= Solution
<Kirchhoff's law of thermal radiation> equates spectral absorptivity and <spectral emissivity> at the same <wavelength> and direction. A surface that is a poor absorber in the thermal <infrared> is also a poor emitter there; a coating cannot be assigned its solar absorption independently of its thermal emission spectrum.
\b[(i) All-wavelength black paint.] It absorbs the sunlight and also radiates maximally at the spacecraft's own <temperature>. With the isothermal-sphere assumptions of (d), the limiting black-surface distance is
$$
\boxed{\frac{R_{\min,\rm black}}r=\frac12\left(\frac{T_S}{T_{\max}}\right)^2=\frac12\left(\frac{6000}{600}\right)^2=50.}
$$
The spacecraft can operate at $R\geq50r$ in this model. A surface already assumed to be a <blackbody> is not improved by another black coating; relative to option (ii), however, this paint permits closer operation.
\b[(ii) Black at short <wavelengths> and grey in the <infrared>.] By <Wien's displacement law>, changing the <black-body radiation> <temperature> from $6000\,\mathrm K$ to $600\,\mathrm K$ shifts the <wavelength> peak from about $0.5\,\mu\mathrm m$ to about $5\,\mu\mathrm m$. Most solar <power> is therefore in the highly absorbing short-wavelength band, whereas most spacecraft emission lies beyond $2\,\mu\mathrm m$, where the <spectral emissivity> is reduced. The coating retains most of the heating but suppresses much of the cooling. Under the intended two-band <grey body> approximation, \b[option (ii) makes the spacecraft hotter at the same distance and requires a larger minimum distance than black paint.]
To express the comparison without inventing a grey absorptivity, let $\bar\alpha_S$ be <radiometric absorptance> averaged over the solar spectrum and $\bar\epsilon_T$ be emissivity averaged over the spacecraft's own thermal spectrum. The <radiative equilibrium> equation becomes
$$
\bar\alpha_S\sigma T_S^4\frac{r^2}{R^2}=4\bar\epsilon_T\sigma T^4,
\qquad
T=T_S\sqrt{\frac r{2R}}\left(\frac{\bar\alpha_S}{\bar\epsilon_T}\right)^{1/4}.
$$
At the failure threshold,
$$
\boxed{\frac{R_{\min}}r=50\sqrt{\frac{\bar\alpha_S}{\bar\epsilon_{600}}}.}
$$
For an ideal two-band coating take a constant <radiometric absorptance> $g<1$ beyond $\lambda_c=2\,\mu\mathrm m$, and unity below it. If $f(T)$ is the fraction of the <power> emitted by a <blackbody> below $\lambda_c$, <Planck's law> gives
$$
f(T)=\frac{15}{\pi^4}\int_{hc/(\lambda_ck_BT)}^\infty\frac{x^3}{e^x-1}\,dx,
\qquad
\bar\alpha_S=g+(1-g)f(T_S),\quad
\bar\epsilon_{600}=g+(1-g)f(600\,\mathrm K).
$$
Since $f(T_S)>f(600\,\mathrm K)$, this model has $\bar\alpha_S>\bar\epsilon_{600}$ and hence $R_{\min}>50r$. The paper supplies neither a numerical value of $g$ nor a detailed spectrum, so no unique numerical distance for the grey paint follows. If its long-wavelength absorption is not approximately grey, the spectrum must be integrated explicitly rather than inferred from a peak alone. A coating designed to improve close approach would instead reflect much of the solar radiation while retaining strong thermal <infrared> emission.
Back to article page
