Thermal radiation from the sun

[raistca]

We know that heat is transferred from the sun to the Earth via electromagnetic radiation, right? Since no other form of heat transfer can occur in a vacuum that must be the case. We also know (?) that electromagnetic radiation does not dissipate over distance (could be wrong here). If that is true, then why is it hotter than Earth on Venus and colder on Mars (atmospheric effects aside)?

In other words, if the energy contained in electromagnetic radiation does not dissipate over distance/time, then Mars should receive the same amount of radiant heat as Earth does (per square mile surface area).

I suppose this could be caused by increased distance between each photon as the radiation travels greater distance, but not sure. Someone please enlighten me because this has been tumbling around my head for days now.

 

[f95toli]

We know that heat is transferred from the sun to the Earth via electromagnetic radiation, right? Since no other form of heat transfer can occur in a vacuum that must be the case. We also know (?) that electromagnetic radiation does not dissipate over distance (could be wrong here). If that is true, then why is it hotter than Earth on Venus and colder on Mars (atmospheric effects aside)?

Mainly because of atmospheric effects...

Also, the distance DOES matter. The amount of energy (well, flux) is the same per solid angle, meaning a body far away will receive less energy. Note that this is a geometric effect, it has nothing as such to do with light (the same would be true for waterwaves, soundwaves etc).

But again, atmospheric effects are VERY important here. The high temperature on Venus is mainly due to rampant greenhouse effects, the gases in the atmosphere has lead to extreme global warming.

 

[Matterwave]

I think there could be some confusion on terminology involved in here. Intensity as defined in the astrophysics sense, is flux per solid angle, and is therefore constant over distance.

However, sometimes in physics, intensity is interchangeable with flux. This definition of intensity is distance dependent: I=P/A, where A is the area over which the wave has now spread.

 

[vlado_skopsko]

I suppose this could be caused by increased distance between each photon as the radiation travels greater distance, but not sure. Someone please enlighten me because this has been tumbling around my head for days now.

Yes, the photons don't lose their energy through distance, their density is decreasing and it goes down inversely proportional to the distance traveled.

 

[gmax137]

Yes, the photons don't lose their energy through distance, their density is decreasing and it goes down inversely proportional to the distance traveled.

Inversely proportional to the square of the distance...

 

[gmax137]

Internet says orbits of venus, earth, and Mars are 108, 150, and 228 million km, respectively. So using the inverse square, venus sees about 1.9 times as much energy per area as Earth does, and Mars sees about 0.4 times. That's at the 'top of the atmosphere'

 

[K^2]

Note that energy radiated back is proportional to the 4th power of temperature, so the equilibrium temperatures, if all else was kept constant, would be decreasing as 1/sqrt(R). However, we see that Venus is significantly hotter, while Mars is significantly colder than that. And that's the effect of the atmosphere.

 

[vlado_skopsko]

Inversely proportional to the square of the distance...

Yes, I meant that. Thank you.

 

[gmax137]

Note that energy radiated back is proportional to the 4th power of temperature, so the equilibrium temperatures, if all else was kept constant, would be decreasing as 1/sqrt(R). However, we see that Venus is significantly hotter, while Mars is significantly colder than that. And that's the effect of the atmosphere.

I don't follow this, can you expand on it a little bit? I don't see how the energy radiated from the planet has anything to do with the distance to the sun. Because the planet's outgoing radiation is essentially all directed to the "dark sky" (at a few degrees K), not so much back to the sun (due to the sun's insignificant apparent area as seen from the planet).

Thanks

 

[D H]

Note that energy radiated back is proportional to the 4th power of temperature, so the equilibrium temperatures, if all else was kept constant, would be decreasing as 1/sqrt(R). However, we see that Venus is significantly hotter, while Mars is significantly colder than that. And that's the effect of the atmosphere.

I don't follow this, can you expand on it a little bit? I don't see how the energy radiated from the planet has anything to do with the distance to the sun.

I'll start by assuming that a planet is a perfect black body. The power absorbed by this blackbody planet in the form of incoming sunlight is

$$P_{\text{absorbed}} = 4\pi R_s^2 \sigma T_s^4 \frac{\pi R_p^2}{4\pi D^2}$$

where R_s is the radius of the Sun, T_s is the Sun's effective black body temperature, R_p is the radius of the planet, D is the distance between the planet and the Sun, and σ is the Stefan-Boltzmann constant.

