1. **State the problem:** Calculate the solar energy flux $S_p$ at Mercury's orbit using the Stefan-Boltzmann law and the inverse square law for radiation spread. 2. **Given data:** - Sun's surface temperature $T = 5800$ K - Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8} \text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-4}$ - Sun's radius $r_s = 700,000$ km = $7.0 \times 10^8$ m - Mercury's average distance from Sun $d_p = 58,000,000$ km = $5.8 \times 10^{10}$ m - Energy flux at Sun's surface $S_s = 63 \times 10^6 \text{ W/m}^2$ 3. **Formula used:** $$S_p = S_s \left(\frac{r_s}{d_p}\right)^2$$ This formula comes from equating total energy emitted by the Sun and energy spread over a sphere of radius $d_p$. 4. **Calculate $S_p$ for Mercury:** $$S_p = 63 \times 10^6 \times \left(\frac{7.0 \times 10^8}{5.8 \times 10^{10}}\right)^2$$ 5. **Simplify the fraction inside the square:** $$\frac{7.0 \times 10^8}{5.8 \times 10^{10}} = \frac{7.0}{5.8} \times 10^{8 - 10} = 1.2069 \times 10^{-2}$$ 6. **Square the fraction:** $$\left(1.2069 \times 10^{-2}\right)^2 = 1.4566 \times 10^{-4}$$ 7. **Calculate $S_p$:** $$S_p = 63 \times 10^6 \times 1.4566 \times 10^{-4} = 63 \times 1.4566 \times 10^{6 - 4} = 63 \times 1.4566 \times 10^{2}$$ 8. **Multiply constants:** $$63 \times 1.4566 = 91.716$$ 9. **Final value:** $$S_p = 91.716 \times 10^{2} = 9.1716 \times 10^{3} \text{ W/m}^2$$ 10. **Answer:** The solar energy flux at Mercury's orbit is approximately $$S_p \approx 9.17 \times 10^{3} \text{ W/m}^2$$ --- **Note:** The problem also mentions $S_{save} = \frac{S_p}{4}$, but since the question only asks for $S_p$, we stop here.