Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

An astronomer is trying to estimate the surface temperature of a star with a radius of $5 \times 10^8\ m$ by modeling it as an ideal blackbody. The astronomer has measured the intensity of radiation due to the star at a distance of $2.5 \times 10^{13}\ m$ and found it to be equal to $0.055\ W/m^2$. Given this information, what is the temperature of the surface of the star?

How do I do this? The hint was to use $I=\sigma T^4$ (where $\sigma = 5.67 \times 10^{-8}$).

As the star radiates energy, the total power that flows through any spherical surface concentric with the star remains constant. The total power flowing through such a surface can be obtained by multiplying the intensity (power per square meter) by the surface area of the sphere.

Using the law of conservation of energy, the surface area of the two spheres (one at the star's surface and one with the radius equal to the observing distance), and the intensity measured by the observer at the observing distance, you should be able to obtain the intensity at the surface of the star.

Don't really understand the 2nd paragraph too ...

share|cite|improve this question

closed as off-topic by Colin McFaul, Kyle Oman, DavePhD, Brandon Enright, John Rennie Jun 12 '14 at 6:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Colin McFaul, Kyle Oman, DavePhD, Brandon Enright, John Rennie
If this question can be reworded to fit the rules in the help center, please edit the question.

Side note: the problem is using the wrong terminology. Flux is power per unit area and is given by $\sigma T^4$ for the surface of a blackbody. Intensity is power per unit area per unit solid angle (they differ by units of steradians, which are often omitted, so it is hard to tell), and so is flux divided by $2\pi$ for an isotropically emitting surface. This doesn't really have any bearing on the problem, except that every instance of "intensity" or "$I$" should be replaced by "flux" or "$F$." – Chris White Apr 13 '13 at 14:47
up vote 3 down vote accepted

The idea is just to make use of the relationship between luminosity (the amount of energy emitted per second from the star - in other words, the power) and flux (the amount of power hitting the surface).

Because, the flux can be modeled as a large number of (imaginary) spherical energetic wavefronts emerging from the star in 3-D space as a function of time. Also, this flux is based on the inverse-square law. So, it decreases with distance (squared). And since the power distributed over every sphere is still the same, both are related by the area of spheres.

$$\mathrm{Flux=\frac{Luminosity}{4\pi r^2}}$$

And since the question says that the astronomer idealizes it as a blackbody (which we always do), we can use the Stefan-Boltzmann equation and say that the flux is $\sigma T^4$...

Edit for confusion: Simply relating the given flux with Stefan-Boltzmann only gives the temperature of the imaginary sphere at the distance at which the flux was measured. First, you can find the luminosity of the star by using the relation above. Then, you need the flux at the surface of the star. Given the radius of star, it can be used to determine how much power is transmitted to the surface through the luminosity. Finally on relating with $\sigma T^4$ gives the answer...

share|cite|improve this answer
Careful: Stefan-Boltzmann says the flux at the object's surface is $\sigma T^4$. – Chris White Apr 13 '13 at 14:41
@ChrisWhite: Oops.. Sorry for that confusion. That's very important for a homework. (Corrected now). But, I think we don't wanna include the parameter surface since flux is already a surface quantity. Should I wanna (necessarily) include that? – Waffle's Crazy Peanut Apr 13 '13 at 14:56
Err, sorry but I dont really get it still, not good at maths/sci ... Using what u said I guess I will have $Flux = \frac{Luminisity}{4 \pi r^2} = \sigma T^4$? So flux is $0.055 W/m^2$. But whats luminosity? What value do I use? $r$ here is distance to star right? – Jiew Meng Apr 15 '13 at 9:05
@JiewMeng: The luminosity is the power (energy per second) emitted by the star. What is being called flux here is the power per area (sometimes called intensity). The flux being measured by the astronomer is the power per area on an imaginary sphere with a radius equal to the distance to the star. However, the flux that is equal to $\sigma T^4$ is the power per area for the spherical surface of the star (as Chris White mentions above). So you want to use the flux measured at Earth, use that to find the star's luminosity, then use the star's radius to find the flux at its surface(cont.) – RecklessReckoner Apr 15 '13 at 20:17
(cont.) That will give you the star's surface temperature using $F_{star} = \sigma T^4$ [I get a bit over 7000 K.] @Crazy Buddy I think what JiewMeng is not getting from this is how the flux has to be figured for the star's radius – RecklessReckoner Apr 15 '13 at 20:18

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