<p>The Sun's radiation spectrum is close to thermal at $T_\odot = 5778\,\mathrm{K}$. That means that that power output, given by the Stefan–Boltzmann law, is around $L = \sigma A T^4$, where $\sigma = 5.6704\times 10^{-8} \,{\mathrm{W}}/{\mathrm{m}^2\mathrm{K}^4}$ is the Stefan–Boltzmann constant and $A$ is its surface area:
$$L_\odot = \sigma(4\pi R_\odot^2)T_\odot^4 = 3.85\times10^{26}\,\mathrm{W}\text{.}$$
However, at the Earth's orbital radius $r = 1\,\mathrm{AU}$, this radiation is spread across a surface area of $4\pi r^2$, so the power incident on an Earth-sized sphere would be just
$$P = L_\odot\frac{\pi R_\oplus^2}{4\pi r^2}\text{,}$$
so approximating the Earth as a spherical blackbody, the equilibrium temperature is determined by its own thermal radiation balanced by this incident radiation:
$$\sigma(4\pi R_\oplus^2)T^4 = P\text{,}$$
which gives a surface temperature of
$$T =  T_\odot\sqrt{\frac{R_\odot}{2 r}} = 279\,\mathrm{K}\text{.}$$
Therefore, no special mechanisms are necessary to prevent the Earth from frying.</p>

<p>In reality, the Earth is not a blackbody because of its albedo and atmosphere, but its actual global temperature is not too far off from this estimate. The average surface temperature is close to $288\,\mathrm{K}$. Earth's atmosphere actually raises the average temperature, not lowers it.</p>

CC BY-SA 3.0
2015-02-24T23:08:05.057Z
2015-02-24T23:08:05.057Z
Stan Liou
<p>Yes, you *can *assign a temperature to the radiation from the Sun. It is approximately a blackbody radiation field with a temperature of 5800 Kelvin. But no, it doesn't "cool" on the way to the Earth.</p>

<p>The radiation field is still a blackbody radiation field with a temperature of 5800K when it reaches the Earth, but its <em>intensity</em> (or power per unit area) has decreased from $6.4 \times 10^{7}\ Wm^{-2}$ at the surface (photosphere) of the Sun, to about $1.4 \times 10^3\ Wm^{-2}$, when it reaches the Earth.</p>

<p>It is this latter figure that is important in terms of how much the Earth is heated, not the temperature of the radiation field, because it determines how much energy per second is fed into the Earth's biosphere. An analogy would be an electric bar heater. The element will be emitting radiation at a temperature of 1000 C, but you don't burst into flames if you sit more than a metre away from it, because the power is distributed over a larger area.</p>

<p>In the absence of an atmosphere, the Earth would achieve an equilibrium temperature, such that it radiated away as much energy as it absorbed. This equilibrium temperature is <em>much</em> lower than the temperature of the Sun's surface because the power per unit area incident on the Earth is much less than the power per unit area emitted at the Sun's surface. The calculation, which I won't bore you with can be found <a href="http://en.wikipedia.org/wiki/Planetary_equilibrium_temperature" rel="nofollow">here</a>.
$$ T = T_{\odot}(1-a)^{1/4}(R_{\odot}/2D)^{1/2},$$
where $T_{\odot}$ is the temperature of the Sun, $R_{\odot}$ is the radius of the Sun and $D$ is the distance between the Earth and the Sun. $a$ is the "albedo" - the fraction of the Sun's light that is reflected back into space.</p>

<p>Putting numbers - $a=0.3$, $L_{\odot}=3.8\times10^{26}\ W$, $R_{\odot}=7\times10^{8}\ m$, $D=1.5\times 10^{11}\ m$; we find $T= 256 K$ or $-16\ C$. Of course the presence of an atmosphere warms us over this simple calculation by some tens of degrees.</p>

CC BY-SA 3.0
2015-02-24T23:18:55.613Z
2015-02-26T20:00:47.437Z
2015-02-26T20:00:47.437Z
ProfRob
<p>What is the temperature of the light that leaves the Sun towards the Earth and does it cool down during it's journey here ? If it doesn't cool down then how come it isn't frying the Earth ? Is it our atmosphere that is preventing it from doing so or does it cool down during it's trip here in space ?</p>

CC BY-SA 3.0
2015-02-26T19:19:53.113Z
The Sun makes black body radiation at 5800 K, but it's far away, so we don't fry. If we were closer, like Mercury, we would fry. If we were further away, like Neptune, we would freeze. So distance is also important. The closer you are, the greater the density of radiation; the further away, the lesser the density. What's actually important is not the black-body energy profile of the radiation, but its density - the amount of energy per square meter per second. We are just in the right spot for the density to be balmy and nice.
Florin Andrei
CC BY-SA 3.0
2015-02-26T19:21:15.967Z
Somewhat related, informative and funny: https://what-if.xkcd.com/115/
IchabodE
CC BY-SA 4.0
2020-09-22T16:41:54.460Z
@KeithThompson The photons have an energy distribution, if it is thermal distribution, it can be associated by a temperate. In the case of the Sun, it is 5870K. Even if they have no thermal distribution, an equilibrial temperature can associated by a photon field (if you put an item into the field, until which temperature does it heat/cool). This temperature is $\approx$ 260K on the orbit of the Earth.
peterh
CC BY-SA 4.0
2020-09-22T17:55:02.927Z
@peterh-ReinstateMonica That's odd. I see your comment flagging me, but I don't see the comment you replied to. Was an old comment of mine deleted? (Not complaining; it's likely that I wrongly said that light doesn't have a temperature, but I don't remember it.)
Keith Thompson
<the-sun>
Temperature of the Sun's rays
Peter U
# Temperature of the Sun's rays

(Asked by: username_0 on 2015-02-24)

What is the temperature of the light that leaves the Sun towards the Earth and does it cool down during it's journey here ? If it doesn't cool down then how come it isn't frying the Earth ? Is it our atmosphere that is preventing it from doing so or does it cool down during it's trip here in space ?

