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Wobbles, part 1

November 19, 2007 · 13 Comments

The question arose recently, how does the total energy received by the earth depend on the eccentricity of earth’s orbit? I’d like to show how this is derived. I’d also like to show the entire impact of eccentricity, precession, and obliquity (the tilt of earth’s axis) on how incoming solar energy is distributed around the planet. But since there seems to be a request for the eccentricity piece, I’ll start with just that.


The intensity of solar energy at a distance of 1 astronomical unit (the average distance from earth to sun) is the solar constant; let’s call it S. This is the energy received by a unit area directly facing sun. If the radius of the earth is R, then the cross-sectional area is \pi R^2, and the total energy intercepted by earth is S \pi R^2. But this energy must be spread over the entire surface of the earth, which has area 4 \pi R^2. So, the solar insolation, when averaged over the entire surface of the earth, when earth is exactly 1 astronomical unit from the sun, is

I = S (\pi R^2) / (4 \pi R^2) = S/4.

Because earth’s orbit is not perfectly circular, the planet is not always exactly 1 astronomical unit from the sun. When earth is at a distance r (in astronomical units), the average insolation is

I = S/(4r^2).

To get the annual average, we need to average this expression throughout the year. This is

J = \frac{1}{T} \int_0^T I ~dt = \frac{1}{T} \int_0^T S/(4r^2) ~dt,

where T is the time required to pass through one orbits. We know from Kepler’s 3rd law that the length of the year is T = 2\pi \sqrt{a^3/\mu}, where \mu is the “gravitational mass” of the sun-earth system. Measuring time in years and distance in astronomical units, the gravitational mass of the sun-earth system is \mu = 4\pi^2, and the semi-major axis of earth’s orbit is a=1 astronomical unit, so the length of time required for earth to traverse one orbit turns out to be 1 year.

Let \lambda be the longitude of the earth along its orbit. Then the rate of longitude change d\lambda / dt is related to the distance r by the conservation of angular momentum

r^2 {d\lambda \over dt} = h,

where h is the specific angular momentum. Working out the mechanics of orbital movement, it’s possible to show that the specific angular momentum is h = \sqrt{\mu a (1-e^2)}.

We now have

dt = {r^2 \over h} ~d\lambda,

and as earth goes through one entire orbit (as time goes from 0 to T), its longitude goes from 0 to 2\pi (radians), so that

J = \frac{1}{T} \int_0^{2\pi} (S/4) ~{d\lambda \over h} = {S \over 4Th} \int_0^{2\pi} d\lambda = {2\pi S \over 4Th}.

Now we can apply the values of T and h, and we are left with

J = S / (4 \sqrt{1-e^2}).

That’s where the factor 1/\sqrt{1-e^2} comes from, due to the eccentricity of earth’s orbit. As eccentricity gets bigger, the annual global average solar insolation increases.

Earth’s eccentricity e varies over time, showing both an approximately 100,000-year cycle and a 400,000-year cycle. At its lowest, it’s nearly 0, but it can get as large as a little over 0.06. When e=0, the average insolation is S/4, but when eccentricity gets as large as 0.06, this increases to 1.0018 ~S/4. The “climate forcing” from solar insolation is about 240 W/m^2 (watts per square meter). An increase due to eccentricity of 0.06 will add an extra 0.43 W/m^2 of climate forcing. That’s considerably less than the additional climate forcing from man-made greenhouse gases.

Changes in precession and obliquity have absolutely no effect on the average solar insolation over the entire globe. But they have a profound impact on the geographical distribution of the incoming solar energy. In particular, they can greatly increase or decrease insolation at high latitudes, where earth’s ice masses reside. This can lead to accumulation or disintegration of ice masses, and that can in turn change the albedo of the planet, earth’s reflectivity. This can cause more of the solar insolation to be absorbed into the climate system or reflected back to space, and hence can strongly impact net climate forcing through albedo change.

In another post (soon!) I’ll show how these come about, and discuss some of the research into how these changes trigger the waxing and waning of ice ages.

Categories: Global Warming · climate change

13 responses so far ↓

  • george // November 19, 2007 at 7:59 pm

    Thanks for the explanation.

  • glorkspangle // November 20, 2007 at 11:22 am

    Is a fixed over geological timescales? This is really two questions:
    1. As the eccentricity changes, what distance parameter of the earth’s orbit is fixed? Semi-major axis (a), semi-minor axis, or some other parameter? There’s probably a trivial conservation of energy answer to this, which I’m too lazy to do.
    2. Are there other variations in the size of the earth’s orbit over geological timescales, and if so how large are they? Tidal deceleration?

    [Response: The energy of an orbit is a function of semi-major axis, so it's "a" that is constant over geologic time scales. I do believe that the planetary orbits are such that this will continue to hold for billions of years both past and future, and that this fact was originally demonstrated by LaPlace.

    Calculations of orbital parameters as much as 50 million yr past and future have been published, and they don't even bother to report "a" -- it's just constant.]

  • Nick Barnes // November 20, 2007 at 12:27 pm

    BTW, “glorkspangle” is me.

  • someareboojums // November 20, 2007 at 7:52 pm

    I had to look at your page source to figure out how you got such beautiful inline LaTeX equations into your post. And I see you’ve known about this since February. Way to make me feel late to the party once again, Tamino.

  • Alexander Ač // November 20, 2007 at 10:27 pm

    of course, changes associated with eccentricity of Earth were just initial in the onset of warming or cooling - the rest is done by feedbacks…

  • S2 // November 21, 2007 at 10:15 pm

    Measuring time in years and distance in astronomical units, the gravitational mass of the sun-earth system is mu = 4pi^2, and the semi-major axis of earth’s orbit is a=1 astronomical unit, so the length of time required for earth to traverse one orbit turns out to be 1 year.

