Solved Problem on Black Body Radiation

a) Determine the rest mass lost per second by the sun in the form of radiation. Data: sun surface temperature: 5700 K; sun diameter: 1.4 × 109 m; Stefan-Boltzmann constant: \( \sigma =5.67\times 10^{-8}\;\frac{\text{W}}{\text{m}^{2}.\text{T}^{4}} \); speed of light in vacuum: 3.0 × 108 m/s;
b) What is the fraction of the rest mass lost each year by the sun from the electromagnetic radiation. Solar mass: 2.0 × 1030 kg.

Problem data:
  • Sun surface temperature:    T = 5700 K;
  • Sun diameter:    D =1.4 × 109 m;
  • Solar mass:    M = 2.0 × 1030 kg;
  • Stefan-Boltzmann constant:    \( \sigma =5.67\times 10^{-8}\;\frac{\text{W}}{\text{m}^{2}.\text{T}^{4}} \);
  • Speed of light in vacuum:    v = 3.0 × 108 m/s.

a) Using the Stefan-Boltzmann Law, spectral radiance is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {R_{T}=\sigma T^{4}} \tag{I} \end{gather} \]
Total radiance is defined as the irradiated power by area unit, we can write
\[ \begin{gather} R_{T}=\frac{P}{A} \tag{II} \end{gather} \]
equating the expressions (I) and (II)
\[ \begin{gather} \sigma T^{4}=\frac{P}{A}\\[5pt] P=A\;\sigma T^{4} \tag{III} \end{gather} \]
Assuming that the sun is spherical, the surface area will be
\[ \begin{gather} A=4\pi r^{2} \tag{IV} \end{gather} \]
The radius of a sphere is half the diameter
\[ \begin{gather} r=\frac{D}{2} \tag{V} \end{gather} \]
substituting the expression (V) into expression (IV) and then in expression (III)
\[ \begin{gather} P=4\pi \left(\frac{D}{2}\right)^{2}\;\sigma T^{4}\\[5pt] P=4\pi \frac{D^{2}}{4}\;\sigma T^{4}\\[5pt] P=\pi \;D^{2}\;\sigma T^{4} \tag{VI} \end{gather} \]
The power is given by the energy differentiation relative to time
\[ \begin{gather} P=\frac{dE}{dt} \tag{VII} \end{gather} \]
the rest energy is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {E=m_{0}c^{2}} \tag{VIII} \end{gather} \]
where m0 is the rest mass, substituting the expressions (VI) and (VIII) into expression (VII)
\[ \begin{gather} \pi D^{2}\sigma T^{4}=\frac{d\left(m_{0}c^{2}\right)}{dt} \end{gather} \]
as the speed of light is constant
\[ \begin{gather} c^{2}\frac{dm_{0}}{dt}=\pi D^{2}\sigma T^{4} \end{gather} \]
thus the variation of the mass will be
\[ \begin{gather} \frac{dm_{0}}{dt}=\frac{\pi D^{2}\sigma T^{4}}{c^{2}} \end{gather} \]
assuming π = 3.14 and substituting the data of the problem
\[ \begin{gather} \frac{dm_{0}}{dt}=\frac{3.14\times (1.4\times 10^{9})^{2}\times (5.67\times 10^{-8})\times (5700)^{4}}{(3\times 10^{8})^{2}}\\[5pt] \frac{dm_{0}}{dt}=\frac{3.14\times 1.96\times 10^{18}\times 5.67\times 10^{-8}\times 1.06\times 10^{15}}{9\times 10^{16}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\frac{dm_{0}}{dt}=4.1\times 10^{9}\ \frac{\text{kg}}{\text{s}}} \end{gather} \]

b) One year in seconds is equal to
\[ \begin{gather} t=1\;\cancel{\text{ano}}\times \;\frac{365\;\cancel{\text{dias}}}{1\;\cancel{\text{ano}}}\times \frac{24\;\cancel{\text{horas}}}{1\;\cancel{\text{dia}}}\times \frac{60\;\cancel{\text{minutos}}}{1\;\cancel{\text{hora}}}\times \frac{60\;\text{segundos}}{1\;\cancel{\text{minuto}}}=31,536,000\;\approx\;3.15\times 10^{7}\text{s} \end{gather} \]
Separating the variables and integrating the expression of the previous item
\[ \begin{gather} \int d{m{\text '}}_{0}=\int 4.1\times 10^{9}dt{\text '} \end{gather} \]
Note: In the expression above m' and t' are dummy variables of integration.

The limits of the integral of the mass is from 0 to m0 (the total lost mass), the integral in time is from 0 to 3.15×107 s (one year in seconds), and moving out the constant 4.1× 109 from the integral
\[ \begin{gather} \int_{{0}}^{m_{0}}d{m{\text '}}_{0}=\int_{{0}}^{3.15\times 10^{7}}4.1\times 10^{9}dt{\text '}\\[5pt] \left.m{\text '}_{0}\;\right|_{0}^{m_{0}}=\left.t{\text '}\;\right|_{0}^{3.15\times 10^{7}}\\[5pt] m_{0}-0=4.1\times 10^{9}\times (3.15\times 10^{7}-0)\\[5pt] m_{0}=4.1\times 10^{9}\times 3.15\times 10^{7}\\[5pt] m_{0}=1.3\times 10^{17}\ \text{kg} \end{gather} \]
so the lost mass fraction each year will be
\[ \begin{gather} \frac{m_{0}}{M}=\frac{1.3\times 10^{17}}{2.0\times 10^{30}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\frac{m_{0}}{M}=6.5\times 10^{-14}} \end{gather} \]