Template talk:Planet Infobox/Earth

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This template was considered for deletion on 2006 October 9. The result of the discussion was to keep..

1 Imperial units
2 Planetary society
3 Circumference
4 Data for physical characteristics

[edit] Imperial units

My reason for adding Imperial units was simple: most of the measurements in the chart are all but totally worthless to the large minority of Wikipedians who are not familiar with the metric system. It does add a slight degree of clutter, but it's not that awful, and it really benefits those of us who have the misfortune to have grown up using the rather ridiculous Imperial system exclusively. The English Wikipedia is not an encyclopedia for non-American English speakers, it's an encyclopedia for all English speakers, and a large percentage of them are American.

As to Wikipedia:Manual of Style (dates and numbers)#Style for numbers, weights, and measures, it states that "It may be helpful to readers to offer the Imperial and metric equivalents, but this should not be done if it reduces the flow of a sentence or otherwise interferes with the quality of the writing." I really don't think that the inclusion of Imperial units significantly reduces the utility of the sidebar to metric users, while it dramatically increases it for Americans; the problem with the Imperial units is aesthetic only. —Simetrical (talk) 02:20, 6 Apr 2005 (UTC)

[edit] Planetary society

The infobox is so long that overwhelms the article. Can the Social section be chopped down to link to a suitable page? We only have a society on one planet so far. (SEWilco 05:32, 12 Apr 2005 (UTC))

[edit] Circumference

User:80.229.231.194 changed the orbital circumference based on some number he got from a web site to

924,375,700 km
(6.179 AU)

However, this circumference p can be calculated from the given semi-major axis a and eccentricity e. So if the number given here before (0.940 Tm) is incorrect, one of those would have to be wrong as well, and I don't think they are.

a = 149.597887 Gm
e = 0.01671022 = $\sqrt{1-a^2/b^2}$

[edit] approximate formula

$p = 2 \pi a \sqrt{1- \frac{e^2}{2}} = 939.886 \mbox{ Gm}$

This is an approximation, but it is actually quite close, especially for a fairly low eccentricity such as this. I'd guess that the first four digits would be accurate, at least. It isn't going to vary by 1.7% from the true value, which it would have to do to get this down to 924 Gm.

[edit] better approximation

Another good approximation: Roger Maertens' YNOT formula, Error < 0.4% (Information taken from: http://users.pandora.be/ronald.rousseau/html/perimeter_formula.html )

p = 4a ( 1 + (1-e²)^y/2)^1/y

with y = ln(2)/ln(pi/2) = 1.5349285356613752020529480451829(y is called the YNOT-constant) and a = 1.52095700 gm, e = 0.0167 Then p = 9.5557883413 gm

A better approximation is Ramanujan's:

$c \approx \pi \left[3(a+b) - \sqrt{(3a+b)(a+3b)}\right]\!\,$

which can also be written as:

$c \approx \pi a \left[ 3 (1+\sqrt{1-e^2}) - \sqrt{(3+ \sqrt{1-e^2})(1+3 \sqrt{1-e^2})} \right] \!\,$

In this case we get 939,885,629.9±0.4 km (using the uncertainty from the semi-major axis and eccentricity).

Urhixidur 04:50, 14 November 2005 (UTC)

That's the "inferior" of his two approximations: Did you see the other one I gave on Gene Nygaard's page (and updated comment)? ~Kaimbridge~ 17:50, 14 November 2005 (UTC)

That may very well be, but it is a harder one to type in. :-) May I suggest you upload Ramanujan's second approximation to the Ellipse article? Also, what we could use are gauges of the errors introduced by the approximations, compared to the actual elliptical integrals: a few lines saying something like « for e between such and such, the (first or second) approximation is accurate to 10^-x parts »...

Urhixidur 01:20, 15 November 2005 (UTC)

I just added them all to the circumference page, including comparison table. I used this UBasic routine:

    5   Point 12 ' ~=~ .1^57
   10   a=10000
   20   b=9966
  100   Oz=acos(B/A)
 1000   V=0:TN=0
 2000   V_o=V:V=V+a*cos(0.5*Oz)^2*.BC(0.5,TN)^2*tan(0.5*Oz)^(4*TN):?Using(,30),V,Left(Str(a*cos(0.5*Oz)^2*.BC(0.5,TN)^2*tan(0.5*Oz)^(4*TN)*10^30),36):If V=V_o Then 3000 Else TN=TN+1:GoTo 2000
 3000   ?:?"|-":?"|";B:?"||";Using(,10)," '''";V;"'''":?"|| '''";Using(,10),a*cos(0.5*Oz)^2*(1+3*tan(0.5*Oz)^4/(10+(4-3*tan(0.5*Oz)^4)^0.5)):?"|| '''";Using(,10),0.5*a*(6*cos(0.5*Oz)^2-((3+cos(Oz))*(1+3*cos(Oz)))^0.5):?"|| '''";Using(,10),a*(0.5*(1+cos(Oz)^1.5))^(1/1.5):?
 4000   List 20
63999   End
64000   .BC(UL_b,RG):V_b=1:For TN_b=1 To RG:V_b=V_b*(UL_b+(1-TN_b))/TN_b:Next:Return(V_b)

