How to calculate how much energy a body gets from a star? Okay, so I could swear that I've seen an equation before for this (which I believe involved the Stephen-Boltzmann constant.) But now that I actually need it, much like my reading glasses I can't seem to find it. Any help would be appreciated as I'm sure this is a fairly easy problem.
 A: A star is modelled pretty well as a blackbody so you use the formula,
$$ P = 4\pi R^2 \sigma T^4$$
to estimate the luminosity and then divide by $4\pi r^2$ to find the flux incident on an object a distance $r$ from the star. 
$$ \frac{dE}{dt dA}=F = \frac{4\pi R^2\sigma T^4}{4\pi r^2}$$

In repsonse to the comment:


*

*$T$ : is the temperature fo the star

*$P$ : power radiated by the star (aka luminosity). 

*$R$ : is the radius of th star. 

*$\sigma$ : I the stefan-boltzmann constant. 

*$F$ : Is the radiant flux at some point a distance $r$ from the star. (Units of $J/(m^2\cdot s )$)

*$dE$ : Increment in energy incident on a area $dA$ during a time $dt$. 



In response to the OP's follow up questions:
The area $dA$ is the area of the object being hit by the light. For instance this might be a "CCD" in a digital camera. 
$dt$ represents how long the are was exposed to the light.
$dE$ is the amount of energy that was delivered to the area during the time interval $dt$. 
In case your not familiar with the notation of differentials the $d$'s represent a change or increment in the quantity the preced. $dt$ is a duration of time, $dE$ is an amount of energy, and $dA$ is a piece of area. 
$$ dE = F \ dA  \ dt = \left( \frac{4\pi R^2\sigma T^4}{4\pi r^2}\right) dA \ dt $$
