The intensity of sunlight at the distance of the Earth's orbit is 380 W/m2. An Earth-orbiting satellite has a solar panel that measures .35 m by 4.86 m, which converts solar energy to electrical energy with an efficiency of 26%. In one hour, how much electrical energy does the panel produce? Assume that the satellite's attitude control jets keep the panel oriented perpendicular to the incoming sunlight.
This is pretty much an exercise in knowing units and dimensional analysis. Watts are in Joules/second. So every second a square with the area(meters^2) of meter^2 receives 380 joules from the sun. In your case the square is the solar panel. So find the area of the solar panel in m^2. If you multiply area times intensity you can see that the meters cancel out and you are left with Watts(J/s). Since you want the Joules received in an hour you again multiply by how many seconds are in an hour. Leaving you with joules. What you have now is the total energy, but your solar panel is only 26% efficient, so just multiply by .26 and you will have your energy.
380 W/m2. An Earth-orbiting satellite has a solar panel that measures .35 m by 4.86 m, which converts solar energy to electrical energy with an efficiency of 26%. In one hour, how much electrical energy does the panel produce? 380 W/m? x .35 m x 4.86 m x 0.26 = 2354 watts energy is watts*time E = 2354 watts x 3600 sec = 847472 Joules or, E = 2354 watt-hours
