Edits to “Statement”, “Solution”, “Answer”
en/12.1.24.md
+6 −5
| @@ -1,12 +1,12 @@ | |||
| ### Statement | |||
| − | $12.1.24 | ||
| + | $12.1.24$ Estimate the maximum size of aluminum dust particles that would move away | ||
| + | from the Sun in outer space under the pressure of solar radiation. | ||
| ### Solution | |||
| − | $12.1.24$ Estimate the maximum size of aluminum dust particles that would move away | ||
| − | from the Sun in outer space under the pressure of solar radiation. | ||
| + | |||
| The power radiated by the Sun is $P$, so the intensity of the radiation at a distance $r$ from the Sun is: | |||
| \begin{equation} | |||
| I = \frac{P}{4 \pi r^2} | |||
| \end{equation} | |||
| Thus, the pressure exerted by the light on the dust particle is: | |||
| \begin{equation} | |||
| \mathcal{P} = 2 \frac{I}{c} = \frac{P}{4 \pi r^2 c} | |||
| \end{equation} | |||
| The corresponding force is: | |||
| \begin{equation} | |||
| F = 2 \pi r_1^2 \mathcal{P} = \frac{P r_1^2}{2 r^2 c} | |||
| \end{equation} | |||
| This force must be at least equal to the gravitational force: | |||
| \begin{equation} | |||
| F = \frac{G M_s m_d}{r^2} \quad \text{where} \quad m_d = \rho \frac{4 \pi r_1^3}{3} | |||
| \end{equation} | |||
| Finally: | |||
| \begin{equation} | |||
| \frac{G M_s m_d}{r^2} = \frac{P r_1^2}{4 r^2 c} \rightarrow r_1 = \frac{3 P}{8 \pi M_s G c \rho} \approx 1 \, \mu\text{m} | |||
| @@ -38,5 +38,6 @@Solution | |||
| \end{equation} | |||
| #### Answer | |||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \begin{equation} | ||
| + | r_1 = \frac{3 P}{8 \pi M_s G c \rho} \approx 1 \, \mu\text{m} | ||
| + | \end{equation} | ||
| @@ -1,12 +1,12 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.24 |
$12.1.24$ Estimate the maximum size of aluminum dust particles that would move away | ||
| from the Sun in outer space under the pressure of solar radiation. | |||
| ### Solution | ### Solution | ||
| $12.1.24$ Estimate the maximum size of aluminum dust particles that would move away | |||
| from the Sun in outer space under the pressure of solar radiation. | |||
| The power radiated by the Sun is $P$, so the intensity of the radiation at a distance $r$ from the Sun is: | The power radiated by the Sun is $P$, so the intensity of the radiation at a distance $r$ from the Sun is: | ||
| \begin{equation} | \begin{equation} | ||
| I = \frac{P}{4 \pi r^2} | I = \frac{P}{4 \pi r^2} | ||
| \end{equation} | \end{equation} | ||
| Thus, the pressure exerted by the light on the dust particle is: | Thus, the pressure exerted by the light on the dust particle is: | ||
| \begin{equation} | \begin{equation} | ||
| \mathcal{P} = 2 \frac{I}{c} = \frac{P}{4 \pi r^2 c} | \mathcal{P} = 2 \frac{I}{c} = \frac{P}{4 \pi r^2 c} | ||
| \end{equation} | \end{equation} | ||
| The corresponding force is: | The corresponding force is: | ||
| \begin{equation} | \begin{equation} | ||
| F = 2 \pi r_1^2 \mathcal{P} = \frac{P r_1^2}{2 r^2 c} | F = 2 \pi r_1^2 \mathcal{P} = \frac{P r_1^2}{2 r^2 c} | ||
| \end{equation} | \end{equation} | ||
| This force must be at least equal to the gravitational force: | This force must be at least equal to the gravitational force: | ||
| \begin{equation} | \begin{equation} | ||
| F = \frac{G M_s m_d}{r^2} \quad \text{where} \quad m_d = \rho \frac{4 \pi r_1^3}{3} | F = \frac{G M_s m_d}{r^2} \quad \text{where} \quad m_d = \rho \frac{4 \pi r_1^3}{3} | ||
| \end{equation} | \end{equation} | ||
| Finally: | Finally: | ||
| \begin{equation} | \begin{equation} | ||
| \frac{G M_s m_d}{r^2} = \frac{P r_1^2}{4 r^2 c} \rightarrow r_1 = \frac{3 P}{8 \pi M_s G c \rho} \approx 1 \, \mu\text{m} | \frac{G M_s m_d}{r^2} = \frac{P r_1^2}{4 r^2 c} \rightarrow r_1 = \frac{3 P}{8 \pi M_s G c \rho} \approx 1 \, \mu\text{m} | ||
| @@ -38,5 +38,6 @@Solution | |||
| \end{equation} | \end{equation} | ||
| #### Answer | #### Answer | ||
| \begin{equation} | |||
| [Insert a concise answer or boxed result] | r_1 = \frac{3 P}{8 \pi M_s G c \rho} \approx 1 \, \mu\text{m} | ||
| \end{equation} | |||