    I struggled with this for a bit, but I think I’ve got it now - mu = 4pi^2 only because we are 1 au from the sun and take 1 year to complete an orbit.

    If we took Mercury or Neptune instead, mu would still approximately equal 4pi^2 provided we set T to the period of the planet and a to it’s semi-major axis.
    (maybe slightly less for Jupiter and Saturn, as mu also equals G(M+m), and the mass (m) of a Gas Giant, while large compared with Earth, is still small compared with the mass of the Sun (M).

    I do hope this doesn’t make me look pedantic - this isn’t my intention - I’m just trying to make sure I understand correctly.

    [Response: Looks right to me.]

  • Thomas Palm // November 22, 2007 at 2:52 pm

    Laplace thought he had proven the stability of the solar system, but he had to do some simplifications to do so, and the terms he neglected has been shown not to be neglectable.

    We now know the solar system is chaotic, but simulations suggests that nothing dramatic should happen in these parts for the next several billion years.

  • Steve Reynolds // November 23, 2007 at 1:12 am

    >Calculations of orbital parameters as much as 50 million yr past and future have been published, and they don’t even bother to report “a” — it’s just constant.

    How can ‘a’ be exactly constant when ‘e’ changes? I’m fairly sure that an orbit with a given ‘a’ with e=0 has a different energy than an orbit with e=1.

    Or do these changes in ‘e’ come from aquiring energy from other bodies?

    [Response: Actually, the value of "a" alone determines the energy of the orbit, "e" has no effect. Changes in "e" come from exchanging angular momentum with other bodies, with no net change in energy.]

  • Andrew Dodds // November 23, 2007 at 10:47 am

    Thomas -

    I thought Mercury was the least stable of the planets orbit-wise, and is a possible candidate for ejection and/or collision with Venus within a billion or so years.

    A collision between Mercury and Venus would probably be a bad thing for Earth, what with us getting bombarded with ejecta.. but I wouldn’t lose too much sleep about it right now.

    http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=60054

  • ron lobeck // November 23, 2007 at 2:07 pm

    It would be much appreciated if you could explain how(if it applies) the inverse square law affects the radiation arriving at the surface. My understanding is that there is about 1 million miles difference in the distances from the sun due to the elliptical orbit. My rudimentary calculations suggest that using the Inverse square law there is a 2% difference in the amount of energy arriving. Is this the case or have I made some error?

    [Response: In the next post (coming in a few days) I'll delve into the details of issues such as this.]

  • Alan D. McIntire // November 24, 2007 at 4:17 pm

    To ron lobeck: You’re right on your inverse square law computation. The discussion wasn’t about insolation at a particular time of year, but the “average” over the course of a full year.- A. McIntire

  • Goffers // November 25, 2007 at 11:01 am

    Off thread I know, but I don’t know whether anybody is still looking at “The Frozen North”

    The recent press release from NASA JPL talks of an upcoming reversal of the Arctic Oscillation. They say also that there is evidence that such changes explain the behaviour of the arctic in recent deades without recourse to AGW.

    However, looking eg at your excellent graph of ice anomaly I don’t see how their story fits in with the actual experienc since 1979. Do you have any comments?

    Also, thank you for CO2 figures on another website. I have now found the NOAA data.

    [Response: Changes in the AO appear to explain *some* of the recently observed changes in arctic. But by no means do they explain either the dramatic (and apparently unprecedented) loss of northern-hemisphere summer sea ice, or the extremely rapid rise in arctic temperatures recently.

    The fault lies with whoever wrote the press release, who apparently tried to make the recent research sound far more dramatic than it really is. This of course is then blown out of proportion by the denialist blogosphere...]

  • John Dodds // June 12, 2008 at 11:13 pm

    From Realclimate
    Re 120, & 126 tamino — 7 June 2008 @ 12:12 PM

    OK Tamino please explain whey my Ephemeris program shows that the earth eccentricity is continuously DECREASING for the last 300 plus years, but the TSI data from IPCC shows a net and fairly regular INCREASE in solar energy coming in.

    Then how can we explain that your equation results in a divide by zero result when eccentricity =1, for a parabolic orbit where the object never returns. A logical answer would be around 1/2. and even less when the orbit is a hyperbola, just grazing the sun at ~1AU., when your equation says the result is negative, whatever that means.

    Could your derivation have lost a minus sign in the integration so the real relationionship is that energy is proportional to 1/(1+e^2) instead of minus. - Just as BPL said originally (ie increasing energy for a decreasing eccentricity)

    [Response: First of all, TSI data don't show a fairly regular increase. The reconstruction of Lean et al. indicates an increase in the first half of this century, but TSI has been stable since about 1950 (as is confirmed by satellite measurements). And Dr. Svalgaard even argues against the TSI increase indicated in the reconstructions.

    Second, TSI measures the output of the sun at 1 astronomical unit, not the energy intercepted by earth.

    The equation gives the solar insolation, averaged throughout the year, for an orbit with varying eccentricity e and *constant* semi-major axis a. A parabolic orbit has energy E=0 so its semi-major axis is a=infinity. An orbit with e=0 and a=1 is *not* a parabola, it's an object falling directly into the sun. Under such circumstances, if we take the solar radiation at distance r as proportional to 1/r^2, then the insolation for such an "orbit" would indeed be infinite. But this doesn't really apply, because once you penetrate the surface of the sun the radiation no longer goes as 1/r^2; the total radiation would be limited by the temperature at the center of the sun (about 27 million K if I recall correctly).

    The formula is not only derived here, it's in many places in the scientific literature. The sign is correct, the equation is correct, I suggest you accept the truth.]

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