Someone please check my math, but here's the way I see it. Because of the earth's rotation, the sun appears to move across the sky from east to west at a rate of 15 degrees per hour. The moon moves in its orbit around the earth from east to west at a rate of just over .5 degrees per hour (13 degrees per day). This means that if you add the speed of the earth's rotation to the moon's orbital speed, then that means that the moon should appear to cross the sky from east to west at a rate of about 15.5 degrees per hour. Does that sound about right?
Yes, and that is why everyday the Moon rises around 52 minutes and 44 seconds later on East. If today you have a full moon exactly on top at midnight, tomorrow it will be at the same spot almost at 00:53am, next day at 1:45am, and so on. The Earth's rotation must be subtracted from any calculation, this is why the best way is to consider the observer sit over the Sun and seeing it as the heliocentric system as it is. When looking from the Sun, doesn't matter if the Earth is rotating or not, the Moon will be orbiting Earth once per 27.3 days Earth's eastward direction, it WILL produce a project a conical shadow with apex at 380,000km, if the shadow hits Earth we call it solar eclipse, it will move eastward, no matter if the Earth is rotating or not.
By the unquestionable fact Earth rotates 15°/h, and the Moon orbital speed of 27.3 days is 1022m/s, its projected shadow speed is always faster (eastward) than Earth's surface speed (463m/s on Equator line), depending on where the umbra hits Earth, a fixed observer on Earth could be under umbra for no more than 7 minutes and 32 seconds.
FE rely on a fixed huge Earth and small Sun/Moon rotating above.
The smaller Umbra reported on RE was 120km in diameter, on FE it will create a problem as I reported in another post including formulas. As the FE Sun and FE Moon has only 48km in diameter and the Sun is 4800km of altitude, the Moon creating the total eclipse must be significative lower than the Sun to projects a 120km umbra total shadow, also the Moon must be larger than 48km in diameter. The final formula I calculated was "D*0.015 + 48km", where "D" is the Moon's distance down below from the Sun in kilometers. The Moon must have 48km plus 15 meters per kilometer far from the Sun, and it can not be very close to the Sun since its shadow (umbra) would cover the entire FE below, not only a spot.
The formula is linear if a little away from the Sun. If the Moon is 4800km down from the Sun, it will on the Earth's ground, so, it needs to be 120km in diameter to cast a 120km diameter shadow, 4800*0.015+48 = 120.
Worse, FE would have a solar eclipse every 27.3 days, since there is no way for the Moon closer and lower than the Sun to project its shadow out of FE, it will hit Earth down below - remember, FE is much larger than Sun/Moon. FE is 40 thousand kilometers in diameter, Sun/Moon is only 48 kilometers in diameter and only 8.33% of FE diameter in altitude. I will produce a model with a graph showing how far horizontally and vertically the Moon from the Sun can be in order to still projecting its shadow over FE, of course it would happen every 27.3 days with low nodal lines, but in real world it is not like that. I would love to read FE optical math details about that.