# Mars has an average surface temperature of about 200K. Pluto has an average surface temperature of about 40K. Which planet emits more energy per square meter of surface area per second? By a factor of how much?

Jul 12, 2016

#### Answer:

Mars emits $625$ times more energy per unit of surface area than Pluto does.

#### Explanation:

It is obvious that a hotter object will emit more black body radiation. Thus, we already know that Mars will emit more energy than Pluto. The only question is by how much.

This problem requires evaluating the energy of the black body radiation emitted by both planets. This energy is described as a function of temperature and the frequency being emitted:

$E \left(\nu , T\right) = \frac{2 {\pi}^{2} \nu}{c} \frac{h \nu}{{e}^{\frac{h \nu}{k T}} - 1}$

Integrating over frequency gives the total power per unit area as a function of temperature:
${\int}_{0}^{\infty} E \left(\nu , T\right) = \frac{{\pi}^{2} c {\left(k T\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}}$

(note that the above equation uses $\overline{h}$, the reduced Planck's constant, rather than $h$. It is difficult to read in Socratic's notation)

Solving for the ratio between the two, then, the result is incredibly simple. If ${T}_{p}$ is Pluto's temperature and ${T}_{m}$ is Mars' temperature then the factor $a$ can be calculated with:

$\frac{{\pi}^{2} c {\left(k {T}_{m}\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}} = a \frac{{\pi}^{2} c {\left(k {T}_{p}\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}}$
$\frac{\cancel{{\pi}^{2} c {k}^{4}}}{\cancel{60 {\left(\overline{h} c\right)}^{3}}} {T}_{m}^{4} = a \frac{\cancel{{\pi}^{2} c {k}^{4}}}{\cancel{60 {\left(\overline{h} c\right)}^{3}}} {T}_{p}^{4}$
${\left({T}_{m} / {T}_{p}\right)}^{4} = a = {\left(\frac{200}{40}\right)}^{4} = {5}^{4} = 625$ times as much