Edits to “Statement”, “Solution”, “Answer”
en/12.1.23.md
+12 −4
| @@ -1,7 +1,12 @@ | |||
| ### Statement | |||
| − | $12.1.23.$ | ||
| + | $12.1.23.$ | ||
| + | What is the pressure of solar radiation on a mirror surface near the Earth | ||
| + | when it normally falls on the mirror? when the radiation falls on the mirror | ||
| + | at an angle of 30◦? The energy flux density of solar radiation is 600W | ||
| + | m2. | ||
| + | |||
| ### Solution | |||
| For a perfect reflector, the radiation pressure on the surface is $p = \dfrac{2I}{c}\cos^2\theta$, where I is the energy flux density, c is the speed of light, and $\theta$ is the angle of incidence measured with respect to the normal. | |||
| With $ I = 600\ \text{W/m}^2 and c = 3.0\times10^8\ \text{m/s}$ | |||
| $\frac{I}{c} = \frac{600}{3.0\times10^8} = 2.0\times10^{-6}\ \text{Pa}$ | |||
| Normal incidence ($\theta = 0^\circ$): | |||
| $p = 2 \times 2.0\times10^{-6} = 4.0\times10^{-6}\ \text{Pa}$ | |||
| Incidence at $30^\circ$ (with respect to the normal): | |||
| @@ -18,10 +23,13 @@Solution | |||
| $p = 2 \times 2.0\times10^{-6} \times \cos^2 30^\circ = 4.0\times10^{-6} \times \left(\frac{\sqrt{3}}{2}\right)^2 = 4.0\times10^{-6} \times 0.75 = 3.0\times10^{-6}\ \text{Pa}$ | |||
| − | If $30^\ | ||
| + | If $30^\circ$ were interpreted as the angle with respect to the surface, it would be $60^\circ$ with respect to the normal and the pressure would be | ||
| − | $ | ||
| + | $1.0\times10^{-6}\ \text{Pa}$ | ||
| + | $\boxed{p_{\text{normal}} = 4\times10^{-6}\ \text{Pa},\qquad p_{30^\circ} = 3\times10^{-6}\ \text{Pa}}$ | ||
| + | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | |||
| + | $\boxed{p_{\text{normal}} = 4\times10^{-6}\ \text{Pa},\qquad p_{30^\circ} = 3\times10^{-6}\ \text{Pa}}$ | ||
| @@ -1,7 +1,12 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.23.$ |
$12.1.23.$ | ||
| What is the pressure of solar radiation on a mirror surface near the Earth | |||
| when it normally falls on the mirror? when the radiation falls on the mirror | |||
| at an angle of 30◦? The energy flux density of solar radiation is 600W | |||
| m2. | |||
| ### Solution | ### Solution | ||
| For a perfect reflector, the radiation pressure on the surface is $p = \dfrac{2I}{c}\cos^2\theta$, where I is the energy flux density, c is the speed of light, and $\theta$ is the angle of incidence measured with respect to the normal. | For a perfect reflector, the radiation pressure on the surface is $p = \dfrac{2I}{c}\cos^2\theta$, where I is the energy flux density, c is the speed of light, and $\theta$ is the angle of incidence measured with respect to the normal. | ||
| With $ I = 600\ \text{W/m}^2 and c = 3.0\times10^8\ \text{m/s}$ | With $ I = 600\ \text{W/m}^2 and c = 3.0\times10^8\ \text{m/s}$ | ||
| $\frac{I}{c} = \frac{600}{3.0\times10^8} = 2.0\times10^{-6}\ \text{Pa}$ | $\frac{I}{c} = \frac{600}{3.0\times10^8} = 2.0\times10^{-6}\ \text{Pa}$ | ||
| Normal incidence ($\theta = 0^\circ$): | Normal incidence ($\theta = 0^\circ$): | ||
| $p = 2 \times 2.0\times10^{-6} = 4.0\times10^{-6}\ \text{Pa}$ | $p = 2 \times 2.0\times10^{-6} = 4.0\times10^{-6}\ \text{Pa}$ | ||
| Incidence at $30^\circ$ (with respect to the normal): | Incidence at $30^\circ$ (with respect to the normal): | ||
| @@ -18,10 +23,13 @@Solution | |||
| $p = 2 \times 2.0\times10^{-6} \times \cos^2 30^\circ = 4.0\times10^{-6} \times \left(\frac{\sqrt{3}}{2}\right)^2 = 4.0\times10^{-6} \times 0.75 = 3.0\times10^{-6}\ \text{Pa}$ | $p = 2 \times 2.0\times10^{-6} \times \cos^2 30^\circ = 4.0\times10^{-6} \times \left(\frac{\sqrt{3}}{2}\right)^2 = 4.0\times10^{-6} \times 0.75 = 3.0\times10^{-6}\ \text{Pa}$ | ||
| If $30^\ |
If $30^\circ$ were interpreted as the angle with respect to the surface, it would be $60^\circ$ with respect to the normal and the pressure would be | ||
| $ |
$1.0\times10^{-6}\ \text{Pa}$ | ||
| $\boxed{p_{\text{normal}} = 4\times10^{-6}\ \text{Pa},\qquad p_{30^\circ} = 3\times10^{-6}\ \text{Pa}}$ | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | |||
| $\boxed{p_{\text{normal}} = 4\times10^{-6}\ \text{Pa},\qquad p_{30^\circ} = 3\times10^{-6}\ \text{Pa}}$ | |||