How much would Earth's orbit have to have decayed to explain the Global Warming data? I've seen many terrestrial based explanations of global warming, but never an orbital one.

  • 4
    $\begingroup$ badastronomy.com/mad/1996/earthburn.html $\endgroup$
    – pentane
    Commented Sep 29, 2016 at 20:30
  • $\begingroup$ Changes in Earth's orbit are important for understanding historical climate information, going back about a million years, but do not explain climate changes in recent decades. $\endgroup$
    – rob
    Commented Sep 30, 2016 at 14:45
  • $\begingroup$ Just for quick notes/references: Earth's orbit ranges from ~1.47 to 1.52 $\times 10^{11}$ meters, or a difference of $5.01 \times 10^{9}$ meters (~311000 miles) over the course of a year [Mathematica v11.0.1 on 09/30/2016]. The difference between Earth's and Venus' average orbital distance is $4.14 \times 10^{10}$ meters (~25.7 million miles or ~0.277 AU). $\endgroup$ Commented Sep 30, 2016 at 15:11

3 Answers 3


In summary: no, this is not a plausible mechanism as large enough orbital changes to account for observed changes would change the length of a year really significantly and we'd notice this.

The right answer to this would be to understand how the solar forcing (the power at the top of the atmosphere from the Sun) depends on orbital radius (easy) and to understand what the climate response is to this (hard). If I have time later I will add a section to this answer which summarises some of that information. But it turns out you can do a nice physicist's answer which is adequate to show that this is absurd (which doesn't mean it's a bad question!).

The simple-minded physicist's answer

Assume the Earth is a, possibly slightly imperfect, black body at a temperature $T$, in a nearly-circular orbit of radius $R$, and is distant from the Sun (which it is). The incoming power from the Sun goes like $R^{-2}$, and the power radiated from the Earth goes like $T^{4}$. From this we immediately get


Where $\alpha$ is to be determined. For Earth we know that $T \approx 287\,\mathrm{K}$ (this is about 14 degrees C) and $R \approx 1.5\times 10^{11}\,\mathrm{m}$. Plugging these in we get $\alpha = 1.11\times 10^{8}$ (in some units I don't want to think about).

(It's worth pointing out at this point that if you simply treat the Earth as a perfect black body at a constant temperature (so it's completely conductive and doesn't have a hotter and colder side) you can work out that what the temperature should be is in fact

$$T =\left(\frac{P_S}{16\pi\sigma R^{2}}\right)^{1/4}$$

where $P_S$ is the power output of the Sun and $\sigma$ is the Stefan-Boltzmann constant. This gives $T\approx 278\,\mathrm{K}$ which is remarkably close: all the planets except Venus are reasonably close in fact. So this justifies the simple-minded approach here: it's not that far wrong.)

So now, rearranging the above formula:

$$R = \left(\frac{\alpha}{T}\right)^{2}$$

And we can plug in our known values and check the right answer comes out, which it does.

So, OK, let's assume 1.5 degrees of warming, which is plausible for current warming vs preindustrial. So now we want to compute $R_w$, the orbital radius for a warmer Earth:

$$\begin{align} R_w &= \left(\frac{1.11\times 10^{8}}{288.5}\right)^{2}\\ &=1.48\times 10^{11}\,\mathrm{m} \end{align}$$

So the orbital radius would need to change by about $2\times 10^{9}\,\mathrm{m}$ to account for this warming.

How much is that? Well, from Kepler we know that the orbital period, $P$ (normally this is $T$ but I've used that) is

$$P = 2\pi \sqrt{\frac{R^{3}}{GM}}$$

where $G$ is Newton's gravitational constant, and $M$ is the mass of the Sun. So $G = 6.67\times 10^{-11}\,\mathrm{N}\mathrm{m}^{2}/\mathrm{kg}^{2}$, and $M = 1.99\times 10^{30}\,\mathrm{kg}$.

Plugging in my original orbital radius gives a year of $3.17\times 10^{7}\,\mathrm{s}$ which is close to the right answer. Plugging in the $R_w$ from above gives a year of $3.11\times 10^{7}\,\mathrm{s}$. So the year length would get shorter by about $6\times 10^{5}\,\mathrm{s}$: about a week.

I am surprised by how large this difference is: I was expecting something easily detectable but reasonably small like an hour or something, not a week. It may be I've made some numerical or other error above.

However the point remains: this isn't a plausible mechanism.

A simple-minded climate modeller's answer

Here is an answer that uses a simple-minded (but not completely implausible) approach based on climate modelling.

