1. Solar Radiation Energy is emitted by the Sun continuously and this energy is not confined to a particular region of space. Although the Sun emits energy over a range of frequencies, it does not emit with the same intensity at each. It follows reasonably closely the “black body” spectrum – ie that of a perfect emitter.

2. The intensity of the radiation from the Sun at any point in space is governed by an inverse square law

3. Intensity of wave is defined as energy passing through unit area in unit time. Unit = W m -2 I = P / 4  r 

4. The above formula is explained by considering a spherical shell at distance r from the source r Surface area of sphere is 4  r 2 Intensity = power / area = total power / total area = P / 4  r 2

5. The power of the sun is 3.86 x 10 26 W and it is approx 1.5 x 10 11 m from Earth. Calculate the average power falling on 1 m 2 of the Earth’s surface. Solar Constant

6. The hotter the object, the higher the peak. The hotter the object, the shorter the peak wavelength (ie radiation is “bluer”) The hotter the object, the greater the area under the graph. This measures total energy emitted. We can also think of the Earth as an emitter – it emits the radiation it receives from the Sun. However, the spectrum it emits is different from the Sun’s because its temperature is different.

7. peak gives us the colour of the star NB the y axes have been adjusted so that the graphs appear the same size.

8. Wien’s Law - used to find the temperature of an emitter such as the Sun or the Earth By looking at the spectrum of radiation emitted from the star, the wavelength at which the intensity of the emitted radiation is a maximum ( peak ) can be found. peak x T = 2.9x 10 -3 m K (a constant) peak gives us the colour of the star; T in kelvin

9. Examples 1.Betelgeuse is a red star. Estimate its peak wavelength and hence its surface temperature. 2.Rigel has a surface temperature of 10 000K. What colour is it? 3.“Men perspire, ladies glow and pigs sweat….” By finding your wavelength determine if you glow or not! (which region of the electromagnetic spectrum do you emit in?)

10. Emissivity The emissivity of a surface measures how close to a perfect emitter t is. A perfect emitter has an emissivity of 1 (100%) We can work out the amount of energy radiated by a surface using the Stefan-Boltzmann law. The total power emitted P = e  A T 4 where e is the emissivity of the surface  is the Stefan-Boltzmann constant =5.7 x 10 -8 Wm -2 K -4 A is the surface area of the emitter T is the temperature of the emitter in kelvin

11. Emissivity values MaterialEmissivity Ice and snow0.94-0.99 Vegetation0.94-0.99 Soil0.75-0.99 Seawater0.98-0.99 Grassland0.95 Barren Land0.93 Forest0.96 Urban0.95

12. The temperature of the Earth also depends on its surface heat capacity C s E = A C s  T where E is the energy incident on the surface A is the area of the surface  T is the change in temperature Note that the surface heat capacity will vary with the nature of the surface. The average value for the 2.1 x 10 8 J m-2 K-1 and is heavily dependent on water’s specific heat capacity.