Just some numbers to help the calculations:

The solar energy perpendicular that hits Earth's ground is 1kW/m².

It is also calculated the average energy all lands receive under direct sunlight, due spherical inclination, to be around 300W/m².

The FE has no steep angles like a sphere has, so lets increase this power to 500W/m² on FE as a logical guess.

If you calculate the average FE area of the FE disc that actually receives solar radiation, you can calculate the solar energy radiated from that particular steradian of the Sun, extrapolate for the whole solar sphere radiance energy.

If the covered FE area is "A m²" and the average energy measured on the ground is 500W/m² then the energy necessary to cover such area is "A x 500W". Considering the Sun a sphere radiating energy all around, you can calculate the exact angle from the FE Sun that covers the stated FE area.

As this area is larger than the 30km (diameter of the FE Sun), more than half of the sphere will be projecting radiation down to the FE area, but on the steep angles the radiation is angled and reduced, so, to facilitate the calculations, lets assume just 1/3 of the FE Sun's "visible disc" is illuminating the FE area. It would represent 1/6 from the whole sphere producing energy hits FE.

I am being good here, since the Isosceles Triangle formed between FE diameter (40000km) and Sun's height (3000km) represents the illumination "cone" that hits FE with 162° of aperture.

It means we could assume the FE Sun produces 6 times more energy that what FE receives, so the FE Sun could produces "6 x A x 500W".

Due the extreme difficult to understand and calculate the illuminated area over FE, considering now, May 10 Central Florida, I can see the Sun from 6:38am to 8:06pm, it is 13h 28min of solar exposure. If 12 hours of sun represents 180° of solar coverage over the equator line (15°/h) with an average power of 500W/m², then 13:28h would represent 193° of a circular line around FE North Pole. That great semicircle means the Sun is illuminating the FE from East to West, reaching the FE Ice Wall, and obviously the FE North Pole.

As only FErs can calculate that, I will only guess a possible **40%** of the total FE disc being illuminated by the solar light.

As the distance from North Pole to equator is 10,000 km, the FE disc radius would be 20,000km, what means a total area of (PI*R²) 1256636000 km² or **1.25 E+9 km²** or **1.25 E+15 m²**. So, 40% of that is **5 E+14 m²**, applying the formula 6 x A x 500W, result in total Sun's energy radiated as **6 x 5 E+14 x 500W** = 15000 E+14 = **1.5 E+18 W **, or, 1500 PetaWatt.

Based on the same calculations, A x 500W would be the energy FE receives directly from the Sun, 5 E+14 x 500W = 2500 E+14 = 2.5 E+17 or 250 PetaWatts.

On RE measurements, we know the Sun pours 174 PetaWatts directly over our planet, 30% of it is reflected back to the universe, the 70% rest is absorbed by clouds, oceans and land. It means around 122 PetaWatts (122 E+15 W). With all the guessing, my calculations above results in 250 PW, what is just double the actual RE calculations.

See, to adjust the solar radiation over FE from 250PW to 122PW, we need to reduce the land area illuminated by the Sun, from 40% to 20% in the calculations above. But with only 20% of area, it would be impossible for the Sun to cover 193° over the circle at 38°N, I would NOT be seeing the Sun raise at 6:38am and set at 8:06pm on Central Florida on May 10, the day light would short than that.

Somebody must explain this to me.