# Talk:Earth's orbit

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## Peak Heat of Summer vs The Longest Day

I came here looking for info on why the summer solstice (longest day of the year) doesn't coincide with the peak heat of summer, which usually is 1-2 months later. Same goes for winter, with the shortest day being in December, I personally would expect that to be the lowest temperature of the year but it's January/February usually. I'm sure there's a good reason for this, I wonder if it's worth adding that to the document.

Answer: The longest day corresponds to the maximum rate of adding heat to the region being illuminated. The accumulated, or integrated, heat is what determines the temperature reached. (96.233.19.135 (talk) 17:19, 11 February 2010 (UTC)) :^)

This is the same reason why it's usually hottest outside at around one or two in the afternoon, rather than at noon. Ljdursi (talk) 18:22, 6 July 2010 (UTC)

Er. Because axial tilt is far more significant in seasonal variations than orbital position. Were it not for this we would not have swapped seasons in the north and south: it would simply be summer everywhere when the earth was nearest the sun. --96.241.156.174 (talk) 04:19, 14 August 2010 (UTC)

The question made no mention of orbital position. Dbfirs 12:27, 9 January 2014 (UTC)

### Solar intensity due to proximity

Ratio between Aphelion and Perihelion: Axial tilt comparison still needed

Request confirmation of math

There is some effect of the distance between the sun and the earth on the radiant energy reaching the earth. There is some measure of heating northern latitudes up in their winter, up to a maximum at Jan 3, and heating southern latitudes up in their summer. Thus, the winters in northern latitudes (all other variables such as warming by ocean currents, etc, removed from consideration), are relatively more mild, and the summers in southern latitudes (variables removed), relatively more extreme. However, the ratio of the aphelion (furthest distance) to the perihelion (nearest) is only 1.033988391 . So, a difference of 3.4%.

Then to calculate what that does to the total radiant energy: Inverse-square law says, "the radiation passing through any unit area is inversely proportional to the square of the distance from the point source" Received light quadruples as the distance halves.

So $E = \frac{I}{d^2}$

E sun energy stays the same so is 1, solving for intensity I, d is distance, using perihelion distance as 1, d (aphelion) = 1.033988391 and d2 = 1.069131994.

Multiply both sides of the equation by 1.069131994, so that I is both multiplied and divided by the same number; this cancels out, and E is now 1.069131994. E is now equal to I, as I is no longer divided by anything. I = 1.069131994.

1.069 times (aphelion intensity) = perihelion intensity.

1 / 1.069131994 = 0.935338204,

.935 times (perihelion intensity) = aphelion intensity.

Axial tilt total radiant energy and/or comparison with the above remains to be found
Anarchangel (talk) 03:50, 16 January 2011 (UTC)

## Solar Day Versus Sidereal Day - Discussion

I calculated a value for the number of mean Solar days in a year from the standard formula for leap years.

A year is a leap year if divisible by 4 but not by 100, unless divisible by 400. That (if I did the math right) leads to the value 365.2425 Solar days per year, which should be pretty close if the leap year rule is right.

Since the Earth's orbit around the Sun "unwinds" one revolution, judging by looking at the Sun, I reason that there must be precisely one more sidereal day in the time it takes to do the orbit, that is, days as determined by looking at when stars cross a certain meridian instead of when the Sun does so. So there should be 366.2425 sidereal days in one Solar year, versus 365.2425 Solar days.

Maybe there's a real astronomer out there who can confirm or deny. 96.233.19.135 (talk) 17:32, 11 February 2010 (UTC)

Answer: There is, approximately, one extra sidereal day in a year. This is because, as you said, as the earth goes around the sun, it rotates the 365 times plus the one extra by virtue of having circled the sun. Now the answer is "approximate" because it depends on what you mean by a year. A sidereal year is the time it takes the earth to rotate the sun relative to a fixed star. However due to precession of the earth's poles (AKA precession of the equinoxes) the axis of the earth precesses about the normal to the ecliptic about once every 26,000 years. Thus if our years were defined in a sidereal sense, 13,000 sidereal years from now the seasons would be reversed, with northern hemisphere summer in December and winter in June. It was precisely this problem that was addressed in the 1500s by the advent of the Gregorian Calendar (and the demise of the Julian Calendar). What we typically call "a year" is not a sidereal year but rather a tropical year, which is the time between successive vernal equinoxes. A tropical year thus preserves the seasons, while a sidereal year preserves the orientation of the stars above, and the difference between the two is roughly one part in 26,000.

