# The Sun treated as a perfect Black Body

• Keiner Nichts
In summary, the Sun emits thermal energy at a constant rate, and as the distance from the Sun to the Earth changes, the intensity of the radiation decreases.
Keiner Nichts

## Homework Statement

At lunch, the Sun's thermal energy incident on the surface of the Earth is 1.4 kW/m^2. Given the radius of the Sun, R, distance from Earth, r, and treating the Sun like a perfect black body, calculate the total intensity of its radiation and determine its temperature.

## The Attempt at a Solution

I have no idea which formula to actually use. I've attempted applying Rayleigh-Jeans and Planck's derivation of it but I simply cannot see how R and r come into play. Perhaps as integer wavelength conditions?

You're given the thermal energy incident on the Earth. How does that play into r, the distance from the Sun to the Earth? What happens to the incident power if r changes?

Knowing that the Sun outputs a constant amount of power, what has to happen as R varies?

Keiner Nichts
I know the given power is proportional to T^4 through the Stefan-Boltzmann law but I was not really taught the formula for how the incident power changes with r. I imagine it would be proportional to (1/r)^2? Also, temperature of the Sun would be proportional to either (1/R)^2 or (1/R)^3 for constant output, I think...

Keiner Nichts said:
I imagine it would be proportional to (1/r)^2?

The inverse-square law, that's right. So why is this law important? What's happening as the distance changes?

Keiner Nichts said:
Also, temperature of the Sun would be proportional to either (1/R)^2 or (1/R)^3 for constant output, I think...

Ah, but why? What's important about the radius of the Sun? Where exactly is the light emitted from?

Keiner Nichts
Oooh, ok, so it's only emitted from the surface. Now...if I were to consider the output as constant, it means that the given value in the beginning, let's call it P, would satisfy: P×4πr^2=P'×4πR^2 where P' is what I need to find. And then through Stefan Boltzmann I divide it by the constant and raise it to the power of 1/4 to find out the temperature at the photosphere. I hope that is correct...

What do you get if you do the math?

Well, for the temperature I got the Sun's temperature at the photosphere (or a very good approximation at least, I got the real value from Wikipedia), so I imagine the total intensity is good as well. Thank you a lot!

## 1. What is a perfect black body?

A perfect black body is an idealized object that absorbs all incoming radiation and reflects none. It also emits radiation at all wavelengths according to its temperature, making it a perfect emitter as well.

## 2. How is the Sun treated as a perfect black body?

The Sun is treated as a perfect black body because it absorbs and emits radiation at all wavelengths, following the laws of thermodynamics and black body radiation. This allows scientists to use black body radiation equations to study the Sun's properties.

## 3. What is the significance of treating the Sun as a perfect black body?

By treating the Sun as a perfect black body, scientists are able to accurately study its temperature, energy output, and other properties without needing to account for the complexities of its atmosphere and surface. This allows for more precise calculations and predictions.

## 4. Are there any limitations to treating the Sun as a perfect black body?

Yes, there are limitations to treating the Sun as a perfect black body. While it is a useful simplification for studying the Sun's overall properties, it does not account for the effects of its magnetic field, surface irregularities, and other factors that can affect its radiation output.

## 5. How does the Sun's status as a perfect black body affect its temperature?

The Sun's status as a perfect black body means that it follows the Stefan-Boltzmann law, which states that the intensity of radiation emitted by a black body is proportional to its absolute temperature raised to the fourth power. This means that the Sun's temperature is directly related to its radiation output.

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