The following model tries to determine how much solar energy can be harvested on a global scale, it doesn't look at what we actually use today for transportation, food, or other things. This is how much we get, whatever we do with it.

The model uses the area presented by the Earth to the Sun at any time, 24/7. For that, I used the area of the disk presented by the Earth, instead of the surface area of the Earth itself, which, because it is a sphere, would receive variable amounts of energy depending on latitude and time of day. Using the surface area of the Earth would require going through multiple use cases, while using the disk that the Earth presents to the sun at any time, allows to determine the global energy received from the sun, regardless of latitude, day-night, equivalent sun-hours or other fancy formulas. This is what we get on a global scale.

The model doesn't tell how we will harvest that energy. Again, this is another problem. Before we decide how we will harvest it, we need to know how much we get.

So here we go...

Solar irradiance = 1400 W/m2 at the top of the atmosphere. This is the amount of solar power on a one square meter area. The atmosphere will absorb and reflect a part, so it is estimated that the average amount of solar power reaching the Earth surface is ~ 1000 Watt / m2.

[ Solar Irradiance Reference ]

Now we can calculate how much square meters the Earth presents to the sun.

Diameter of Earth = 12,000 km = 12*10^6 meters

Radius = 6*10^6 meters.

Surface presented to the Sun = PI * (6*10^6)^2 = 113*10^12 m2.

The total amount of energy reaching the Earth surface is:

113*10^12 m2 * 1000 Watts/m2 = 113*10^15 Watts.

About 70% of the Earth surface is covered by water, so 30% remains. Of those 30% land area, we can assume that covering 1% of the land area with solar panels would be a maximum practical ratio, so 0.3% of the total energy may be harvested.

113*10^15 * .3% = 34*10^13 Watts

This is the amount of electrical power we receive from the Sun over 1% of the land area.

We are 6 billion people, so we need to share this power.

34*10^13 / 6*10(9) = 56 KWatts per person

Energy is power multiplied by time.

56*10^3 Watts * 24 hours ~ 1.3 MWh / day / person.

This is the solar allowance each of us can use before we start depleting resources.

Now this is before conversion into useful energy. The average conversion efficiency is about 20% for electricity (best case).

[ Photovoltaic Efficiency ]

Converting all of our solar energy allowance to electricity, we would have:

**1.3 MWh * 0.2 = 271 KWh / day / person**

Now lets see where we are today in the United States, as far as energy usage per capita.

Total energy usage of USA in 2005 = 29*10(15) Wh

US population = 300 Millions = 300*10(6)

**Energy usage per capita in 2005 in the US = 100 MWh/year = 280 KWh / day / person.**

[ US Energy Usage Reference ]

We can fulfill our electrical needs if we cover 1% of the land surface of the Earth with solar panels.