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I'm looking for a qualitative understanding. Suppose I put an object which was at $0$K half way between earth and sun. Initially the object sucks up more photons and emits fewer photons. After some time it reaches equilibrium. Now would the temperature of this object stay at $2.7$K ?

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No. That is the temperature it would reach if it was exposed only to the cosmic microwave background. In other words, if it was far away from any stars or other sources of light.

The sun is very hot, and the light released from it reflects that. Your object will get rather a lot hotter than $2.7~\rm K$, as it reaches equilibrium with the sun's light, not with the cosmic microwave background.

The actual temperature it reaches depends on a lot of factors, like how much light it reflects and how much of its surface area is illuminated. If it is a perfect blackbody, with half of its surface area illuminated normally (in other words, it's a flat disk), and the whole object reaches thermal equilibrium (i.e. there are no temperature gradients in it), then the temperature it reaches is given by:

$$ T_{\rm eq}=T_{\rm sun}\left({r^2\over2R^2}\right)^{1\over4}\approx470~\rm K$$

where $R$ is the distance from the sun, and $r$ is the radius of the sun.

If you change the assumptions I've made (for instance, assume that there is a large temperature gradient, so the sunward side is much hotter), this can change quite a bit. Specifically, that assumption changes the temperature to about $560~\rm K$, as it gets rid of the factor of $2$ in the denominator.

Different assumptions on shape and thermal conductivity can change that factor anywhere from $0$ (imagine a long, thin cylinder, with just the end illuminated) to $1$ (a perfect insulator will only heat up where it is illuminated, and so won't lose any energy to radiation from non-illuminated parts).

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  • $\begingroup$ Oh I see ty! Is 2.7K the temperature when there are no hot objects near by ? Does that mean temperature decreases gradually as we move away from the sun and reaches 2.7K as we reach pluto maybe ? $\endgroup$ – AgentS Mar 9 '18 at 6:43
  • $\begingroup$ Pluto is around $40~\rm K$ at its coldest, so even that's not far enough. But yes, generally that's the idea. $\endgroup$ – Chris Mar 9 '18 at 6:55
  • $\begingroup$ Gotcha! So when astronomers say 'space' most of the time they don't really mean 400km away from earth. They mean much far away, the vast space between stars. Thanks again :) $\endgroup$ – AgentS Mar 9 '18 at 7:07
  • $\begingroup$ what temperature would the object get if it was placed half way between the sun and earth tho? $\endgroup$ – pentane Mar 10 '18 at 7:53
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    $\begingroup$ @pentane It comes from a straightforward application of the Stefan-Boltzmann law. I was assuming a flat disk so the cross-sectional area is half the surface area, while that link is assuming a spherical object, where the ratio is $4$ instead. The $2$ in the denominator can vary anywhere between zero (imagine a long, thin cylinder with just the end illuminated) to one (any shape, if the object is a good insulator, so only the illuminated side gets hot). $\endgroup$ – Chris Mar 10 '18 at 19:48
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Earth's upper atmosphere has a temperature of about 260 kelvin, which is in equilibrium with the amount of radiation it receives from the Sun. The Earth radiates as much as it absorbs.

An object at half the distance would receive four times as much sunlight. In equilibrium, its thermal radiation would be four times as large. According to Stefan-Boltzmann's law, thermal radiation is proportional to the fourth power of absolute temperature, $T^4$. So the object would have temperature $\sqrt[4]{4}= \sqrt{2}$ as high, about $1.41 \times 260 \approx 370$ kelvin.

(This is what happened to Icaros the son of Daidalos. He flew halfway to the Sun, but then the wax on the wings started to melt.)

