The Department of Energy estimates 4,038 billion kiloWatt-hours of electricity were expended in the United States in 2005. The flux of the Sun at the top of the Earth’s atmosphere is about 1 kiloWatt per square meter. From these two numbers it is easy to estimate how much area in solar panels would be needed to provide the current electrical needs of the entire country.

There are 8766 hours in an average year, so the average electricity consumption in 2005 was 460 million kiloWatts (4,038 kW-hours/8766 hours). For the United States we only have sunlight available for about half the time (12 hours out of 24), and it does not shine directly on the ground due to our moderate latitude. So let’s knock that kiloWatt per square meter of sunlight in half for the nighttime, and in half again for the oblique angle of the sunlight, and we get 250 Watts per square meter. With 10 per cent efficiency of solar panels we would get 25 Watts (0.025 kW) per square meter of photovoltaic cells. That means it would take 18 billion square meters (460 million kW/0.025 kW) of photovoltaic cells to supply all of the country’s current electrical usage. 18 billion square meters is a big number, but the United States is also big. If we put that in one large solar panel farm it would have to be 134 km by 134 km, or about 80 miles on a side. This is a remarkable number. It means that with current technology we could get all our electrical power needs with relatively small chunks of real estate.