--~Kaimbridge~ 20:23, 17 November 2005 (UTC)

[edit] exact formula

$p = 4a \int_0^\frac{\pi}{2} \sqrt{1-e^2 \sin^2 \theta} \,d\theta$

where $E = \int_0^\frac{\pi}{2} \sqrt{1-e^2 \sin^2 \theta} \,d\theta$ is the complete elliptic integral of the second kind.

This integral cannot be expressed in terms of elementary functions, and needs to be calculated by numerical methods. Ages ago, I used to have this programmed into a calculator, but I'm not up to figuring out how to do it again now. From the table in the CRC handbook, an imprecise value, but accurate to this many digits, for this problem should be E = 1.5707 or possibly more (sin^-1 e is just less than 1°, and E is π/2 = 1.5808 for 0° and 1.5807 for 1°), which cannot be enough different to get this down to 924 Gm.

Since I'm a little rusty at this, I hope my formulas are correct and my calculations didn't include any errors. If not, at least there should be enough here for someone to correct me if I got it wrong, but since my numbers agree with what was here before, there is a pretty good chance I'm correct and that web page is just wrong. Gene Nygaard 15:45, 18 August 2005 (UTC)

Just to clarify: the lowest value E could have, given the CRC table values, is 1.57065. Substituting that into p = 4aE gives us a lower limit of 939.86 Gm for the perimeter. So I'd wager pretty heavy that all the digits are correct if we express it as 939.9 Gm. Gene Nygaard 15:57, 18 August 2005 (UTC)

[edit] Where the web site may have gone wrong

My guess is that the web site made this calculation by using

$p = 2 \pi a \sqrt{1- \frac{e}{2}}$ rather than : $p = 2 \pi a \sqrt{1- \frac{e^2}{2}}$

That formula would give 924.310 Gm for the values given here. Minor variations in the a and e values would likely give the result they got. So now the only guestion is to check for sure which formula is correct, for this e as used here. Gene Nygaard 16:14, 18 August 2005 (UTC)

[edit] Too much precision?

The astronomical unit is only good to 30 m, and the JPL value for the semi-major axis only has 8 digits after the period (i.e., it is 1.00000011). Combining these two uncertainties gives a semi-major axis precision of about 750 m. Accepting the extra two digits for the semi-major axis (...01124) gives a precision of 37.5 m. What's the source for the extra semi-major axis digits?

Secondly, note that the circumference bandied about in the infobox is heliocentric; if you look at the Earth's sidereal orbit, you'll get a considerably different answer, since the semi-major axis shrinks by nearly 450 km! Urhixidur 01:32, 15 November 2005 (UTC)

Good point. I think we should try to give these numbers with explicit error estimates. Fredrik 06:49, 15 November 2005 (UTC)

Everything appears based on the aphelion and perihelion. On NASA's page, they give the aphelion ("a") as 152,100,000 km and the perihelion ("b") as 147,100,000 km, which gives a median of 149,600,000 km, but for the "Average Orbit Distance" (which is the "Semi-major axis"), they give 149,597,890 km, which corresponds to the aphelion and perihelion values used here. The semi-major axis ("A") equals .5*[a+b] and the semi-minor axis ("B")—as prescribed here— is [ab]^.5 (which equals a*cos{asin{orbital eccentricity}}), so, here, A equals 149,597,887.5 km and B equals 149,576,999.826 km, which provides an orb.ecc'ty of 0.0167102192, matching NASA's 0.01671022 (whereas rounding B to 149,577,000 km gives 0.0167101496).

Same thing with the AU values: AU is given as 149,597,870.691 km ± 0.030 km. So, A÷AU can vary from 1.00000011216 to 1.00000011256, giving an average of 1.00000011236 AU.

Since we are supposed to be a reference, I would think that we should be giving the most reasonably precise, distinguishable values as possible, letting the user then round to whatever they want. ~Kaimbridge~ 13:45, 15 November 2005 (UTC)

[edit] Data for physical characteristics

Somewhere in this template, it needs to mention the sources for the various measurements. This is especially important because they seem to disagree often: see for example several possible values at [1].

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