Physics 214 October 24, 2007

Intensity of sunlight at surface of the earth: 1 kW/m2 (1 kilowatt per square meter) Intensity of sunlight at “top” of atmosphere: 1.37 kW/m2 Approximately 20% iof incoming energy is scattered in the atmosphere and 80% is trans- mitted to surface. Of the scattered light about half reaches surface so the total light from the sky is about 10% of the direct sunlight. The effective thickness of the atmosphere is about 5 miles or 7 km. For comparison a typical light bulb dissipates 100 watts (=0.1 kW) of energy of which 10% is in the form of visible light. For comparison a diet of 2000 calories per day corresponds to 8.4 106 joules of energy. There are 8.64 104 seconds in a day. 1 watt equals 1 joule per second. The rate of energy dissipation by a person is then about 100 watts = 0.1 kW.

Illumination of a surface is related to Energy per unit time per unit area. Consider a “point” source of light (like a lightbulb). How does the illumination of a surafce fall off with distance? If you draw a sphere with radius R1 around the bulb all the energy must pass througth this sphere. If you draw a larger radius (R2) sphere all the energy must pass through that sphere. Think about a small sheet of paper at distance R1 or at distance R2. At the larger distance a smaller fraction of all the energy falls on the sheet of paper. This is because it is a smaller fraction of the area of the sphere. So what is the area of a sphere? Area = 4πR2. The amount of light falling an piece of paper then decreases inversely as the square of the distance from the light source. For example if the paper is moved from 5 meters out to 15 meters from the lightbulb only 1/9 as much light energy falls on it. area of paper The actual rate at which energy falls on the sheet of paper is I0 4πR2 where I0 is the rate at which energy is emitted from the lightbulb and R is the distance from the bulb.

How bright is Moonlight? The rate at which energy reaches the moon’s surface is the same as that reaching the earth’s upper atmosphere (the moon has no atmosphere). Call this I0, the rate of energy reaching the surface per square meter. The earth and the moon are at about the same distance from the sun since the distance to the sun is very much larger than the earth-moon distance. i Call the radius of the moon, r and the diameter of the moon, d = 2r. Assume that the moon is a flat disk of area πr2 and that it reflects 10% of the light hitting it. Then the total rate at which energy leaves the moon’s surface (facing the earth) is Im = 2 π 2 0.1 I0 πr = 0.1 I0 4 d . How much of this enery falls on 1 square meter of earth? Call the distance between the moon and the earth, R, the radius of the moon’s orbit. All of the light leaving the disk of the moon Im, covers a half sphere of radius R. So the amount falling on a square meter on 1 the earth is Ie = Im 2πR2 . ³ ´2 1 d Then Ie = 0.1 I0 8 R Both direct sunlight and moonlight have to penetrate the earth’s atmosphere so since we are just comparing these we don’t have to take the atmosphere into account.

The Angle subtended by the moon’s diameter. This is very close to 1/2 degree. In radian meaure 1/2 degree is about 1/120 radians. This angle in radians is d/R = 1/120. The ratio of moonlight to sunlight is

2 −6 Ie/I0 = 0.1 (1/8) (1/120) = 0.9 10

So moonlight is about 1 millionth as bright as sunlight.

Angles measured in Radians

R

θ s

R θ is the angle subtended by s s=R θ The angle is measured in Radians.

Notice that if s is the full circle it equals 2πR and the angle θ is 2π. So we see that 360 degrees equals 2π radiuan. One radian is approximately 60 degrees.