The US used about 3.9 x 10^12 KWH of power in 2019. The average terrestrial photovoltaic modules are close to 20% efficiency. Sunlight reaches the Earth with power of about 1kW/m^2. Hours per day for the southern half of the US on average, is 5 peak solar hours.

P(needed) = P(out) x All deratings x Area needed

3.9 x 10^12 kWH = (1kWH/m^2 x 5 hrs/day x 0.2 x 365 days) x (0.9(inverter loss) x0.89(temperature derating) x 0.93(dust and dirt derating) x 0.95(panel mismatch and wiring loss)) x Area needed

(3.9 x 10^12 kWH)/(1kWH/m^2 x 365days x 0.9 x 0.89 x 0.93 x 0.95) = Area needed per year

3.9 x 10^12/(365 x 0.7076835 m^2/yr) = 1.51 x 10^10m^2/yr (1km^2/10^6m^2)

Area needed= 1.51 x 10^4km^2/yr = 0.386 x 1.51 x 10^4 miles^2 per year = 5,828.6 square miles

USA is 3,797,000 square miles so what percent of US needs to be covered with solar? 0.15 percent