Only 20kw per square meter on the surface of the sun ? How come it is so low ?
We receive about 1kw of sunlight per square meter on Earth, and earth is 149M km from the sun. From napkin math, it should rather be ~45MW/sqm on the sun to receive 1kw/sqm on Earth (surface of the sphere of radius 149M km divided by surface of the sun gives ~45000, so 1 watt from the sun becomes 1/45000 watt when it reaches the Earth)
Where am I wrong ?
Because that's the irradiance at the Solar Orbiter's closest approach (well, more like 17.5kW, hence getting on for).
It's pretty amazing that you can have a spacecraft in nearly 20x direct sunlight, permanently and still have it actually work.
Your calculations are incorrect. Use common sense, models, and first principles. Light point source irradiance is E = P/4πr², so inverse square law. It's 1361 W/m² at Earth's distance of 1.5e11 m. Solar Orbiter dips down to 4.2e10 m. ¼ the distance,
Total solar power output = 4 * π * (1.5e11 [m])² * 1361 [W] = 3.85e26 W/m²
Sun's "surface" irradiance = TSPO / (4 * π * (6.96e8 [m])²) = 6.32e9 W/m²
At Solar Orbiter's perihelion, assuming the distance from the Sun's point center rather than the Sun's surface = TSPO / (4 * π * (4.2e10 [m])²) = 1.74e4 W/m².
^ Except for Earth's irradiance and the distances, these are theoretical rough values rather than observed ones because reality is messier than simplified models.
The real issue was that I didn't get that you were talking about Solar Orbiter, I thought you were saying that the irradiance of the sun was 20kW/m2, which seemed low to me, but I didn't even know the word "irradiance" so I didn't know what to type on Google to check it. Thanks for your detailed calculus :)
It's basic algebra. Calculus would involve derivatives or integrals.
https://www.merriam-webster.com/dictionary/calculus
The usual sense involves integration and derivation but look at senses 2 & 4. It also means any calculation.