Dust grain heated by a star: equilibrium temperature

Finding the equilibrium temperature of a dust grain heated by a star, by considering the balance of absorbed and emitted radiation. We first treat the grain as a black body, then consider how the model might be made more realistic.
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About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
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[@ananthanarayananr9176]

Hi Dr Ben. I have a question. At 2:25 where you multiply the cross section area of the dust, why do you do that ? Why not multiply by the hemispherical area of the dust particle on which the heat would be incident ?
Also thank you for your videos. They are always unique in the problems you choose to tackle. Never too simple nor too complicated :)

[@Jhfm1793]

I believe that has to do with the fact that that far away from the star, we can assume that no radation will be reaching the surface of the dust particle at an angle, so only the shadow area is used in the calculation as a projection of the area of incidence of the heat.

[@ananthanarayananr9176]

@@Jhfm1793 oh i get it now. Thanks for the clarification

[@DrBenYelverton]

@ananthanarayananr9176 Thanks for your kind words! Using the area of the hemisphere would be equivalent to assuming that the stellar radiation is incident on the dust grain perpendicular to its surface at all points. This isn't the case because the radiation's intensity vector is essentially constant, pointing to the right in the diagram in the video. What we really want is the flux of the intensity vector through the hemispherical surface, which involves a dot product between the surface normal and the intensity vector. Mathematically, you could work this out by using Stokes' theorem to note that the flux of intensity through the hemisphere is the same as the flux through the cross-section because both surfaces are bounded by the same curve (the equator of the dust grain). Intuitively, think of the fact that the dust grain will leave a shadow whose size is equal to its own cross-section, rather than equal to the surface area of the hemisphere, and all of the radiation that would otherwise have been in the shadow must have been absorbed.

[@ggxsky4811]

Will the answer remain same for a static body and an orbiting body ( if the size isnt negligible)