The first notion you need here is that of radiative forcing: this is essentially the difference in the insolation (the energy coming in from the Sun at the top of the atmosphere) and the energy being radiated back out at the top of the atmosphere. At equilibrium it needs to be zero: in practice it is not zero because the system is not at equilibrium (there is weather, seasons and so on), but unless the long-term average of it is zero then the system will warm or cool over the long-term ('long-term' here means decades: anything shorter than a decade is likely to be swamped by the chaos which we call 'weather').

Everything is measured in watts per square metre: radiative forcing is in fact the energy flux per unit area at TOA, averaged over the planet's surface for our purposes (obviously a climate model would want to not do that averaging but to average over whatever the grid size of the model is).

One component of the radiative forcing -- and the only one we're going to care about -- is the solar forcing, which is the incoming radiation from the Sun with the radiation immediately reflected subtracted from it. I'll call this $F_S$. There is a simple formula for this:

$$F_S = S\frac{1-A}{4}$$

where $S$ is the solar constant, which is the flux crossing the Earth's orbit ($S$ is what we are going to tweak), and $A$ is albedo (I don't know what the usual symbol for this is). The factor of $4$ is because the Earth is a sphere: one half is not getting any sunlight at all, and the average over the lit hemisphere is only a half of the incoming energy (this is easy to work out). For the Earth $A \approx 0.3$, so we get

$$F_S = \frac{0.7}{4}S$$

So that's the solar forcing.

The next thing to know about is the climate sensitivity: this how the surface temperature (which is what we care about) depends on changes in the forcing. Because all of the changes are rather small, it turns out to be sufficient to express this as a linear relationship: there are obviously higher-order terms but they don't matter. I always find this a bit surprising, but it's actually perfectly well justified. So we can write an equation like

$$\Delta T = \lambda \Delta F$$

Where $\Delta T$ is temperature change, and $\Delta F$ is change in forcing.

Now you'll need to take my word for it that a value of $\lambda$ of about $0.8$ is reasonable: this is a value which gives about $3$ degrees of warming for doubling of atmospheric carbon dioxide, which is somewhere around reasonable. However climate sensitivity is something we don't completely know.

So again, let's assume one and a half degrees of warming over preindustrial climate. And we get

$$\begin{align} \Delta F &= \frac{1.5}{0.8}\\ &\approx 1.9\,\mathrm{W}/\mathrm{m}^{2} \end{align}$$

And now we will assume that all of this comes from the Earth's orbit changing: nothing else changes. This assumption should fill you with fear, but all we're after is a ball-park figure remember. So what we have is

$$\Delta F_S \approx 1.9\,\mathrm{W}/\mathrm{m}^{2}$$

So now we can just replicate some of the stuff we did for the black-body model:

$$S = \frac{\alpha}{R^{2}}$$

And we know that $S_0 = 1370\,\mathrm{W}/\mathrm{m}^{2}$, and $R_0 = 1.5\times 10^{11}$ (I am using $0$ suffices for 'initially', and note that this $\alpha$ is not the same as the $\alpha$ in the previous section, sorry). So we can calculate $\alpha$:

$$\begin{align} \alpha &= 1370\times\left(1.5\times 10^{11}\right)^{2}\\ &\approx 3.1\times 10^{25} \end{align}$$

And now we know that $S = F_S\times 4/0.7$, so

$$\begin{align} \Delta S &= 1.9\times 4/ 0.7\\ &\approx 11\,\mathrm{W}/\mathrm{m}^{2} \end{align}$$

And finally we can calculate $R_w$:

$$\begin{align} R_w &= \sqrt{\frac{\alpha}{S + \Delta S}}\\ &= \sqrt{\frac{3.1\times 10^{25}}{1370 + 11}}\\ &= 1.49\times 10^{11}\,\mathrm{m} \end{align}$$

Which gives a difference of about $1\times 10^{9}\,\mathrm{m}$ -- about half what the black-body approach gave, but clearly comparable with it -- corresponding to a difference in year of a few days.

So again, this isn't a plausible mechanism.


I have been very sloppy about significant figures and occasionally sloppy about units (omitting some inconveniently complicated ones). Sorry. I would, of course, welcome corrections!

All the numbers were scraped from Wikipedia or some other search engine: I think they're reasonable but they may not be completely.

The climate-modelling approach contains two enormous omissions which will make climate-modelling people (full disclosure: I work somewhere that does climate modelling) go white.

