Consider the following question:

enter image description here

Power emitted by the star via the Stefan Boltzmann law is as follows:

$$4\pi\sigma T_s^4 r_s^2 $$ However the power will drop over distance and so we can say that at the position of the planet we will observe a flux incident upon the planet as

$$ \frac{4\pi\sigma T_s^4 r_s^2}{4\pi d^2} = \frac{\sigma T_s^4 r_s^2}{ d^2} $$

We can equate the above to $$ \sigma T_p^4 $$

Then we can deduce that $$ T_p = T_s \sqrt(\frac{r_s}{d} ) $$

However it seems that a factor of 2 is not present and I am unsure where to get it from or how to justify its existence. ANy help would be appreciated.

  • $\begingroup$ Just guessing: at each moment, the planet absorbs radiation over a hemisphere, but it emits radiation over its whole surface. $\endgroup$ – PM 2Ring Jan 25 at 0:10
  • 1
    $\begingroup$ BTW, you'll get a better reception for your question if you transcribe the text from that image to actual text. $\endgroup$ – PM 2Ring Jan 25 at 0:12
  • 1
    $\begingroup$ @DavidAbraham Not true that the power drops with distance. It is the flux through a given area (exposed area of the planet) that implies lesser energy reaching the planet. $\endgroup$ – KV18 Jan 25 at 0:36

enter image description here

Here, the term $\pi r_p^2$ is the area of cross section through which the radiation from the star actually hits. The flux you calculated for the star's radiation as it travels a distance $d$ ( i.e. $\frac{4\pi r_s^2\sigma T_s^4}{4\pi d^2}$ has to take consideration of that area as well.

The equality is true as the planet is at thermal equilibrium with the radiation from the star such that the energy emitted by the planet is equal to that it absorbs.

In response to your comment:

It's not correct to even take the entire area of a sphere if you possibly could as the radiation only hits the exposed area (the side facing the star - which would literally be half the surface area of the sphere). But the star would likely be far away, not really very close that the radiation hits a majority part of the exposed surface. $\pi r_p^2$ is a reasonable approximation to that exposed surface. Imagine if you were on the star itself looking down on the planet, to see a circle - which you would approximate as being the surface area. Also as the radiation is travelling a long distance, those radiations would be essentially like parallel lines striking the surface.

  • 1
    $\begingroup$ Could you elaborate more on where this cross section comes from? As in why do we use the area of a circle rather than the area of a sphere, since the planet is a sphere. $\endgroup$ – David Abraham Jan 25 at 0:59
  • $\begingroup$ @DavidAbraham I hope what I explained is clear. $\endgroup$ – KV18 Jan 25 at 16:14
  • $\begingroup$ @DavidAbraham If it is clear, could you accept the answer? $\endgroup$ – KV18 Jan 27 at 22:04

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.