# Treating stars and planets as a blackbody determine the temperature of a planet that has EM radiation incident upon it from a star

Consider the following question:

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.

• Just guessing: at each moment, the planet absorbs radiation over a hemisphere, but it emits radiation over its whole surface. – PM 2Ring Jan 25 '19 at 0:10
• BTW, you'll get a better reception for your question if you transcribe the text from that image to actual text. – PM 2Ring Jan 25 '19 at 0:12
• @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. – Karthik Jan 25 '19 at 0:36

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.
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.