(a) Determine the power of radiation from the Sun by noting...

(a) Determine the power of radiation from the Sun by noting that the intensity of the radiation at the distance of Earth is $1370 \mathrm{W} / \mathrm{m}^{2}$. Hint: That intensity will be found everywhere on a spherical surface with radius equal to that of Earth's orbit. (b) Assuming that the Sun's temperature is 5780 K and that its emissivity is 1, find its radius.

Key Concepts

Blackbody Radiation

Blackbody radiation refers to the idealized spectral distribution of radiation emitted by an object in thermal equilibrium. The Sun is approximated as a blackbody radiator, which allows the use of established laws relating temperature to radiated energy, thus enabling the determination of physical properties like radius.

Inverse-Square Law

This fundamental law of physics explains that the intensity of radiation from a point source decreases proportionally to the square of the distance from the source. It is crucial when relating the observed intensity at a given distance to the total power output, as the energy spreads uniformly over the surface of a sphere whose area grows as the square of the radius.

Luminosity

Luminosity is the total amount of energy radiated per unit time by an astronomical object, such as a star. In the context of the problem, the luminosity of the Sun is calculated by relating the measured intensity at Earth’s distance to the total energy output by considering the spherical surface area over which the energy is distributed.

Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the power emitted per unit area of a blackbody is proportional to the fourth power of its temperature. This law provides a direct way to calculate the total radiated energy from a spherical body, such as the Sun, when its surface temperature is known, thereby linking the luminosity to the radius and temperature.

Emissivity

Emissivity is a measure of a material's ability to emit radiation relative to that of an ideal blackbody. An emissivity of 1 implies that the object radiates energy as perfectly as a blackbody. In the problem, assuming an emissivity of 1 for the Sun simplifies calculations by ensuring that the ideal blackbody formulas apply directly.

Transcript

00:01 Now in this question we have the sun, emitting radiation in all directions.

00:07 And we're given that the intensity of the radiation that is heating from the earth, from the sun, the intensity over here is 1370 watts per meter square.

00:26 So what we're going to do is to find the total power of radiation that is emitted by the sun.

00:31 And we know that the power of the radiation, as it moves further out, it gets spread.

00:38 It spreads out evenly in a uniform sphere.

00:43 So for the radiation that actually hits the earth, it has actually been spread over an entire sphere with radius that is equal to the distance from the sun to the earth, right? in a circular sphere with center at the center of the sun.

01:09 So given that this is the intensity, we can find what is the total power by multiplying the intensity by the total area.

01:16 And this area is just the surface area of the entire sun, sorry, the entire spherical object.

01:32 Which we have described earlier.

01:33 So this is just i times 4 pi r square.

01:38 So what is this distance? this distance, we can approximate this to just be the distance between the sun and the earth itself, because the distance is actually much further than what is drawn over here.

01:58 So this distance are approximately 150 .14 million.

02:11 Meters, so it goes to 1 .5014 times 10 power 11 meters...

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