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  • $\begingroup$ Clever! 370K.. that's just 97C. Since moon has no atmosphere, would the temperature on moon be around 260K too ? $\endgroup$ – AgentS Mar 9 '18 at 7:20
  • $\begingroup$ Google says 100C on moon during day time. May be it is not reaching equilibrium.. $\endgroup$ – AgentS Mar 9 '18 at 7:21
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    $\begingroup$ @rsadhvika The moon has a huge difference between the day side and the night side because rotation is slow and there is no atmosphere to average things out. Some craters at the pole are permanently in the shadow. Temperature of the rock there can be as low as 30 kelvin. $\endgroup$ – Pieter Mar 9 '18 at 7:26
  • $\begingroup$ The moon is closer to a perfect blackbody than the earth, and the hottest parts are relatively close to thermal equilibrium with the sun's light. So you'd probably get closer to the right answer using the moon's temperature, rather than the earth's upper atmosphere. $\endgroup$ – Chris Mar 9 '18 at 18:51
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One can also approach this quantitatively. Let us assume the object to be a sphere of radius $r$ with emissivity $\epsilon$ and fast heat conduction. It will receive energy from the sun and the CMB, and radiate away heat depending on its temperature. Its internal temperature will change so that energy is conserved, and this will tell us how it will heat up.

The solar input is equal to the sunlight that hits the object. The total amount of sunlight will be the light crossing the area $\pi r^2$ (while the lit surface is half of the surface, near the terminator line it gets spread thinly). If the sun has luminosity $L$ and the object is at a distance $R$ the total power is $$P_{sun}=\epsilon\frac{L}{4\pi R^2}\pi r^2$$ Here the emissivity plays a role: if the object is shiny it will reflect some of the energy.

The CMB input comes from all directions (we ignore the shading of the sun, it is not large in the sky) and is blackbody radiation of temperature $T_{CMB}=2.7$ K. We get power $$P_{CMB}=4\pi r^2 \epsilon \sigma T^4_{CMB}$$

The object radiates heat back into space with the full surface, $$P_{rad}=4\pi r^2 \epsilon \sigma T^4$$ depending on its temperature.

Putting all this together, the amount of energy gained (or lost) per unit of time is $$P_{sun}+P_{CMB}-P_{rad} = \epsilon \pi r^2 \left (\frac{L}{4\pi R^2} + 4\sigma(T^4_{CMB}-T^4)\right )$$

This already tells us a lot of things. Emissivity determines how much the object participates in the energy exchange. A shiny object will change in temperature more slowly than a black one. The rate of energy exchange depends on the area. Calculating the size of the terms in the parenthesis will show that the first one is large (hundreds of Watt per square meter), the CMB one small (0.000003 W/m$^2$) and that the temperature will hence have an equilibrium when the solar term is equal and opposite to the object thermal term, or $L/4\pi R^2 = 4\sigma T^4$, which gives $$T_{balance}=\left [\frac{L}{16 \pi \sigma R^2}\right ]^{1/4}.$$ Temperatures will decline as ${(R^{-2})}^{1/4}=1/\sqrt{R}$ as we go outwards. The equilibrium temperature is also independent of object size $r$.

Just to complete the model, how does $T$ change if the object is not in thermal equilibrium? Let us assume it has specific heat capacity $C_V$ and mass $4\pi \rho r^3/3$, where $\rho$ is density. Then the internal heat store will be $4\pi \rho r^3 C_V T/3$, and plugging this all together as a differential equation: $$\frac{d}{dt}\left [ 4\pi \rho r^3 C_V T/3 \right ] = \epsilon \pi r^2 \left (\frac{L}{4\pi R^2} + 4\sigma(T^4_{CMB}-T^4)\right ).$$ Since the object is assumed to not change size or heat capacity (the last one is only approximately true, and I am very interested in the complications that happen at very low temperatures), we can simplify to: $$\frac{dT}{dt} = \left [\frac{3\epsilon}{4 \rho r C_V} \right] \left (\frac{L}{4\pi R^2} + 4\sigma(T^4_{CMB}-T^4)\right ).$$ The bracket term is basically the thermal inertia, factors that determine how fast things cool down or heat up. It is inversely proportional to radius: big things react slowly. Similarly, dense things and things with high heat capacity respond slowly.

Solving the differential equation is a bit overkill (it is basically $T'=a-bT^4$, and one can get the implicit solution for time $t = C+\int \frac{dT}{a-bT^4}$ - the integral is unfortunately messy but in principle analytically tractable). If the object also conducts heat slowly we would have to model internal heat diffusion - doable with the heat equation in spherical coordinates, but requiring a numerical solution.

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