  • Climate sensitivity, my $\lambda$, is a number we just don't know very well, and it's terribly important as it makes the difference between we-are-all-going-to-die and only-poor-people-will-die. The number I have chosen is I think uncontroversial (except among people being paid to say otherwise).
  • I just assumed that the change in radiative forcing was all due to the Sun. If that was true then we could all go home, because we know in practice (since we measure it) that $S$ is rather stable, as the Earth is in fact not spiraling into the Sun (The Clash not withstanding). In fact computing the forcing, $F$ is hugely complicated, and it's more-or-less what climate models do, more-or-less well. As an example, if $F_S$ were to go up, then you end up with more water-vapour in the atmosphere (as it heats), and this changes albedo (as some of it will be clouds) and also is a powerful greenhouse gas in its own right. So you are immediately dropped into some hugely nonlinear fluid-dynamics problem. And this is just one mechanism that we at least understand in principle. That's why climate modelling isn't simple, and matters.
  • 1
    $\begingroup$ Fascinating - despite your admitted "sloppiness" with the math, I'm intrigued by the process you used to determine this. Slightly related: I remember that episode of Futurama where all the robots had to blow exhaust (out of their hind quarters!) in unison, all pointed toward the Sun, to push Earth's orbit outward to the tune of making the year one week longer, in an attempt to counter global warming. ;-) $\endgroup$
    – pr1268
    Commented Sep 30, 2016 at 10:34
  • $\begingroup$ @pri268: do you have a reference to the Futurama episode? I've failed to watch them all, but if they were adjusting the year by a week it looks like someone had done the maths! $\endgroup$
    – user107153
    Commented Sep 30, 2016 at 13:07
  • $\begingroup$ The unit you don't want to think about is obviously "SI units" $\endgroup$ Commented Sep 30, 2016 at 17:04
  • $\begingroup$ To quickly compute the physicist's numerical value, you can simply say ΔT/T=.5%, so ΔR/R=1% (since it's quadratic in T) and ΔP/P=1.5% so 5-6 days $\endgroup$ Commented Sep 30, 2016 at 17:11
  • $\begingroup$ @tfb: Season 4 Episode 8. imdb.com/title/tt0584437 en.wikipedia.org/wiki/Crimes_of_the_Hot $\endgroup$
    – pr1268
    Commented Sep 30, 2016 at 17:17

It's winter in the Northern Hemisphere and we're at our closest point to the Sun. Closest? Yes, you read that right. Closest. For northerners, the winter solstice has just passed. But the truth is, on January 3, 2007, Earth reaches perihelion, its closest point to the Sun in its yearly orbit around our star.

Earth reaches perihelion on January 3, 2007 (Figure C). The Earth-Sun distance will be 147,093,602 km. Aphelion, the greatest distance from the Sun, occurs on July 7, 2007, when the Earth-Sun distance will be 152,097,053 km.

So from this it is evident that it is not distance that controls the average temperature in the hemispheres, certainly the winter/summer change is much larger than the tiny with respect to it, global warming observed .

That is why no model has been developed on the lines "the earth may be nearing the sun". It is obvious that the temperatures observed have to do with a lot of other factors affecting the atmosphere and the surface, like albedo etc and not just the distance to the sun. The answer by tfb shows roughly that our calendars would have changed had such a thing occurred.

Have a look at this site to see what is causing the seasons.


If our planet went out of orbit, we would have been fried or frozen. There is no possible way with out us all dying to explain global warming. for every mile forward our planet goes, it goes 1/9 of a inch towards the sun, and if we went 1/8 of an inch, we would be fried. if we went 1/10 of an inch, we would freeze.

  • $\begingroup$ it's precisely this tiny fraction of movement that makes me wonder, afterall there is a myriad of forces acting on the planet, and it's orbit is not at all smooth and consistent $\endgroup$
    – SumGuy
    Commented Sep 29, 2016 at 23:11
  • $\begingroup$ i dont believe earths orbit decayed. its just Chloroflourocarbons from 20 years ago in hair sprays and other 90's junk. $\endgroup$
    – muellwm
    Commented Sep 29, 2016 at 23:16
  • $\begingroup$ as far as a reference, it was from a book i read last year, though i should double check on the fractions. it was either it is right now 1/9, or 1/8. but either way it takes a minute amount of change in the orbital to seriously affect this planet. we are in the perfect position $\endgroup$
    – muellwm
    Commented Sep 29, 2016 at 23:20
  • $\begingroup$ The numbers you give here can't be correct. Earth's orbit has a non-zero eccentricity, and the radial difference between its closest and furthest points from the Sun is much more than one mile. $\endgroup$
    – HDE 226868
    Commented Sep 30, 2016 at 1:18
  • $\begingroup$ my question isn't "did this cause global warming" and it isn't "did it move" but is "how far would it have to have gotten closer to match the data on global warming" it's a physics question, not a history question :D $\endgroup$
    – SumGuy
    Commented Sep 30, 2016 at 1:45

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