# How much change in Earth's orbital distance to change average temperature by 1°C?

How much closer/farther would the Earth need to be to/from the Sun to effect a $$1 \sideset{^{\circ}}{}{\mathrm{C}}$$ increase/decrease in average temperature?

• This question is not well defined as all else "being equal" is more or less impossible. Even a minute change of 1 degree modifier the cycle of water, the amount of ice and permafrost, and the earth average albedo. Keep in mind that seasons are a result of a strong modulation of the energy flow due to oblicuiylty of the earth axis (with respect to ecliptic plane) and not of the non circular shape of the orbit which modulates the earth-sun distance. – Jhor Mar 5 at 7:34
• If you have a Sun-Planet distance to temperature relationship you can compute this. A very rough and probably wrong estimate could be the mean of all planets in the solar system. This is assuming a lot of things which will obviously be wrong on a 1 degree precision. – Bonsay Mar 5 at 11:13
• Note that a 1 degree average change at the equator may mean a quite different average change at the poles (and all points in between). Likewise 1 degree average in summer may not be 1 degree in winter. – StephenG Mar 5 at 12:49
• @Jhor all else being equal means keeping the earth the way it is now and just changing the average distance from the sun. if water cycle etc. changes, that's part of the calculation – michael Mar 5 at 18:49
• As I said and that other comment emphasize, one degree increase while keeping the climate unchanged is allmost impossible, and climate change can induce much larger temperature change that ca – Jhor Mar 5 at 22:55

## 2 Answers

If you want to do this properly you have to take into account the Earth system, which is, to put it mildly, extremely complicated, and probably off-topic for Physics SE.

But being physicists we like oversimple approximations, especially when they give answers which are approximately correct. One such is to treat a planet as a perfect black body, and use this to calculate how hot it should be at a given distance from the Sun.

First of all, given the known top-of-atmosphere flux of power from the Sun, $$S$$, we can, by integrating over a sphere whose radius, $$R$$, is the Earth's orbital radius (semimajor axis), compute the total power output of the Sun:

$$P_S = 4\pi R^2 S$$

We know $$S \approx 1360\,\mathrm{Wm^{-2}}$$ and $$R \approx 1.50\times 10^{11}\,\mathrm{m}$$ and this gives $$P_S \approx 3.85\times 10^{26}\,\mathrm{W}$$.

Now we can use the black-body formula and the above expression for the flux in terms of $$P_S$$ and remember that only half of the planet is illuminated to compute the predicted temperature of a perfect black-body planet:

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

And we can check this for Earth, and we get $$T \approx 278\,\mathrm{K}$$: this is too cold but is clearly in the ballpark (it is entertaining to do this for Venus, where the answer is not in the ballpark). So the approximation is not hopeless.

We can invert the above expression to get $$R$$ as a function of $$T$$:

$$R =\frac{1}{T^2}\sqrt{\frac{P_S}{16\pi\sigma}}$$

And then we can just plug in numbers to this: for an increase of $$\Delta T = 1\,\mathrm{K}$$ the $$\Delta R \approx -1\times 10^9\,\mathrm{m}$$, and for a similar decrease, $$\Delta R \approx 1\times 10^9\,\mathrm{m}$$: this is about a million km, or about $$1/138\,\mathrm{au}$$: it's rather under one percent of $$R$$ in other words.

I will emphasize once again: this is insanely oversimplified: you can't treat planets with atmospheres, life, oceans &c this way and expect to get reasonable answers: look, again, at Venus (which has only some of these things) for an example.

(And, just in case, no, climate change is not being caused by the Earth moving closer to the Sun!)

• I fear to have detected some errors is your answer. First, owing to the definition of solar constant (your S), you should multiply it by the area of the disk πR^2 and not that of the sphere 4πR^2.. This factor 4 is not a big issue, for an order of magnitude, and further more it is exactly compensated by another factor 4 in the second formula. (To be continued) – Jhor Mar 6 at 14:54
• A much more serious mistake (or misunderstanding) is the use of R : the new radius that we are looking for is the orbital radius and not the earth radius. Hence the answer is very strange, an at least does not answer the OP question, and is moreover unconsistant with your final comment. The radius evaluated in my answer is 1/150 UA, which is of the same order of magnitude of the difference between the minor and major axes of the earth's orbit, that i already feel quite small, and yours in so smaller... – Jhor Mar 6 at 14:54
• @Jhor: the power output of the sun is the integral of $S$ over the surface of a sphere whose radius is the radius of the Earth's orbit, $R$. Throughout the question I have used $R$ as the orbital radius of the Earth not its radius, which does not enter into the calculation (small planets have the same surface temperatures as large ones in this approximation). However I have clearly got km and m mixed up: I'll fix that. – tfb Mar 6 at 15:06
• OK, i didn't pay attention to the "total power output of the Sun:". I don't see the reason to introduce it, but with this definition the fist displayed equation obviously seems correct. Nevertheless don't clearly understand the reasoning leading to your second equation. I guess that you don't mean T_S as written below the numerical value of P_S, an the cross section of the sphere disappear as both the incoming power and the emitted one ae proportional to it ? – Jhor Mar 6 at 15:42
• I've fixed it: it was fine in fact (I think I had a $T_S$ in the text when I meant $P_S$ somewhere but all the maths was fine). But Mathematica was giving me some incomprehensible answer which I blindly trusted even though it was obviously insane (10km, seriously?). The answer now agrees with yours, which is what it should do. The reason for the general expression in terms of $P_S$ is you can plug, say, Mars into it (it's reasonably OK for all the planets but Venus in fact). – tfb Mar 6 at 16:14

