Signal to noise
In order to investigate the feasibility of our project, we needed to calculate the size of the signal, and the quantity of noise which will interfere with this signal. The rate at which the Sun's surface emits light was found to be 6.3*107 Wm-2, which means that, the Sun's surface area being around 6*1018 m2, the total power of the light released is about 3.8*1026 W. At the Earth's distance from the Sun of around 1.5*108 km, this is spread over a spherical area of 2.8*1017 km2, of which the area of the Earth facing the Sun takes up approximately 1.3*108 km2. The quantity of light from the Sun the Earth receives is therefore 2.2*109 times less than that emitted by the Sun, and so equal to 1.7*1017 W. As the average albedo of the Earth is approximately 0.367, the quantity of light reflected from Earth is 6.2*1016 W. At the satellite's distance from the centre of the Earth, of around 7 Mm, this light would be spread over 308 Mm2, or around 3.1*1014 m2. The intensity of light from the entire Earth, taking up a field of view of around 120o, would therefore be 200 Wm-2. The collector, with an assumed area of 1 m2 and field of view of 30o, would pick up 1/16 of this, so 12.5 W. With the light we will be looking at, with a wavelength of around 700nm, each photon carries 2.8*10-19 J, so the signal will consist of 2.1*1019 photons per second. Not all of this will be in the range of wavelengths our sensor is capable of detecting, but as approximately half of the Sun's light is between 500 and 700nm in wavelength, the signal size will be large enough to be useful.
The main sources of noise which will interfere with this signal are the light from stars other than Sol and the light from other planets, which will have different features from Sol's light and so could confuse results. The total light from other stars can be calculated from the number of stars and their apparent magnitude from Earth in relation to that of the Sun. Using the frequency with which different magnitudes occur from http://www.nso.edu/PR/answerbook/magnitude.html, the total light from other stars at the Earth is the quantity of light from the Sun * 10-8. This is equal to 1.7*109 W, which means that around 8.5*108 W of light from other stars reach each hemisphere of the Earth, so affecting the signal. As this is around 2*108 times less than the intensity of the light reaching Earth, stars other than Sol will not interfere greatly with the signal. The satellite's orbit being sun synchronous with a fixed time during daylight hours, only the light reflecting from those planets closer to the Sun than Earth will affect the signal. Venus, with a maximum apparent magnitude of -4.6, and Mercury, with an apparent magnitude of -1.9, compared to the Sun's apparent magnitude of -26.7, will have a maximum brightness equal to the Sun's divided by roughly 2.522.1 and 2.524.8 respectively, or the Sun's brightness *1.6*10-9 and *1.4*10-10, respectively. This means that the light at the Earth from both is, at the most possible, about that of the Sun at Earth *1.7*10-9, or 2.9*108 W. As this is clearly far smaller even than the noise from other stars, planets will not have an effect.