Sunlight on one square metre
Hold a flat plate outside the atmosphere, facing the Sun, at the Earth's average distance from it. The power of the sunlight falling on each square metre of the plate is called the solar constant \(S\). Satellites have measured it for decades: \(S = 1361\,\text{W/m}^2\). This lesson uses that value, the Earth-Sun distance \(d = 1.5 \times 10^{11}\,\text{m}\), the Sun's radius \(R = 6.96 \times 10^8\,\text{m}\) and \(\sigma = 5.67 \times 10^{-8}\,\text{W/(m}^2\,\text{K}^4)\).
Some of that sunlight is reflected and absorbed by the air and the clouds. On a clear day at noon, about \(1000\,\text{W/m}^2\) reaches a surface on the ground that faces the Sun. That is still a great deal: the sunlight on two square metres of a roof is as much as an electric heater gives.
The same power on a bigger sphere
The Sun radiates equally in all directions. Imagine a huge sphere centred on the Sun, with the Earth's orbit on its surface. All of the Sun's power crosses this sphere, since there is nowhere else for it to go. The sphere has area \(4\pi d^2\), and each square metre of it receives \(S\). So the Sun's total power, its luminosity, is \(P = 4\pi d^2 S\).
Put in the numbers: \(P = 4\pi \times (1.5 \times 10^{11})^2 \times 1361\) \(= 3.85 \times 10^{26}\,\text{W}\). The power per square metre falls as \(1/d^2\), the inverse-square law. Mars is 1.52 times as far as the Earth, so it gets \(1361/1.52^2 = 589\,\text{W/m}^2\). Mercury, at 0.39 of the Earth's distance, gets about \(8950\,\text{W/m}^2\).
The Earth catches only a small part of this. Facing the Sun, it blocks a disc of radius \(R_E = 6.4 \times 10^6\,\text{m}\), the Earth's radius. The disc has area \(\pi R_E^2\), so the Earth receives \(\pi R_E^2 S = \pi \times (6.4 \times 10^6)^2 \times 1361\) \(= 1.75 \times 10^{17}\,\text{W}\). That is less than one part in two billion of the Sun's output.
Why is the Sun's total power equal to \(4\pi d^2 S\), where \(d\) is the Earth-Sun distance?
Jupiter is about 5.2 times as far from the Sun as the Earth. How much sunlight falls on each square metre facing the Sun there, in W/m²? Take \(S = 1361\,\text{W/m}^2\) at the Earth.
Taking the Sun's temperature
The Sun's power leaves from its surface, a sphere of radius \(R\). If the Sun radiates as a black body at temperature \(T\), the Stefan-Boltzmann law gives that power as \(\sigma (4\pi R^2) T^4\). The same power arrives at the Earth's distance as \(4\pi d^2 S\). Setting the two equal gives the temperature.
\[ \sigma\,(4\pi R^2)\,T^4 = 4\pi d^2 S \]
Why. the power leaving the Sun's surface equals the power crossing the sphere of radius \(d\).
\[ T^4 = \dfrac{S\,d^2}{\sigma R^2} \]
Why. cancel \(4\pi\) and divide by \(\sigma R^2\).
\[ T^4 = \dfrac{1361 \times (215.5)^2}{5.67 \times 10^{-8}} = 1.11 \times 10^{15}\,\text{K}^4 \]
Why. \(d/R = 1.5 \times 10^{11}/6.96 \times 10^8\) \(= 215.5\), so \((d/R)^2 = 46\,400\).
\[ T = 5780\,\text{K} \]
Read it. take the fourth root, a square root twice: \(\sqrt{1.11 \times 10^{15}} = 3.34 \times 10^7\), and \(\sqrt{3.34 \times 10^7} = 5780\).
This gives 5778 K. Wien's law, from the peak of the Sun's spectrum at 500 nm, gave 5800 K. Two different laws, one from the total power and one from the colour, give almost the same temperature. That agreement is strong evidence that the Sun radiates very nearly as a black body.
Suppose a star had the same surface temperature as the Sun but twice its radius, and sat at the Sun's distance from us. What would the power per square metre at the Earth be?
The Earth's balance
The Earth absorbs sunlight all the time, yet on average it does not keep warming. So it must radiate away as much as it absorbs. It absorbs on the disc it presents to the Sun, of area \(\pi R_E^2\). It radiates from its whole surface, of area \(4\pi R_E^2\), day side and night side, into cold space.
