IPhO 1992 · Theory problem 3

A Satellite in Sunshine

Download PDFIPhO 1992

In this problem you will calculate the temperature of a space satellite. The satellite is assumed to be a sphere with a diameter of 1 m. All of the satellite remains at a uniform temperature. All of the spherical surface of the satellite is coated with the same kind of coating. The satellite is located near the earth but is not in the earth's shadow.

The surface temperature of the sun (its blackbody surface temperature) \(T_{\mathrm{sun}} = 6000\ \mathrm{K}\) and its radius is \(6.96 \times 10^8\ \mathrm{m}\). The distance between the sun and the earth is \(1.5 \times 10^{11}\ \mathrm{m}\). The sunlight heats the satellite to a temperature at which the blackbody emission from the satellite equals the power absorbed from the sunlight. The power per unit area emitted by a black body is given by Stefan-Boltzmann's law \(P = \sigma T^4\) where \(\sigma\) is the universal constant \(5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}\). In the first approximation, you can assume that both the sun and the satellite absorb all electromagnetic radiation incident upon them.

1) Find an expression for the temperature \(T\) of the satellite. What is the numerical value of this temperature?

2) The blackbody radiation spectrum \(u(T,f)\) of a body at temperature \(T\) obeys Planck's radiation law

\[ u(T,f)\,df = \frac{8\pi k^4T^4}{c^3h^3} \frac{\eta^3\,d\eta}{e^\eta-1} \]

where \(\eta = hf/kT\) and \(u(T,f)df\) is the energy density of the electromagnetic radiation in a frequency interval \([f,f+df]\). In the equation \(h = 6.6 \times 10^{-34}\ \mathrm{J\,s}\) is Planck's constant, \(k = 1.4 \times 10^{-23}\ \mathrm{J\,K^{-1}}\) is Boltzmann's constant and \(c = 3.0 \times 10^8\ \mathrm{m\,s^{-1}}\) is the speed of light.

The blackbody spectrum, integrated over all frequencies \(f\) and directions of emission, gives the total radiated power per unit area \(P = \sigma T^4\) as expressed in the Stefan-Boltzmann law given above.

\[ \sigma = \frac{2\pi^5k^4}{15c^2h^3} \]

The figure shows the normalized spectrum

\[ \frac{c^3h^3}{8\pi k^4}\frac{u(T,f)}{T^4} \]

as a function of \(\eta\).

In many applications it is necessary to keep the satellite as cool as possible. To cool the satellite, engineers use a reflective coating that reflects light above a cut-off frequency but does not prevent heat radiation at lower frequency from escaping. Assume that this (sharp) cut-off frequency corresponds to \(hf/k = 1200\ \mathrm{K}\).

What is the new equilibrium temperature of the satellite? The exact answer is not needed. Therefore, do not perform any tedious integrations; make approximations where necessary. The integral over the entire range is

\[ \int_0^\infty \frac{\eta^3\,d\eta}{e^\eta-1} = \frac{\pi^4}{15} \]

and the maximum of \(\eta^3/(e^\eta-1)\) occurs at \(\eta = 2.82\). For small \(\eta\) you can expand the exponential function as \(e^\eta \simeq 1+\eta\).

3) If we now have a real satellite, with extending solar panels that generate electricity, the dissipated heat in the electronics inside the satellite acts as an extra source of heat. Assuming that the power of the internal heat source is 1 kW, what is the equilibrium temperature of the satellite in case 2 above?

4) A manufacturer advertises a special paint in the following way:

"This paint will reflect more than 90% of all incoming radiation (both visible light and infrared) but it will radiate at all frequencies (visible light and infrared) as a black body, thus removing lots of heat from the satellite. Thus the paint will help keep the satellite as cool as possible."

Can such paint exist? Why or why not?

5) What properties should a coating have in order to raise the temperature of a spherical body similar to that of the satellite considered here above the temperature calculated in 1)?

Figure 1

Official English text of the problem.