So when you compare sidereal days to solar years, you are about right. If you compare sidereal days to sidereal years, you are exactly right. But if you compare sidereal days to tropical years, you are off by a very little bit (about one part in 26,000). Joyride47 (talk) 22:18, 5 March 2010 (UTC)Joyride47

Follow-up: The first sentence of the article says "the Earth's orbit is the motion of the Earth around the Sun, at an average distance of about 150 million kilometers, every 365.242199 mean solar days[dubious – discuss] (1 sidereal year)." The "dubious" tag's explanation on the edit page is: "dubious|reason=the number quoted is for the wrong kind of year". Actually the text is correct as is: the article orbital period says "When mentioned without further qualification in astronomy [orbital period] refers to the sidereal period of an astronomical object, which is calculated with respect to the stars".
Thus a solar day (rotation relative to the sun) is on average 24 hours; it takes 365.242199 of these to orbit the sun once in the sense of returning to the same position relative to the stars. So I'm removing the "dubious" tag. I'll put in a footnote explaining this. Duoduoduo (talk) 18:25, 26 June 2010 (UTC)

## Earth's orbit's perimeter

How long is the Earth's orbit's perimeter? Newone (talk) 03:44, 8 April 2010 (UTC)

Interesting question. The article ellipse says that a good approximation for the circumference is
$C\approx\pi\left(a+b\right)\left(1+\frac{3\left(\frac{a-b}{a+b}\right)^2}{10+\sqrt{4-3\left(\frac{a-b}{a+b}\right)^2}}\right)\!\,$ with semi-major axis a and semi-minor axis b. From p. 17 of Bakich, Michael E., The Cambrigde Planetary Handbook, Cambridge Univ. Press, 2000, I find the distance from the sun to the perihelion is 147,085,800 km and the distance from the sun to the aphelion is 152,104,980 km; adding these gives a major axis of 299,190,780 km and so a semi-major axis of 149,595,390 km. Page 18 of the same book gives Earth's orbital eccentricity e as 0.0167, and the Wikipedia article semi-major axis gives the formula $b = a \sqrt{1-e^2},\,$. So I compute b = 149,574,528 km. Using these values of a and b in the approximate circumference formula gives the circumference of Earth's orbit as
Circumference ≈ 939,870,018 km.
This of course contains rounding error, and I haven't checked my arithmetic. But it looks right: Earth's orbit is almost circular (the eccentricity of 0.0167 is very low), and a typical round number for the average distance from the sun is 93 million miles; for a circle, this would give a circumference of 2×π×radius = 584,336,234 miles, and assuming one km equals .62 miles the circumference in kilometers would be 584,336,234 ÷ .62 = 942,477,797 km, which is pretty close to the above calculation. Duoduoduo (talk) 19:31, 26 June 2010 (UTC)
Very close. There are exact formulas for the perimeter, using infinite series. Using one that I found in Meeus's Astronomical Algorithms, and a semi-major axis of one astronomical unit, and eccentricity of .0167, I get 939,885,788 km. Saros136 (talk) 06:56, 17 July 2010 (UTC)
Another clever refinement: calculating the mean eccentricity for 2010 using the formula from Meeus, decreases the figure by 34.5 km. Saros136 (talk) 07:03, 17 July 2010 (UTC)
The semi major is not exactly 1 au. Using the more accurate number increases my last by 957 km, to 939,886,710 km. Saros136 (talk) 07:13, 17 July 2010 (UTC)

With an eccentricity of just 0.0167 the earth's orbit is fairly close to circular. I believe all of the illustrations here exaggerate the shape significantly. An exaggeration might be useful for illustrative purposes, but accurate illustrations should be presented first. Ideally if a more parabolic shape is required the orbit of another body (such as mercury) could be used in order to avoid deception. --96.241.156.174 (talk) 04:18, 14 August 2010 (UTC)

## Incorrect figure given for sidereal year

If you review the article Year, you will find that a sidereal year (the time it takes for the earth to revolve around the sun one time, with respect to the (hypothetically) fixed stars, is 365.256363 days. This article claims that the sidereal year is 365.242199 days; this, however, is actually the tropical year, which is not technically related to the earth's revolution around the sun, but is instead simply the axial precession of earth, with respect to the position of the sun (as opposed to with respect to the fixed stars, which is a very different figure of approximately 25,770 years). I am about to edit the introduction of the article to give the correct figure. —Preceding unsigned comment added by Lionboy-Renae (talkcontribs) 23:47, 30 November 2010 (UTC)