We can make a very crude order of magnitude of the required extra energy $$δE$$ as such : the mass of atmosphere, oceans, and polar ice, are respectively about 50, 1500 and 30 Ekg ( 1 Ekg = one exakilogram = $$10^{18}$$ kg). Assuming specific heat capacities or 1, 4, and 2 kJ/K/kg , neglecting the temperature change in the earth itself, an increase of temperature by 1K will correspond to an energy of $$δE\sim (50×1 + 1500× 4 +30× 2) 10^{21} \sim 6 ×10^{24}$$ J. To the which should be added the enthalpy of fusion of one part of polar ices. Assuming 1/3 of ice melt, ant with the latent heat of ~300 kJ/kg, this results in $$\sim 30× 300/3 10^{21} = 3 10^{24}$$ J, for a total of about $$δE~10^{25}$$ J. Notice that this estimate neglects the heating of the rocks, and the melting and heating of the permafrost, which, among others, could significantly increase the total required energy.

On the other side, the average power flow received from the Sun is presently of about $$1 \mathrm{kW/m^2}$$, and multiplying by the cross section $$π R_t^2$$, one get a total power of about $$170$$ PW ( 1 PW = one petawatt = $$10^{15}$$ W). The other sources of heating can be neglected in this simple calculation

Now, can we convert the above calculated energy into a power ? This would require a sophisticated model, which will be very sensitive to climate change, and that can't be undertaken here. But let's us make a few comment about it. One key parameter is the albedo coefficient which measures the fraction of power which is immediately reflected toward space without heating the earth. It is presently of about 35%, but will decrease significantly if the surface covered by clouds or by ice is reduced. One part of the absorbed energy is re-emitted, and the relationship between radiated power and an temperature can be modeled. But the problem is in fact much more complicated, as a significant part of the incoming radiation is absorbed by atmosphere (see Wikipedia). Similarly a part of the re-emitted radiation is absorbed or reflected back y the atmosphere and clouds : this part corresponds to the so called greenhouse effect, evaluated presently to about 5% of the incoming power. It was originally a benefic natural effect, but has dramatically increased by human activities and will increase again a lot, because heating the earth, atmosphere and ocean, displaces many subtle equilibrium. Among others, we can neither know with sufficient precision how much the excess CO2 will be captured by ocean, nor how much methane will be released in atmosphere by the melting of permafrost.

Any way, the short answer is that we can not convert this energy into power.

But this is not required. If climate change is ignored, and the present parameters of earth are used, and one assume that the present energy's balance is at equilibrium. In this restrictive context, we can use the relationship between radiated power and temperature stated for a black body by the Stefan law ($$P \propto T^4$$). Earth is clearly not a black body, and the "radiated power" is not the real power sent back to space, but if "all else being equal" (especially same albedo, same greenhouse effect, etc. which is completely unrealistic) the unknown coefficient must be the same in the two considered situations.

Hence the calculation is straightforward: the relative increase of power is about $$δP/P = 4 δT / T$$, and as the power flow is proportional to $$1/D^2$$, $$δP =-2 δD/D$$. As the black body model of earth (once included all side effect) corresponds to an effective temperature close to 300K, we get : $$\frac{δD}{D} \approx -2 \frac{δT}{T} \approx -1/150$$ As $$D=1 \mathrm{UA}=150\times 10^6$$ km, we get $$δD \approx -10^6$$ km.

This value could seem very large, but considering the eccentricity of the Earth's orbit $$e \approx 0.016$$, we observe that it is only $$50$$ times larger than the difference of the major axis and minor axis of the orbit.

Finally the energy can be used to evaluate the the characteristic time for reaching the new equilibrium : $$τ_c= δE/δP = 4 δE/P x T/δE \sim 10$$ days, which is again unrealistic.