Новое решение
en/14.2.14.md
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| + | ### Statement | ||
| + | |||
| + | $14.2.14.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Velocity of each rocket from the station | ||
| + | |||
| + | The station receives frequencies $\nu_1$ and $\nu_2 $from sources that are approaching. The Doppler effect formula for direct approach$ (\beta = v/c) $is: | ||
| + | |||
| + | $\nu_i = \nu_0 \sqrt{\frac{1+\beta_i}{1-\beta_i}}, \qquad i=1,2$ | ||
| + | |||
| + | Solving for $\beta_i$ | ||
| + | |||
| + | $\left(\frac{\nu_i}{\nu_0}\right)^2 = \frac{1+\beta_i}{1-\beta_i} | ||
| + | \quad\Longrightarrow\quad | ||
| + | \beta_i = \frac{(\nu_i/\nu_0)^2 - 1}{(\nu_i/\nu_0)^2 + 1}$ | ||
| + | |||
| + | In terms of velocity: | ||
| + | |||
| + | $v_i = c\,\frac{\nu_i^2 - \nu_0^2}{\nu_i^2 + \nu_0^2}$ | ||
| + | |||
| + | Velocity difference in the station's frame | ||
| + | |||
| + | The rockets move along the same line toward the station. The speed with which they approach each other, measured in the station's frame, is simply the difference of their velocities (if one is faster than the other) or the sum (if they come from opposite directions). We assume they move in the same direction (one behind the other): | ||
| + | |||
| + | $\Delta v = v_1 - v_2$ | ||
| + | |||
| + | Substituting the expressions for $v_1$ and $v_2$ | ||
| + | |||
| + | $\Delta v = c\left( \frac{\nu_1^2 - \nu_0^2}{\nu_1^2 + \nu_0^2} - \frac{\nu_2^2 - \nu_0^2}{\nu_2^2 + \nu_0^2} \right)$ | ||
| + | |||
| + | Simplification | ||
| + | |||
| + | We put a common denominator: | ||
| + | |||
| + | $\Delta v = c\,\frac{(\nu_1^2 - \nu_0^2)(\nu_2^2 + \nu_0^2) - (\nu_2^2 - \nu_0^2)(\nu_1^2 + \nu_0^2)}{(\nu_1^2 + \nu_0^2)(\nu_2^2 + \nu_0^2)}$ | ||
| + | |||
| + | Expanding the numerator: | ||
| + | |||
| + | $\begin{aligned} | ||
| + | &(\nu_1^2 - \nu_0^2)(\nu_2^2 + \nu_0^2) - (\nu_2^2 - \nu_0^2)(\nu_1^2 + \nu_0^2) \\ | ||
| + | &= (\nu_1^2\nu_2^2 + \nu_1^2\nu_0^2 - \nu_0^2\nu_2^2 - \nu_0^4) - (\nu_1^2\nu_2^2 + \nu_2^2\nu_0^2 - \nu_0^2\nu_1^2 - \nu_0^4) \\ | ||
| + | &= \cancel{\nu_1^2\nu_2^2} + \nu_1^2\nu_0^2 - \nu_0^2\nu_2^2 - \nu_0^4 - \cancel{\nu_1^2\nu_2^2} - \nu_2^2\nu_0^2 + \nu_0^2\nu_1^2 + \nu_0^4 \\ | ||
| + | &= 2\nu_0^2(\nu_1^2 - \nu_2^2). | ||
| + | \end{aligned}$ | ||
| + | |||
| + | The denominator is: | ||
| + | |||
| + | $(\nu_1^2 + \nu_0^2)(\nu_2^2 + \nu_0^2)$ | ||
| + | |||
| + | Therefore: | ||
| + | |||
| + | $\Delta v = c\,\frac{2\nu_0^2(\nu_1^2 - \nu_2^2)}{(\nu_1^2 + \nu_0^2)(\nu_2^2 + \nu_0^2)}$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $14.2.14.$ [Insert the problem statement] | |||
| ### Solution | |||
| Velocity of each rocket from the station | |||
| The station receives frequencies $\nu_1$ and $\nu_2 $from sources that are approaching. The Doppler effect formula for direct approach$ (\beta = v/c) $is: | |||
| $\nu_i = \nu_0 \sqrt{\frac{1+\beta_i}{1-\beta_i}}, \qquad i=1,2$ | |||
| Solving for $\beta_i$ | |||
| $\left(\frac{\nu_i}{\nu_0}\right)^2 = \frac{1+\beta_i}{1-\beta_i} | |||
| \quad\Longrightarrow\quad | |||
| \beta_i = \frac{(\nu_i/\nu_0)^2 - 1}{(\nu_i/\nu_0)^2 + 1}$ | |||
| In terms of velocity: | |||
| $v_i = c\,\frac{\nu_i^2 - \nu_0^2}{\nu_i^2 + \nu_0^2}$ | |||
| Velocity difference in the station's frame | |||
| The rockets move along the same line toward the station. The speed with which they approach each other, measured in the station's frame, is simply the difference of their velocities (if one is faster than the other) or the sum (if they come from opposite directions). We assume they move in the same direction (one behind the other): | |||
| $\Delta v = v_1 - v_2$ | |||
| Substituting the expressions for $v_1$ and $v_2$ | |||
| $\Delta v = c\left( \frac{\nu_1^2 - \nu_0^2}{\nu_1^2 + \nu_0^2} - \frac{\nu_2^2 - \nu_0^2}{\nu_2^2 + \nu_0^2} \right)$ | |||
| Simplification | |||
| We put a common denominator: | |||
| $\Delta v = c\,\frac{(\nu_1^2 - \nu_0^2)(\nu_2^2 + \nu_0^2) - (\nu_2^2 - \nu_0^2)(\nu_1^2 + \nu_0^2)}{(\nu_1^2 + \nu_0^2)(\nu_2^2 + \nu_0^2)}$ | |||
| Expanding the numerator: | |||
| $\begin{aligned} | |||
| &(\nu_1^2 - \nu_0^2)(\nu_2^2 + \nu_0^2) - (\nu_2^2 - \nu_0^2)(\nu_1^2 + \nu_0^2) \\ | |||
| &= (\nu_1^2\nu_2^2 + \nu_1^2\nu_0^2 - \nu_0^2\nu_2^2 - \nu_0^4) - (\nu_1^2\nu_2^2 + \nu_2^2\nu_0^2 - \nu_0^2\nu_1^2 - \nu_0^4) \\ | |||
| &= \cancel{\nu_1^2\nu_2^2} + \nu_1^2\nu_0^2 - \nu_0^2\nu_2^2 - \nu_0^4 - \cancel{\nu_1^2\nu_2^2} - \nu_2^2\nu_0^2 + \nu_0^2\nu_1^2 + \nu_0^4 \\ | |||
| &= 2\nu_0^2(\nu_1^2 - \nu_2^2). | |||
| \end{aligned}$ | |||
| The denominator is: | |||
| $(\nu_1^2 + \nu_0^2)(\nu_2^2 + \nu_0^2)$ | |||
| Therefore: | |||
| $\Delta v = c\,\frac{2\nu_0^2(\nu_1^2 - \nu_2^2)}{(\nu_1^2 + \nu_0^2)(\nu_2^2 + \nu_0^2)}$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||