**Tags:** the-sun

> **Comments:**
> **username_1**: The Sun makes black body radiation at 5800 K, but it's far away, so we don't fry. If we were closer, like Mercury, we would fry. If we were further away, like Neptune, we would freeze. So distance is also important. The closer you are, the greater the density of radiation; the further away, the lesser the density. What's actually important is not the black-body energy profile of the radiation, but its density - the amount of energy per square meter per second. We are just in the right spot for the density to be balmy and nice.  
> **username_2**: Somewhat related, informative and funny: https://what-if.xkcd.com/115/  
> **username_3**: @KeithThompson The photons have an energy distribution, if it is thermal distribution, it can be associated by a temperate. In the case of the Sun, it is 5870K. Even if they have no thermal distribution, an equilibrial temperature can associated by a photon field (if you put an item into the field, until which temperature does it heat/cool). This temperature is $\approx$ 260K on the orbit of the Earth.  
> **username_4**: @username_3-ReinstateMonica That's odd. I see your comment flagging me, but I don't see the comment you replied to. Was an old comment of mine deleted? (Not complaining; it's likely that I wrongly said that light doesn't have a temperature, but I don't remember it.)  

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## 2 answers

### Answer by username_5 on 2015-02-24. Score: 1

The Sun's radiation spectrum is close to thermal at $T\_\odot = 5778\,\mathrm{K}$. That means that that power output, given by the Stefan–Boltzmann law, is around $L = \sigma A T^4$, where $\sigma = 5.6704\times 10^{-8} \,{\mathrm{W}}/{\mathrm{m}^2\mathrm{K}^4}$ is the Stefan–Boltzmann constant and $A$ is its surface area:
$$L\_\odot = \sigma(4\pi R\_\odot^2)T\_\odot^4 = 3.85\times10^{26}\,\mathrm{W}\text{.}$$
However, at the Earth's orbital radius $r = 1\,\mathrm{AU}$, this radiation is spread across a surface area of $4\pi r^2$, so the power incident on an Earth-sized sphere would be just
$$P = L\_\odot\frac{\pi R\_\oplus^2}{4\pi r^2}\text{,}$$
so approximating the Earth as a spherical blackbody, the equilibrium temperature is determined by its own thermal radiation balanced by this incident radiation:
$$\sigma(4\pi R\_\oplus^2)T^4 = P\text{,}$$
which gives a surface temperature of
$$T = T\_\odot\sqrt{\frac{R\_\odot}{2 r}} = 279\,\mathrm{K}\text{.}$$
Therefore, no special mechanisms are necessary to prevent the Earth from frying.

In reality, the Earth is not a blackbody because of its albedo and atmosphere, but its actual global temperature is not too far off from this estimate. The average surface temperature is close to $288\,\mathrm{K}$. Earth's atmosphere actually raises the average temperature, not lowers it.


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### Answer by username_6 on 2015-02-24. Score: 1

Yes, you \*can \*assign a temperature to the radiation from the Sun. It is approximately a blackbody radiation field with a temperature of 5800 Kelvin. But no, it doesn't "cool" on the way to the Earth.

The radiation field is still a blackbody radiation field with a temperature of 5800K when it reaches the Earth, but its *intensity* (or power per unit area) has decreased from $6.4 \times 10^{7}\ Wm^{-2}$ at the surface (photosphere) of the Sun, to about $1.4 \times 10^3\ Wm^{-2}$, when it reaches the Earth.

It is this latter figure that is important in terms of how much the Earth is heated, not the temperature of the radiation field, because it determines how much energy per second is fed into the Earth's biosphere. An analogy would be an electric bar heater. The element will be emitting radiation at a temperature of 1000 C, but you don't burst into flames if you sit more than a metre away from it, because the power is distributed over a larger area.

In the absence of an atmosphere, the Earth would achieve an equilibrium temperature, such that it radiated away as much energy as it absorbed. This equilibrium temperature is *much* lower than the temperature of the Sun's surface because the power per unit area incident on the Earth is much less than the power per unit area emitted at the Sun's surface. The calculation, which I won't bore you with can be found [here](http://en.wikipedia.org/wiki/Planetary_equilibrium_temperature).
$$ T = T\_{\odot}(1-a)^{1/4}(R\_{\odot}/2D)^{1/2},$$
where $T\_{\odot}$ is the temperature of the Sun, $R\_{\odot}$ is the radius of the Sun and $D$ is the distance between the Earth and the Sun. $a$ is the "albedo" - the fraction of the Sun's light that is reflected back into space.

Putting numbers - $a=0.3$, $L\_{\odot}=3.8\times10^{26}\ W$, $R\_{\odot}=7\times10^{8}\ m$, $D=1.5\times 10^{11}\ m$; we find $T= 256 K$ or $-16\ C$. Of course the presence of an atmosphere warms us over this simple calculation by some tens of degrees.


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