Almost all the Earth's energy (other than a bit of geothermal energy, some naturally occurring nuclear decay, and a few cosmic rays) comes from the Sun. The Sun is predominantly a ball of fiery hydrogen plasma with a surface temperature of about 5505 °C (or 5778 K). The energy it produces comes from the nuclear fusion of that hydrogen into helium within its core (the core being defined as the region of the interior that is less than 0.7Rs from the centre, where Rs is the distance from the centre of the Sun to its outer edge). This fusion process gives off electromagnetic radiation with a range of frequencies, but which are mainly in the visible portion of the electromagnetic spectrum. It is the force of this radiation pushing out as it tries to escape that stops the Sun collapsing into a black hole under its own gravitational force. It also takes up to a year for many of these photons (particles) of light to actually escape. By contrast it only takes a further 8 minutes and 24 seconds for them to reach Earth.
The energy released by the Sun is 3.9 x 1026 joules per second, or a power of P = 3.9 x 1026 W. This is dispersed uniformly in all directions so that by the time it reaches Earth (at a distance r = 1.519 x 1011 m from the Sun), it has spread out (due the inverse-square law) over a sphere in space of area A = 2.9 x 1023 m2 (= 4πr2). This means that the intensity of the radiation that we receive is P/A, or 1361 W/m2. This is the power per unit area reaching Earth at the top of its atmosphere, and it still comprises the full range of different frequencies that were emitted from the Sun, each with a different relative intensity, as illustrated below in Fig. 12.1.
Fig. 12.1: The electromagnetic radiation output of the Sun (yellow) and the radiation received at ground-level on Earth (red).
What Fig. 12.1 shows is that the Sun's radiation output closely (but not exactly) resembles that from a black body of temperature 5250 °C (black curve), and that not all this radiation arrives at ground-level on Earth. Some is blocked by the atmosphere and reflected back into space.
In physics a black body is defined as an object that absorbs all the electromagnetic radiation that falls on its surface. It also has the property that when it is in thermal equilibrium with its surroundings at a constant thermodynamic temperature (i.e. the temperature measured on the kelvin scale), then it will emit heat in the form of electromagnetic radiation across a continuous range of frequencies. The distribution or spectrum of those frequencies is known as the black body spectrum, and it always has the normalized functional form shown in Fig. 12.2 below.
Fig. 12.2: The black body frequency spectrum.
The curve in Fig. 12.2 is described by the general equation
(12.1)
where a = 15/π4 is a scaling constant that ensures that the area under the curve is 1, and the parameter x depends on the thermodynamic temperature T (in kelvin) and the electromagnetic frequency (ν) as follows:
(12.2)
The other terms are Planck's constant, h, and the Boltzmann constant, k. By using the normalized variable x in Eq. 12.1 instead of ν and T, the equation can be applied universally. It is also worth noting that as λν = c, where λ is the wavelength of the radiation and c is the speed of light, it follows that x = b/Tλ where b = 14387.9 µm.K.
The actual intensity of the radiation emitted by a black body is then given by Planck's law as follows
(12.3)
where B(ν,T) represents the power per unit area of the surface emitted per steradian of solid angle. The total power emitted by a unit area of the surface, I(T), is therefore the sum of B(ν,T) over all frequencies and solid angles (this adds a factor of 2π for the half sphere the emission is flowing into).
(12.4)
We have already defined in Eq. 12.1 that
(12.5)
so it follows that
(12.6)
The result is known as the Stefan-Boltzmann law and σ = 5.67 x 10-8 Wm-2K-4 is the Stefan-Boltzmann constant. This law represents the power that will be emitted from a perfect black body at a temperature T. This therefore represents the maximum power that any real object at that temperature can emit. In reality the emission is always less because no object behaves like a perfect black body, except a black hole.
What this demonstrates is that the power density output of a black body is not only proportional to the fourth power of the temperature, but it depends only on temperature. All the other quantities in σ are fundamental constants that cannot be changed. From this law we can therefore work out the maximum power output of the Sun. The surface temperature is 5778 K, so from Eq. 12.6 the power density of the emitted radiation is 63.2 MW/m2. As the radius of the Sun is about 700,000 km, the surface area is 6.16 x 1018 m2. So the total power output is 3.9 x 1026 W, in agreement with the value at the top of this post.