Answers

  1. With solar radius \(R_\odot\) and Sun–satellite distance \(D\), \[ \boxed{T_1=T_{\mathrm{sun}}\sqrt{\frac{R_\odot}{2D}} \simeq 289\ \mathrm{K}.} \]

  2. Using the specified small-\(\eta\) approximation, \[ \boxed{T_2\simeq T_1 \left[\frac{5}{\pi^4} \left(\frac{1200\ \mathrm{K}}{T_{\mathrm{sun}}}\right)^3\right]^{1/4} \simeq 41\ \mathrm{K}.} \]

  3. Accounting for the coating’s suppression of high-frequency thermal emission, \[ \boxed{T_3\simeq 3.2\times10^2\ \mathrm{K} \quad\text{(approximately \(317\ \mathrm{K}\)).}} \]

  4. No. Kirchhoff’s law requires equal absorptivity and emissivity at each frequency. A surface reflecting more than \(90\%\) cannot simultaneously emit as a blackbody at those frequencies.

  5. The coating should absorb solar radiation strongly and emit thermal infrared radiation weakly: high absorptivity at solar frequencies and high reflectivity at the satellite’s lower thermal frequencies.

Solution

Let the satellite radius be \(a=0.50\ \mathrm{m}\), the solar radius be \(R_\odot\), and the Sun–satellite distance be \(D\). We use the stated radiation-balance model, neglecting radiation from Earth and other heat exchanges.

1) Blackbody satellite

The Sun’s luminosity is

\[ L_\odot=4\pi R_\odot^2\sigma T_{\mathrm{sun}}^4. \]

Consequently, the solar irradiance at the satellite is

\[ S=\frac{L_\odot}{4\pi D^2} =\sigma T_{\mathrm{sun}}^4\left(\frac{R_\odot}{D}\right)^2. \]

The satellite intercepts sunlight over its projected area \(\pi a^2\), while its entire surface \(4\pi a^2\) emits radiation. Thermal equilibrium therefore requires

\[ \pi a^2S=4\pi a^2\sigma T_1^4. \]

Thus

\[ T_1^4=\frac14 T_{\mathrm{sun}}^4 \left(\frac{R_\odot}{D}\right)^2, \qquad \boxed{T_1=T_{\mathrm{sun}}\sqrt{\frac{R_\odot}{2D}}.} \]

Numerically,

\[ T_1 =6000\sqrt{\frac{6.96\times10^8}{2(1.5\times10^{11})}} \ \mathrm{K} =\boxed{289\ \mathrm{K}}. \]

The satellite’s radius cancels because both absorbed and emitted powers scale as \(a^2\).

2) Satellite with the frequency-selective coating

Write

\[ \Theta=\frac{hf_c}{k}=1200\ \mathrm{K}, \]

where \(f_c\) is the cutoff frequency. The ideal coated satellite absorbs and emits as a blackbody below \(f_c\), while reflection suppresses both absorption and emission above \(f_c\). This equality of spectral absorptivity and emissivity is Kirchhoff’s law.

Define the fraction of blackbody power below a dimensionless cutoff \(x\) by

\[ F(x)=\frac{15}{\pi^4} \int_0^x\frac{\eta^3}{e^\eta-1}\,d\eta. \]

For a body at temperature \(T\), the relevant cutoff is \(x=\Theta/T\). The equilibrium equation is therefore

\[ \pi a^2\sigma T_{\mathrm{sun}}^4 \left(\frac{R_\odot}{D}\right)^2 F\left(\frac{\Theta}{T_{\mathrm{sun}}}\right) = 4\pi a^2\sigma T_2^4 F\left(\frac{\Theta}{T_2}\right). \]

Using part 1,

\[ T_2^4F\left(\frac{\Theta}{T_2}\right) = T_1^4F\left(\frac{\Theta}{T_{\mathrm{sun}}}\right). \]

For sunlight,

\[ x_\odot=\frac{1200}{6000}=0.20. \]

Throughout this small interval, \(e^\eta-1\simeq\eta\), so

\[ F(0.20)\simeq \frac{15}{\pi^4}\int_0^{0.20}\eta^2\,d\eta = \frac{5(0.20)^3}{\pi^4} \simeq4.11\times10^{-4}. \]

Thus only about \(0.041\%\) of the incident solar power is absorbed.

We initially assume that the cooled satellite emits almost all its thermal radiation below the cutoff, so \(F(\Theta/T_2)\simeq1\). Then

\[ T_2\simeq T_1[F(0.20)]^{1/4} =289(4.11\times10^{-4})^{1/4}\ \mathrm{K}, \]

giving

\[ \boxed{T_2\simeq41\ \mathrm{K}.} \]

This assumption is self-consistent: the satellite’s cutoff is

\[ \frac{\Theta}{T_2}\simeq\frac{1200}{41}\simeq29, \]

far beyond the spectral maximum at \(\eta=2.82\). The excluded high-frequency thermal tail is negligible.

