Okay, nowhere near done yet, but here's a little preamble.

The problem considers an Earth-like planet orbiting a Sun-like star ("Earth Mk II") with a geostationary moon with radius 1/5 of earth radius (=1274.2 km). The question is where and when eclipses will occur due to the geostationary moon. Just to consider - as an estimate, geostationary orbit is about a tenth of the distance to the moon (from Earth, according to my physics teacher) and the real moon has radius about 0.273*r(e), which suggests this geostationary moon will appear some seven times larger in the sky than the real moon. Since we know that the real moon appears about as large as the sun in the sky, we can say that the proposed moon will appear about seven times larger than the sun in the sky.

There are a number of ways this problem could be approached; it has been suggested in this thread that the cone of light (or rather, shadow) formed by the moon be considered through time. However, this method brings about a lot of more complex maths and deals with more difficult 3D geometry. It is a more elegant solution to consider the position of celestial bodies in the sky.

We need, fundamentally, two pieces of information: the position of the sun and the position of the moon. The position of these things to an observer can be considered as two co-ordinates: elevation and azimuth (the direction of the sun on the horizontal plane). Equations found on Wikipedia show how these two co-ordinates vary according to terrestrial co-ordinates, time of year and time of day. Although these formulae need no new derivation, they are the more complex part as they vary so much. The moon is much simpler as it remains in the same position always (ignoring any long term changes) except for when the observer's position changes. It should be noted here that we need only consider the "eclipse hemisphere" (the hemisphere about the point directly under the moon) as the other hemisphere will never see the moon or be eclipsed.

So far, I have identified the formulae needed to consider the sun's location in the sky, and have derived a formula for the moon's elevation based on latitude. I now need to derive a formula for the moon's azimuth. Then, I need to find where they coincide, keeping in mind the formulae only show the centre of the bodies' locations. There are two routes here: using a spreadsheet is arguably superior, calculating locations for many times and locations and searching for overlaps within the necessary degree of accuracy to account for size. However, I don't like spreadsheets, so will get a general impression by plotting graphs instead.

Edit:

For the sun:

sin_θ__(s)_*=**cos_h_cos_δ_cos_ϕ*+*sin_δ_sin_ϕ*

where *θ*_(s)_ is the solar elevation,

delta is declination (a type of co-ordinate used on the celestial sphere),

phi is latitude on the earth,

h is the hour angle (which can be found using h=15(t-12), where t is the time on the 24 hour clock; it is a measure of longitude where every hour is 15 degrees)

The declination can be found using this formula to a good approximation:

*δ=-**arcsin*[0.39779[/I]cos(0.98565[[/I]*N+10]***+***1.914sin*(0.98565[[/I]_N-2]))]

where N is the number of days since January 1st (decimals can be used to maintain accuracy)

cos_ϕ__(s)**=_(sin_δ_cos_ϕ-_cos_h_cos_δ_sin_ϕ)**/cos__θ(*s)* Where *ϕ*(s)_ is the azimuth angle (less than 180 if h<0, more than 180 if h>0)

For the moon:

the moon will stay in the same place above the eclipse hemisphere at all times. Therefore, it's position in the sky varies only with the position (longitude and latitude) of the observer. By considering this, we can derive the following formulae:

*θ(***m)**=*arctan*(**(35786+6371**[1-[/I]cos_ϕ_])/6371sinϕ__)_

for lunar elevation. Phi is latitude, again - consider that the latitude gives us a ring of locations on the planetary surface whose centre is in line with (i.e. under) the centre of mass of the moon. Therefore, every location on this ring has a common angle (and distance) from the horizontal to the moon.

*ϕ*_(m__)=180+λ_

This gives us lunar azimuth. Lambda is longitude which will be measured as positive or negative from north up to 180 degrees, which allows for only one formula. This one is pretty straightforward - we can imagine everything squashed into 2D space so that we are on a circle whose centre is the centre of mass of the moon. This allows us to derive this formula which converts angle to the point at which the observer stands to angle to the moon. Each longitude gives one quarter-ring of points (since we differentiate between positive and negative and we reject everything in the other hemisphere).

We can therefore plot position on a graph w.r.t. time and run a number of test terrestrial co-ordinates to see what happens.____