Using these same equations we can now work out how hot the Earth will become when illuminated by the Sun's rays. We have already determined that the radiation intensity at the top of the atmosphere is ITOA = 1361 W/m2. The amount of power captured by the Earth will therefore be ITOA multiplied by the cross-sectional area of the Earth πRE2 where RE is the Earth's radius. But, as the Earth spins on its axis, this energy will need to heat up the entire surface area of the sphere, i.e. an area of 4πRE2. So the mean power density striking the Earth's surface will be a quarter of 1361 W/m2, or 340.25 W/m2.
This power will heat up the surface of the Earth until the point is reached where the temperature of the surface is high enough to ensure that the power that the Earth re-emits through the Stefan-Boltzmann law balances the incoming power from the Sun. This will happen when the surface temperature Ts is high enough so that σTs4 = 340.25 W/m2. This equates to a temperature of Ts = 278 K or 5 °C. This is not the mean temperature of planet Earth, though. It is what the mean temperature would be if the Earth had no atmosphere and the surface were completely absorbant and black. Except that this neglects the effects of the curvature of the Earth and its thermal conductivity. The former would mean the temperature at the poles would barely rise above 3 K (-270 °C), and that at the equator it could be as high as 393K or 120 °C in the daytime. The thermal conductivity of the surface would, on the other hand, mitigate these extremes.
In reality much of the incoming radiation is reflected by the atmosphere, the clouds, and the surface. This is the origin of the difference in the red and yellow areas in Fig. 12.1. The net result is that only about 161 W/m2 gets absorbed by the Earth's surface, so according to Eq. 12.6 the surface temperature should not rise above 231 K, or -42 °C. However, this isn't true either. The mean temperature is actually closer to 289 K, or +16 °C. This is where the Greenhouse Effect comes into play.
While the atmosphere stops much of the incoming radiation getting in, it also stops much of the longer wavelength radiation emitted by the surface (and acting as a black body at 231K) getting out. Because the surface of the Earth is at a much lower temperature than the Sun, most of the radiation it emits will be at a much lower frequency, or longer wavelength. Some of this radiation then gets reflected back by the atmosphere, and in particular by the greenhouse gases, carbon dioxide and water vapour.
If the fraction reflected back we denote by f, and the reflected power is Ir, then it follows that Ir = f IT, where IT is the total power emitted from the surface. But as Ir is now incident on the Earth's surface, it too must be absorbed and add to the heating. It therefore follows that
(12.7)
where Io = 161 W/m2 is the original power absorbed by the Earth's surface. Following some elementary algebra we then obtain the result
(12.8)
As f is a number between 0 and 1.0, it follows that IT will be greater than Io. That in essence is the Greenhouse Effect with IT being the power emitted from the Earth's surface with the Greenhouse Effect included, and Io being the power emitted with it omitted. The only debate is what the value of f is, and by how much it is changing. Based on a surface temperature of 289 K, it would appear that f = 0.593; in other words 59.3% of the emitted power is reflected back. It is this value that I will look at in more detail next time when I discuss the Earth's energy budget.
On a final note, I will now go back to the emission spectra in Fig.12.1 and Fig. 12.2. There is an inconsistency here in that the first is a function of wavelength (λ), while the second (and most of the subsequent equations) is a function of frequency (ν). Does this matter? Well yes, particularly if you are wanting to calculate the total energy, or find the emission peak.
The peak or maximum of the curve in Fig. 12.2 occurs when x = 2.821, which means that for any object at a constant thermodynamic temperature, T, the frequency at which the maximum amount of radiation is emitted from that object, νmax, is determined by the following formula:
(12.9)
For the Sun with a surface temperature of 5778 K, the peak frequency is 3.39 x 1014 Hz or 339 THz.
For those who may have forgotten their basic high-school physics, I remind you that frequency and wavelength are related by the equation λν = c, where c is the wave speed. For electromagnetic radiation c is the speed of light and it is always constant and the same for all wavelengths and frequencies. This means as the wavelength increases, the frequency decreases, and visa versa. It also means that as the speed of light is 3 x 108 m/s, the frequency νmax corresponds to a wavelength of 885 nm. Yet the peak in the solar spectrum in Fig. 12.1 is at 500 nm. So why are these two numbers not the same?