## Proposed addition of Meteor Showers reference

Might be useful to include a reference to meteor showers occurring when Earth's orbit crosses the debris stream from comets. I thought including it in the "Events in the orbit" section might be appropriate, though the nature of that section's current content might make the addition seem out of place. Perhaps a separate section? Thoughts? Gmporr (talk) 03:17, 13 December 2012 (UTC)

## commons:File:SunMercuryVenusEarthMoon-Rockets full-court.png

I do not consider this an improvement to the article. A lot of completely irrelevant stuff would only confuse the reader. Incnis Mrsi (talk) 16:53, 15 January 2013 (UTC)

## Sahara Desert

A recent BBC TV programme on the Sahara Desert, narrated by David Attenborough, claimed that the formation of the Sahara Desert occurred rather suddenly about 5,000 years ago, partly as a result of changes in the Earth's orbit. I have found a few sources on the web to support this, which is apparently a very recent research finding, but I have not seen any clear explanation of what orbital changes are involved. Most of the orbital changes, like the Milankovitch cycle and precession, are slow and periodical, so I am curious how such a recent change could have such a sudden and dramatic effect. Does anyone know?86.180.93.32 (talk) 18:32, 1 February 2013 (UTC) - Please ignore this question of mine: the subject is covered in an article on North African Climate Cycles. But I will leave the comment here in case anyone else is curious about the subject.86.180.93.32 (talk) 18:40, 1 February 2013 (UTC)

Crap. Milankovitch cycles and Axial precession has nothing to do with Earth's orbit. Even time scales are different: a significant change in Earth's orbit occurs only in ~ 105 years. Incnis Mrsi (talk) 19:48, 1 February 2013 (UTC)

## Abbreviations look sloppy

Is there a reason in the orbital characteristics section some of the names are presented as sloppy abbreviations? For instance, "avg_speed". This is an English language encyclopedia, not a programmer's guide to handy variable names. I am going to make the changes if there are no objections. 129.63.129.196 (talk) 18:53, 26 July 2013 (UTC)

## AU is not average distance

The first sentence gives the value of the Astronomical Unit to the nearest 10 km but then incorrectly claims it is the average distance to the Sun. The AU is a rough approximation to the semi-major axis a, but this is the distance from Earth not to the Sun but rather to the geometric center of Earth's orbit, which is not the average distance even in the first digit after the decimal point let alone the fifth!

The distance from Earth to the Sun varies in the course of a year from a(1 − e) at perihelion to a(1 + e) at aphelion. To get an idea of the statistics for distance to the Sun over periods longer than a year I downloaded ephemerides from NASA JPL HORIZONS for the period 1950-2010, sampled at 6-hour intervals and using the Earth-Moon barycenter rather than Earth itself to iron out the fluctuations from the Moon (the average should be much the same but the standard deviation between years may be a bit smaller). I averaged the distance for each of the 60 years, and then for each decade tabulated the mean and standard deviation of the ten average distances for that decade, as follows.

_ MEAN _ _ _ STD DEV
149618792 _ _ 333
149618914 _ _ 430
149618957 _ _ 462
149618913 _ _ 434
149618867 _ _ 508
149618871 _ _ 398

Given this variability it would appear that if one is going to give the average distance in the first sentence, anything more accurate than 149.619 million km creates a false impression of the precision of "average distance". But even with that limited precision the AU falls short of the average distance by 102 20 thousand km. Vaughan Pratt (talk) 22:17, 21 December 2013 (UTC)

There are different numbers that are average. In an ellipse, the average over times is greater than the semimajor axis. Because of the low eccentricites the two are much alike. I think almost all, if not all, of the numbers given with 8-12 significant figures should get rounded off. Saros136 (talk) 07:41, 9 July 2014 (UTC)

## Longitude of ascending node

The article uses the value of -11.26 degrees for the longitude of the ascending node in J2000. The only independent (i.e. not Wikipedia-based) source I've been able to find for this is Dave Williams' Earth Fact Sheet---the other cited source has disappeared. NASA JPL however gives it as 0 degrees, which makes more sense. (How on earth could it be as large as -11.26 degrees?)

Anyone have any strong objections to going with NASA JPL's value of 0? Or alternatively an explanation of the difference between LAN in J2000 and Vernal Equinox in J2000?

