For !: Solar Intensity is 1.366 KW/m^2 at 1AU. At 10AU, it'd be 1/100th of that (1/10^2) or 0.01366KW/m^2. The power available is given by P=I*A*cos(theta) where theta is the amount you're off from normal. If you're perfectly perpendicular, the numbers run out to be (0.01366KW/m^2)*(1000m^2*pi)=42914.15565KW. This is, of course, assuming there is nothing in between then sun and the collector, which in reality there are 7-8 planets (mercury through jupiter, saturn occasionally). If you want to take that into account, you'd need Kepler's Laws and Orbital Motion. Which I don't want to do. For 2: it's the same principle, upping the area. 107285389.1KW.
Will need more research for hydrogen fusion, but you can use the mass-energy equivalence of 1kg= (approx) 8.98755e16J for a rough estimate.
---
The Master said: "It is all in vain! I have never yet seen a man who can perceive his own faults and bring the charge home against himself."
>Analects: Book V, Chaper XXVI
Will need more research for hydrogen fusion, but you can use the mass-energy equivalence of 1kg= (approx) 8.98755e16J for a rough estimate.
---
The Master said: "It is all in vain! I have never yet seen a man who can perceive his own faults and bring the charge home against himself."
>Analects: Book V, Chaper XXVI