To model the light levels entering a pinhole camera (to a first order approximation), the following calculations must be made:

First the intensity of solar flux at Earth is roughly 1370W.m^{-2}.

The Earth presents a circular face to the solar radiation. Some of it is reflected directly (the fraction being determined by the Earth's albedo; A = 0.367) the rest being absorbed by the Earth,and then re-emitted as a black body spectrum of a different temperature.

Soalr radiation peaks in the optical band, and in the Earth's in the microwave. It can be assumed that the contribution to the optical component of the spectrum from the Earth's blackbody spectrum will be negligable (Dr Steward). The light reflected by the atmosphere (albedo) takes the form of the Sun's Planck function.

The proportion of stellar flux that is optical is 46%.

So, firstly, the total flux reflected by the Earth's albedo is calculated:

Optical Flux = (Flux in x Area reflecting x Albedo x Optical fraction)/(Area of surface at satellite altitude)

The satellite is assumed to be in LEO (500km) and the reflected flux is radiated hemispherically.

Optical flux = (1372 x pi x (6370 x 10^{3})^{2} x 0.367 x 0.46) / (2 x pi x ((6370+500) x 10^{3})^{2}),

Optical flux = 99.57 W.m^{-2}.

As the optical spectrum peaks in the green region, and green light has an average wavelength of 510nm, the photon flux can now be found:

E_{photon} = hc / lamda,

E_{photon} = (6.626 x 10^{-34} x 3 x 10^{8}) / (510 x 10^{-9}),

E_{photon} = 3.898 x 10^{-19} Joules.

By dividing the total energy flux, by the energy per photon, the approximate optical photon flux can be shown to be:

F_{photon} = F_{energy}/E_{photon},

F_{photon} = 99.57/(3.898 x 10^{-19}),

F_{photon} = 2.55453 x 1020 photons.s^{-1}.m^{-2}.