1
$\begingroup$

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 ?

$\endgroup$
  • 5
    $\begingroup$ Strictly speaking light doesn't have a temperature. $\endgroup$ – Keith Thompson Feb 24 '15 at 22:56
  • $\begingroup$ 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. $\endgroup$ – Florin Andrei Feb 26 '15 at 19:19
  • $\begingroup$ Somewhat related, informative and funny: what-if.xkcd.com/115 $\endgroup$ – IchabodE Feb 26 '15 at 19:21
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$
1
$\begingroup$

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. $$ T = T_{\odot}(1-a)^{1/4}(R_{\odot}/2D)^{1/2},$$ 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.

| improve this answer | |
$\endgroup$
0
$\begingroup$

An attempt to clarify/answer both this question and your Heat from other Stars

Temperature is a property of matter, not of light. As others have noted, you can assign a temperature to light, but I think that just confuses the issue.

It's true the (mostly hydrogen and helium) atoms at the Sun's surface are about 10,000F.

However, these atoms don't fly through space and then strike Earth. Even with their high temperature, most of them can't escape the Sun's gravity and remain on the Sun. The few that do escape form the "solar wind", which has virtually no effect on the Earth's temperature.

What travels though space is electromagnetic radiation (light), which is measured by intensity and frequency. Light is not a physical object, and doesn't have a temperature in the traditional sense of the word.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I can just ignore the use of Farenheit, but there is a real physical difference between how radiation fields at high temperatures behave/interact with matter compared with radiation fields at lower temperature but with exactly the same power per unit area. These differences are not greatly relevant to this question, other than that a hotter radiation field (peaking in the UV) would not penetrate the Earth's atmosphere. But it is confusing and incorrect that both you and @KeithThompson have said that light does not have a temperature - it frequently does. $\endgroup$ – Rob Jeffries Feb 26 '15 at 19:55
  • $\begingroup$ For a stark limiting case, see photon gas. That's not the situation here, but it illustrates quite plainly the danger of absolute matter/light distinctions on this issue. $\endgroup$ – Stan Liou Feb 26 '15 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.