# Temperature of a sphere heating up in the sun (average emissivity of the sphere unknown)

I'm stuck at the following exercise:

Imagine an aluminum sphere that is painted matte white with an emissivity of $$\epsilon=0.10$$ at visible wavelengths and $$\epsilon=0.95$$ at wavelengths longer than $$5\text{ }\mu\text{m}$$. This sphere is orbiting the sun receiving a power flux of $$R_{sun}=1330\text{ Wm}^{-2}$$. Find the steady-state temperature of the sphere. Find the peak wavelength of the radiated energy and determine which value of $$\epsilon$$ should be used to calculate the radiated energy.

Now first I realized that the sphere with Radius $$r$$ only receives the radiation of the sun on an area of $$\pi r^2$$ while being able to emmit radiation over its whole area of $$4\pi r^2$$. So the power flux emitted by the sphere should be a quarter of the power flux it receives:

$$R_{sphere}=332.5\text{ Wm}^{-2}$$

I don't get much further than that though. I woud like to continue calculating the temperature, for example with $$R=\epsilon_{avg}\sigma T^4$$, but the average emissivity is dependent on the temperature and I don't even know $$\epsilon$$ for most wavelengths, so that's not really an option. The only other relevant equation that I know is Planck's law:

$$\frac{dR}{d\lambda}=\frac{2\pi hc^2}{\lambda^5}\frac{\epsilon}{\exp(hc/k\lambda T)-1}$$

But $$\epsilon$$ is only given for visible light and wavelengths longer than 5 micrometers. I thought about approximating the gap between those with a linear function like this:

$$\epsilon=\left\{\begin{matrix}0.1&0<\lambda<750nm\\0.2\lambda/\mu m-0.05&750nm<\lambda<5\mu m\\0.95&5\mu m<\lambda\end{matrix}\right.$$

but even with that I don't know how I could use Planck's law to calculate the temperature. Any ideas are greatly appreciated.

• "So the power flux emitted by the sphere should be a quarter of the power flux it receives:" Why do you say this? If this is an equilibrium situation then power in = power out. – Rob Jeffries Dec 13 '19 at 11:02
• @RobJeffries Yes, the power output is equal to the power input. However, the power flux is in the units of power per area. We have given the power flux of the sun, with which we can calculate the power received by the sphere by calculating $P=R_{sun}\cdot\pi r^2$. And now, the sphere radiates that exact same power in all directions, so over an area of $4\pi r^2$, leading to a power flux of $R_{sphere}=P / 4\pi r^2 = R_{sun} / 4$. Of course, this power flux decreases the further you move away from the sphere. I'm giving the value at the surface. – Keno Dec 13 '19 at 12:37
• I think that KM's answer addresses this nicely, but if you found a better solution you can always add an additional answer to your own question. You can also see how early spacecraft handled this problem in practice in Are black and white stripes any better than uniform gray for thermal control? and also Why were Europe's first few satellites so stylish? Why the pronounced alternating white and black stripes? – uhoh Sep 2 at 3:49

I think the idea is that there is very little overlap in wavelength between the Sun's and the sphere's radiation curves (maybe you need to provide an argument for this), so you can treat them separately. You need to equate (taking account of the factor of four) $R=\epsilon_{avg}\sigma T^4$ for the Sun's temperature and the optical emissivity and for the sphere's (unknown) temperature and infrared emissivity .