The light entering the pinhole camera is assumed to do so at a maximum of 45^{o} from the normal. Therefore, it is easy to calculate the ground coverage of such a camera, and therefore the resolution per pixel.

Altitude of the satellite in LEO is assumed to be 500km.

For ease, due to the relative size of area covered and the surface of the Earth, the area is assumed to be flat and circular.

The radius of the surface covered can be found using trigonometry (sorry for the lack of diagram):

tan(45) = opposite/adjacent, where adjacent = 500km and opposite = radius.

tan(45) = R/(500 x 10^{3})

R = (500 x 10^{3}) x tan(45)

R = 500km

Therefore the area of ground covered is:

A_{covered} = pi x R^{2}

A_{covered} = pi x (500 x 10^{3})^{2}

A_{covered} = 7.854 x 10^{11} m^{2}

## Ground Coverage in Solid Angle

The equation for solid angle is:

ΔΩ = (Δa) / r^{2}

ΔΩ = 7.854 x 10^{11}m^{2} / ((500 x 10^{3})^{2})

**ΔΩ = 3.14 steradians**

## Resolution

Still based on the STAR1000BK7:

Knowing the ground coverage, and the number of image pixels in the array, the resolution of the optical camera can be easily found:

Image Pixels = 1024 x 1024

Ground coverage = 1000km (at circle diameter)

Therefore, assuming an area of 1000km to be the width and height of the field of view, the resolution per pixel in each direction (height and width) is as follows:

Resolution_{height} = 1000 / 1024

**Resolution _{height} = 0.977 km/pixel**

Because the array of pixels is square, the resolution is the same in both directions. Therefore, the resolution of each pixel in an area is **0.954 km ^{2}/pixel**.