and made it more readable. I've also added some examples there too. Still needs some cites, which I'll do next. Arianewiki1 (talk) 13:51, 20 January 2017 (UTC)

Someone tell him this revert is wrong, please
I tried to combine two nearly redundant and widely separated sections that obfuscate the idea that M1+M2 is approximately equal to the larger mass when there is a big difference. Only to get reverted by User:GliderMaven, claiming the confused version is more ... educational. What do you think? Wnt (talk) 23:55, 31 March 2016 (UTC)


 * The point is that the more general result is usually taught after the more specific one; IRC, and I may not, Newton's Principia did it that way for example; and most derivations of the general result use the specific result for the proof.GliderMaven (talk) 00:39, 1 April 2016 (UTC)

Kepler's 3'rd law
Why isn't this mentioned at all in this article? It is linked from [low earth orbit] as such. Chris2crawford (talk) 11:52, 5 August 2019 (UTC)


 * Who is/was Keller? AstroLynx (talk) 12:23, 5 August 2019 (UTC)
 * Do you mean Kepler? JustinTime55 (talk) 20:38, 5 August 2019 (UTC)

Correct, but confusing
I deleted the following sentence:

"Note that the orbital period is independent of size: for a scale model it would be the same, when densities are the same, as M scales linearly with a3 (see also Orbit § Scaling in gravity)."

This is technically correct (if you understand what "independent of size" means), but likely to be confusing to anybody who isn't already very familiar with the subject. This could be corrected by adding explanation of exactly what is meant by "independent of size", but since this is a fact that isn't particularly useful, better to leave it out: Geoffrey.landis (talk) 15:58, 10 January 2023 (UTC)

Related Periods section is encrusted garbage
The sidereal and synodic period links from the info boxes for the planets lead to this article. The section Related Periods with the definitions for the period types has evidently become cancerously encrusted with dross and nonsense from repeated editing. For example, there are three paragraphs in which synodic period appears in bold print. The first two are confusing, inaccurate, and inelegant. The third, in its own section, is better. I have a degree in physics and astronomy and even I find this section confusing, how much more so the average interested reader. It needs a complete rewrite. — ★ Parsa ☞ talk 16:30, 11 February 2024 (UTC)