Treat the Earth first as a black body that absorbs all the sunlight. In the steady state, \(\pi R_E^2 S = 4\pi R_E^2\,\sigma T^4\). The radius cancels, and \(T^4 = S/(4\sigma)\). So \(T = (1361/(4 \times 5.67 \times 10^{-8}))^{1/4}\) \(= 278\,\text{K}\), about 5 °C. Any planet at the Earth's distance, large or small, would settle at the same temperature.
The Earth reflects about 30 % of the sunlight falling on it, from clouds, ice and deserts. Find its steady temperature, treating it as a black body for the radiation it emits. Compare with the measured average surface temperature, about 288 K.
Absorbed power: \(0.7 \times \pi R_E^2 S\). Radiated power: \(4\pi R_E^2\,\sigma T^4\).
\(0.7\,\pi R_E^2 S = 4\pi R_E^2\,\sigma T^4\), so \(T = \left(\dfrac{0.7\,S}{4\sigma}\right)^{1/4}\).
\(T^4 = \dfrac{0.7 \times 1361}{4 \times 5.67 \times 10^{-8}}\) \(= 4.20 \times 10^9\), so \(T = 255\,\text{K}\), about \(-18\,°\text{C}\).
The balance gives 255 K, but the measured average is 288 K, 33 K warmer. The difference is the greenhouse effect of the atmosphere.
Why is the real Earth 33 K warmer? At 255 K the Earth radiates in the far infrared, with its peak near \(2.9 \times 10^{-3}/255 = 11\,\mu\text{m}\). Water vapour and carbon dioxide let the short-wavelength sunlight through, but absorb much of this infrared. They radiate part of it back down, so the ground must be warmer to push the same power out to space. This is the greenhouse effect. It is the clear desert sky of Lesson 1 in reverse: a cloudy, humid sky returns more.
Mars is 1.52 times as far from the Sun as the Earth. A black Earth would settle at \(278\,\text{K}\). Ignoring reflection and atmosphere, what temperature would a black Mars settle at, in kelvin?
A black planet twice the Earth's radius orbits at the Earth's distance. Compared with a black Earth, its steady temperature is
A panel on the roof
A rooftop solar panel of area \(1.7\,\text{m}^2\) faces the Sun at noon on a clear day. About \(1000\,\text{W/m}^2\) falls on it, so it receives about \(1700\,\text{W}\). A good silicon panel turns about 20 % of this into electricity, \(340\,\text{W}\). The remaining \(1360\,\text{W}\) becomes heat in the panel.
That heat must leave the panel by radiation and by the moving air, the subjects of this chapter. So a panel on a hot roof runs well above the air temperature. A hot panel is less efficient. So installers leave a gap below each panel, and air flowing behind it removes the heat.
A flat roof of a house in Jaipur is \(10\,\text{m}\) by \(10\,\text{m}\). Roughly how much solar power falls on it at noon on a clear day in May?
In 1838 the French physicist Claude Pouillet built a pyrheliometer, a flat can of water with a blackened face turned to the Sun. He timed how fast the water warmed and calculated the power of the sunlight on it. His value, about 1230 watts per square metre, was the first sound measurement of the solar constant. It was low only because he could not measure from above the air.
Millions of Indian homes now carry solar panels on their roofs, many helped by government schemes. A 3 kW system of about eight panels needs a roof area of roughly \(20\,\text{m}^2\). On a clear day it gives about 12 to 15 units of electricity, enough for a family's lights, fans and a refrigerator. The solar constant and the angle of the Sun set the limit on what any roof can collect.
- Using the Earth's whole surface, \(4\pi R_E^2\), for the sunlight it catches.The Earth catches sunlight on a disc, \(\pi R_E^2\). It radiates from the whole sphere, \(4\pi R_E^2\).
- Using the Sun's radius in \(4\pi d^2 S\).\(S\) is measured at the Earth's distance, so the sphere has radius \(d\), the Earth-Sun distance.
- Forgetting that a planet's radius cancels.Absorption and emission both grow as \(R^2\), so the steady temperature does not depend on size.
- Taking the solar constant as the sunlight at the ground.\(S = 1361\,\text{W/m}^2\) is above the air. At the ground, at noon on a clear day, it is about \(1000\,\text{W/m}^2\).