# Talk:Orbital period

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## Untitled

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.

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),

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 (talkcontribs) 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 m3/(kg•s2). The units of a3 are canceled by m3, while the units of M1 + M2 are canceled by kg. This leaves s2, 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: [1] 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)