The value of 0 would appear to be further confirmed at http://www.amsat.org/amsat/keps/kepmodel.html which says "So finally, the vernal equinox is nothing more than the ascending node of the Sun's orbit. The Sun's orbit has RAAN = 0 simply because we've defined the Sun's ascending node as the place from which all ascending nodes are measured." That's what I would have thought. What else could it mean? Vaughan Pratt (talk) 23:29, 22 December 2013 (UTC)

## Keplerian Elements

The Earth's orbit is described in Keplerian elements, a, e, i, etc. Are these averaged elements? Have perturbations due to Jupiter's gravity, say, been detected? Do the elements also display averaged rates or are they chaotic averaging out to zero? If the orbit is inclined, where does it cross through the solar reference plane? Is the same solar coordinate system used to describe the orbital elements of the other planets? Virgil H. Soule (talk) 21:44, 26 December 2013 (UTC)

Very slightly different Keplerian elements are given for different periods of time, and typically are best fit to the more detailed orbit over that period, with longer periods giving worse fits hence less accurate results when using those elements. The detailed orbit is calculated taking all the significant influences (Jupiter and Venus are the biggest but by no means the only ones) into account. Since the Moon has an obvious and large impact it is often preferable to give the orbit of the Earth-Moon barycenter (EMB) instead of Earth itself. Typically the reference plane is Earth's orbit (best fit), aka the ecliptic, in which there is no concept of Earth's orbit crossing the reference plane since it is the plane. Other planets however do cross the ecliptic. Vaughan Pratt (talk) 17:25, 30 December 2013 (UTC)

## Distance from the sun

I thought the earth was something like 93,000,000 miles from the sun but the article makes no mention of what I always thought was this basic fact. I mean, what's going on here? Isn't there supposed to be useful information on this site? To avoid pointless arguing please spare us all from political arguments in your responses. Thanks. — Preceding unsigned comment added by 71.83.58.170 (talk) 08:54, 9 January 2014 (UTC)

I've moved your question moved to the bottom of the page where new questions normally go.
I agree that 93 million miles is the best-known figure, but some editors will not allow us to put Imperial or customary units in our astronomy articles. (I think they are wrong, but they think they are right.) ... later note: The article does have miles for aphelion, perihelion and semi-major axis, so perhaps we will be allowed to put "about 93 million miles" in the lead? Dbfirs 12:27, 9 January 2014 (UTC)
... later ... I see that C Logan has added the distance in miles. I assure him/her that it's not just Americans who are familiar with miles. We'll see whether the "metric police" revert the edit. I hope they don't. Dbfirs 06:41, 3 April 2014 (UTC)

## distances

The method of using elements and assuming a Keplerian orbit has limits. With five figures, the numbers agree with the EMB averages (from Solex) and VSOP87 and Standish's best fit (the source here). You could make the case for six significant figures, but definitely not seven. Saros136 (talk) 06:45, 26 July 2014 (UTC)

## Those aren't mean solar days.

Length of the sidereal year cited does indeed come from the 1994 VSOP87 paper by Simon et. al. as some sources say. The paper did not present the value; it was derived from the VSOP87 elements presented. To derive a sidereal year from elements takes the rate of change of longitude, from that fixed to the J2000 equinox. The unit of time is the Julian millennium of 365250 days of 86,400 s of uniformly flowing time. Not the mean solar day .And get the rate of change at one instant, in days per units of angular change (such as degrees) , by dividing 365250 date by that rate. Then multiply by 360 (if using degrees) to get the days required for the longitude to change by that much.

VSOP87 quantities are calculated as polynomials in powers of T (time) There is a constant+terms in T,T^2,T^3, etc. When T=0, the rate of change is equal to the coefficient of the first degree term. They have it in arc seconds. 1295977422.83429 of them. I want to use degrees, converting to them (dividing by 3600) I get degrees. Dividing this into 365250 gives the days per degree and multiplying by 360 gives the result of 365.256363004 days. Note this equivalent to dividing the number of arcseconds into 473,364,000,000.

Getting the anomalistic year is similar, but the mean anomaly is used. That is equal to rate of change of L -pi (the longitude of perihelion). Saros136 (talk) 05:43, 27 July 2014 (UTC)

## 1.5 Gm ?

"The Hill sphere (gravitational sphere of influence) of the Earth is about 1.5 Gm (or 1,500,000 kilometers) in radius."

Is this text correct, or should it be fixed, I do not know, so I ask? — Preceding unsigned comment added by 46.10.229.1 (talk) 08:59, 15 September 2014 (UTC)