1 Watt = 1 J/s = 10^7 erg/s
1GW = 10^9 J/s
the area of the sun is
4pi*(7*10^5 km)^2 = 6.2*10^12 km^2
So the solar output is
4*10^17 GW/6.2*10^12km^2
= 6.452*10^4 GW/km^2
So, now the question is: how much of that reaches the earth? I guess you need to use the inverse-square law using the distance from the sun.
The luminosity of the Sun is 4*10^33 erg/s, and its radius is 7*10^10 cm.
You are tasked with building a solar power plant in the Arizona desert, using solar panels with 10% efficiency. How large an area (km^2) must your solar panels cover to match the power output of a large nuclear powerplant (about a GigaWatt)? Please enter your answer in units of km^2
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