The power emitted by a blackbody planet is

$$P_{\text{emitted}} = 4\pi R_p^2 \sigma T_p^4$$

The emitted power will equal the absorbed power when the body is at equilibrium temperature. Thus for an ideal blackbody planet,

$$T_p = T_s\sqrt{\frac{R_s}{2D}}$$

When applied to Venus, Earth, and Mars the above yields blackbody temperatures of 54.5°C, 5.5°C, and -47.4°C.

The above equation does not account for albedo. Accounting for albedo, but nothing else, results in the following modified equilibrium temperature equation:

$$T_p = T_s\sqrt{\frac{R_s}{2D}\sqrt{1-\alpha}}$$

Using albedos of 0.75, 0.29, and 0.16 respectlvely for Venus, Earth, and Mars yields albedo-adjusted equilibrium temperatures of -41.5°C, -17.3°C, and -57.0°C. Compare that to the mean surface temperatures of 460°C, 14°C, and -63°C. The albedo-adjusted temperature is pretty close for Mars, wrong for Earth, and very very wrong for Venus -- but maybe not. The mean cloud top temperature of Venus is -33°C, so in that light the -41.5°C figure looks pretty good.

 

[gmax137]

I'll start by assuming that a planet is a perfect black body. The power absorbed by this blackbody planet in the form of incoming sunlight is

$$P_{\text{absorbed}} = 4\pi R_s^2 \sigma T_s^4 \frac{\pi\R_p^2}{r\pi D^2}$$

where R_s is the radius of the Sun, T_s is the Sun's effective black body temperature, R_p is the radius of the planet, D is the distance between the planet and the Sun, and σ is the Stefan-Boltzmann constant.

The power emitted by a blackbody planet is

$$P_{\text{emitted}} = 4\pi R_p^2 \sigma T_p^4$$

The emitted power will equal the absorbed power when the body is at equilibrium temperature. Thus for an ideal blackbody planet,

$$T_p = T_s\sqrt{\frac{R_s}{2D}}$$

When applied to Venus, Earth, and Mars the above yields blackbody temperatures of 54.5°C, 5.5°C, and -47.4°C.

The above equation does not account for albedo. Accounting for albedo, but nothing else, results in the following modified equilibrium temperature equation:

$$T_p = T_s\sqrt{\frac{R_s}{2D}\sqrt{1-\alpha}}$$

Using albedos of 0.75, 0.29, and 0.16 respectlvely for Venus, Earth, and Mars yields albedo-adjusted equilibrium temperatures of -41.5°C, -17.3°C, and -57.0°C. Compare that to the mean surface temperatures of 460°C, 14°C, and -63°C. The albedo-adjusted temperature is pretty close for Mars, wrong for Earth, and very very wrong for Venus -- but maybe not. The mean cloud top temperature of Venus is -33°C, so in that light the -41.5°C figure looks pretty good.

Thank you very much - I believe when you wrote

$$P_{\text{absorbed}} = 4\pi R_s^2 \sigma T_s^4 \frac{\pi\R_p^2}{r\pi D^2}$$

What you meant was

$$P_{\text{absorbed}} = 4\pi R_s^2 \sigma T_s^4 \frac{R_p^2}{4 D^2}$$

Then the rest follows nicely. That last fraction term is the ratio of the planet's projected area to the surface area of the sphere centered at the sun, with the planet's orbital radius.

Very nice, thanks !

 

[D H]

Thank you very much - I believe when you wrote

$$P_{\text{absorbed}} = 4\pi R_s^2 \sigma T_s^4 \frac{\pi\R_p^2}{r\pi D^2}$$ ...

:uhh: Oops.

What I meant was

$$P_{\text{absorbed}} = 4\pi R_s^2 \sigma T_s^4 \frac{\pi R_p^2}{4 \pi D^2}$$

I also corrected this error in post #10.

Addendum:
The leading part of the right-hand side, $4\pi R_s^2 \sigma T_s^4$, is the radiated power from the Sun. This is just the Stefan-Boltzmann law. The trailing part, the fraction $(\pi R_p^2)/(4 \pi D^2)$ is the portion of that solar radiation that is absorbed by the planet.