Originally Posted by altaskier

Long answer: the insolation rate above the earth's atmosphere is around 1.4 kiloWatts per square meter, and after absoroption in the atmosphere the rate in the desert southwest can approach 15 kiloWatt-hours per day so assuming 15 hour days let's say 1 kiloWatt per square meter is practical. Typical solar panels around 10% efficient, so we're down to 0.1 kiloWatt delivered per square meter. The wing area of a 737 is around 125 square meters, and for guesstimation purposes let's say we can cover 100 square meters with solar panels so we can deliver around 10 kiloWatts. In one hour, you get (1e4 joules/sec)*(3600 sec)=3.6e7 joules or 36 megajoules. The energy content of Jet A is around 43 megajoules per kilogram, and the thermodynamic (Carnot) efficiency of turning that to work is around a third, so one hour of energy from the solar panels would be equal to the energy gained by burning around 2 kilograms (or around half a gallon) of jet fuel.