Well the crux of the problem is this. The theoretical black body spectrum in Fig. 12.2 with its peak at 2.821 in dimensionless units is a scaled version of Eq. 12.3. The quantity B(ν,T) represents the power density for each unit of frequency. When you integrate the area under the curve you get the total power. Now if you plot B(ν,T) as function of wavelength (λ) you are basically plotting the function below
(12.10)
For the case T = 5778 K this will still have a peak at λ = 885 nm, but when you integrate under this curve with respect to λ you do not get the result in Eq. 12.6. In other words, the area under this curve is not equal to the total power density. Instead you need to integrate with respect to the frequency and that means making the following transformation:
(12.11)
Note the switch of limits in Eq. 12.11 (zero to infinity and the reverse). The first switch occurs because the limits are now for λ and not ν. The second switch is because the differential of ν with respect to λ yields -c/λ2 and removing the minus sign reverses the limits again. The result is that the equivalent spectral function for wavelength will be
(12.12)
If the function D(λ,T) is integrated over all wavelengths, the total power density is obtained.
(12.13)
Because the index of 1/λ in the function D(λ,T) is different from the index of ν in B(ν,T), it follows that the peaks in each spectra will be at different wavelengths because these are now inherently different equations. We can demonstrate this by defining a new dimensionless parameter y such that
(12.14)
where x is the same term as was introduced in Eq. 12.2. It then follows from Eq. 12.1, and using the differential result from Eq. 12.11, that we can define a second dimensionless term for λ that has the form
(12.15)
where again a = 15/π4. The area under this curve does indeed equal unity as before
(12.16)
which means that the function D(λ,T) can be rewritten to incorporate q(y) as follows:
(12.17)
If the function q(y) is plotted as a function of y, the result is the curve shown in Fig. 12.3 below.
Fig. 12.3: The black body wavelength spectrum.
The peak or maximum of the curve in Fig. 12.3 occurs when y = 0.201, which means that for any object at a constant thermodynamic temperature, T, the wavelength at which the maximum amount of radiation is emitted from that object, λmax, is determined by the following formula:
(12.18)
This in turn leads to Wien's displacement law, which states that the product of the black body temperature, T, and the peak wavelength at that temperature, λmax, is always equal to the same constant. In other words
(12.19)
The value of Wλ is 2.90 x 10-3 Km. That means that when the temperature of the object is 5778 K, the peak wavelength will be 501 nm. This is why the peak of the solar spectrum in Fig. 12.1 is in the visible region of the spectrum corresponding to yellow light. Then, if the temperature decreases, so the peak wavelength will move to higher wavelengths as shown in Fig. 12.4 below.
Fig. 12.4: The black body wavelength spectrum at different temperatures.
However, it is not just the wavelength spectrum that obeys Wien's displacement law; the frequency spectrum does as well, it is just that the constant Wν is different. Substituting c/λmax for νmax in Eq. 12.9 results in Eq. 12.20.
(12.20)
The value of Wν is 5.11 x 10-3 Km. This means that when the temperature of an object is 5778 K, the peak wavelength for its thermal emission will be at 884 nm. But while the position of the peak may change, what does not change is the amount of power found in a particular spectral range, even if that range appears at different places in Fig. 12.2 and Fig. 12.3.
For example, visible light is generally found between wavelengths of 400 nm and 800nm. In the case of the solar spectrum, the 400 nm radiation will have an equivalent x value of 6.236 according to Eq. 12.2 and a y value of 0.160 according to Eq. 12.14. For 800 nm radiation the values are x = 3.118 and y = 0.321. If the area under the curve in Fig. 12.2 between x = 3.118 and x = 6.236 is measured, it turns out to be 0.460 or 46.0% of the total area under the curve. If we do the same for the curve in Fig. 12.3 between y = 0.160 and y = 0.321 then we get the same result. Perhaps more interestingly, and relevant is, if you do the same for a 289 K spectrum for wavelengths between 13.89 μm and 16.13 μm you get an answer of 10.5%, but that is another story.
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