Based on my calculations, we cannot see the Earth from the ISS. Obviously it's wrong. Why?

Question

What started as a fun exercise really annoys me because I cannot see where I got it wrong.

I initially wanted to see how many photons hit a pixel of a camera on the ISS pointed at the Earth - but I kept getting less than one at 1/250 shutter speed which would be invisible so now I'm just trying to get a number that makes some sense. Even the Earth as seen from the Moon is 40 times brighter than the Moon is from Earth, and the Moon can be captured at 1/250 shutter speed with cameras with the same pixel size I'm using here (5µm). What did I do wrong?

Method

My general method is that all the light a pixel collected during the exposure time comes from the projection of that pixel on the scene.

And that light here originates from the Sun, hits the Earth (which is assumed as a shell) and gets diffused or scattered evenly in a hemisphere. Here I basically take 1340W/m², multiply it by the pixel's projected surface on the Earth, multiply it by the albedo of 0.3 and 10% which is approximately the power fraction of the visible light in the Sun's radiation.

The pixel in the detector collects a tiny fraction of the incident power as the ratio of its own surface over the surface of the aforementioned hemisphere originating on the surface of the Earth and which has the altitude of the ISS as its radius. I already get powers in the 10^-17 Watt here which is fishy.

Then I just convert the collected power into equivalent flow of photons using the Energy equivalence formula (same formula, but in power instead of energy), and finally multiply that flow by the exposure time.

And I get 0.9 photons, or in other words none, which is impossible because there exist tons of pictures taken by normal cameras and they definitely do not have "zero e-" dark current. Interestingly it doesn't change with altitude. Not sure this is correct either, at least it could make sense since the projected pixel and with it the power source grows when we're further away.

Any idea?

Answer 1

Let $A_p$ be the pixel size, $D$ the lens aperture diameter, $f$ the focal length, $r$ the Earth-Moon distance, and $M$ the exitance of the Earth's surface in the visible ($40\text{ W/m}^2$ according to your numbers).

The projected pixel area on the Earth's surface is $A_p' = (r/f)^2A_p.$ Note that all the power radiated from this area and collected by the entrance pupil of the lens falls on the pixel (neglecting transmission losses). This power is $$P_p=MA_p'\frac{\pi D^2/4}{2\pi r^2}=MA_p\frac{D^2}{8f^2}=\frac{MA_p}{8(f/\#)^2}$$ where $f/\#$ is the f-number of the lens. This also shows that the distance between the lens and the object being imaged doesn't matter. This is expected, because the greater the distance, the greater the region of the object that each pixel collects light from. Diffuse objects don't get less bright in the image as we move away from them, they just take up fewer pixels on the sensor. Just as moving away from a sheet of paper doesn't make it seem dimmer to your eyes.

Using $A_p=(5\text{ μm})^2$ and $f/\# = 1.8$, we get $P_p=3.9\times10^{-11}\text { W}$. Each visible photon has an energy of roughly $2.5\text{ eV}=4.0\times10^{-19}\text{ J}$ on average, so this corresponds to a photon flux of $9.8\times10^7\text{ photons/s}$. A 1/250 s exposure results in $3.9\times10^5$ photons falling on the pixel.

We have to consider the lens because an imaging system is not just a sensor out in the open. You may not think the lens "collection factor" is that important, but there is an enormous difference between the aperture area and that of a pixel.