3) Addition of a \(1\ \mathrm{kW}\) internal heat source

The emitting area is now needed explicitly:

\[ A=4\pi a^2=\pi\ \mathrm{m^2}. \]

Before applying the coating, the absorbed solar power was

\[ Q_{\odot,0}=A\sigma T_1^4 \simeq1.24\times10^3\ \mathrm{W}. \]

With the coating, it becomes

\[ Q_\odot=Q_{\odot,0}F(0.20) \simeq(1.24\times10^3)(4.11\times10^{-4}) \simeq0.51\ \mathrm{W}. \]

This is negligible compared with the internal heat source \(Q_{\mathrm{int}}=1000\ \mathrm{W}\). The balance is

\[ \boxed{ A\sigma T_3^4F\left(\frac{1200\ \mathrm{K}}{T_3}\right) =1000\ \mathrm{W}+Q_\odot. } \]

Here we cannot set \(F=1\). Even an unrestricted blackbody would need the temperature

\[ \left(\frac{1000\ \mathrm{W}}{A\sigma}\right)^{1/4} \simeq274\ \mathrm{K}, \]

for which the cutoff would already lie at \(\eta_c\simeq4.38\). A significant part of the thermal spectrum is above that cutoff, so the coated satellite must become hotter.

We can estimate the required fraction directly from the supplied graph. Near \(\eta_c=3.8\), the area under the curve is approximately

\[ \int_0^{3.8}\frac{\eta^3}{e^\eta-1}\,d\eta\simeq3.6. \]

Since the total area is \(\pi^4/15\simeq6.49\),

\[ F(3.8)\simeq\frac{3.6}{6.49}\simeq0.55. \]

Substitution gives

\[ T_3\simeq \left[ \frac{1000.5\ \mathrm{W}} {(\pi\ \mathrm{m^2})(5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}) (3.6/6.49)} \right]^{1/4} \simeq317\ \mathrm{K}. \]

Its corresponding cutoff is \(1200/317\simeq3.79\), confirming the graphical estimate. Therefore,

\[ \boxed{T_3\simeq3.2\times10^2\ \mathrm{K}.} \]

4) The advertised paint

Kirchhoff’s law of thermal radiation states that, at a given frequency and for corresponding directions,

\[ \epsilon_f=\alpha_f, \]

where \(\epsilon_f\) is emissivity and \(\alpha_f\) is absorptivity.

Energy conservation gives

\[ \alpha_f+\rho_f+\tau_f=1, \]

where \(\rho_f\) and \(\tau_f\) are reflectivity and transmissivity. Consequently, if \(\rho_f>0.90\),

\[ \alpha_f<0.10 \quad\Longrightarrow\quad \epsilon_f<0.10. \]

Blackbody emission would instead require \(\epsilon_f=1\).

Thus the claimed combination of high reflection and blackbody thermal emission at the same frequencies is impossible for a passive paint. It would violate Kirchhoff’s law and detailed thermal balance.

5) A coating that raises the temperature

The coating should have the opposite spectral selectivity to that in part 2:

  • Strong absorption at the relatively high frequencies carrying most solar radiation.
  • Weak emission, and correspondingly strong reflection, at the lower infrared frequencies carrying most of the satellite’s thermal radiation.

Let \(\bar\alpha_\odot\) be absorptivity averaged over the solar spectrum, and let \(\bar\epsilon_T\) be emissivity averaged over the satellite’s blackbody spectrum at temperature \(T\). The balance without internal heating is

\[ \pi a^2\bar\alpha_\odot S = 4\pi a^2\bar\epsilon_T\sigma T^4. \]

Hence

\[ \boxed{ T=T_1\left(\frac{\bar\alpha_\odot}{\bar\epsilon_T}\right)^{1/4}. } \]

The equilibrium temperature exceeds \(T_1\) when

\[ \boxed{\bar\alpha_\odot>\bar\epsilon_T.} \]

This is compatible with Kirchhoff’s law because the two averages weight different frequency ranges. A frequency-independent grey coating would have equal absorptivity and emissivity and would leave the equilibrium temperature unchanged.

The solution was checked against the official solution.