Chapter 6, Problem 91A

To determine

The speed of the laser beam relative to the fixed observer. Also, the speed of the missile relative to the fixed observer.

Answer to Problem 91A

The speed of the laser beam relative to the fixed observer is $3\times {10}^8\text{m/s}$ .

The speed of the missile relative to the fixed observer is $2.40\times {10}^8\text{m/s}$ .

Explanation of Solution

Given:

The velocity of the rocket is $v_{\text{A}}=11\text{km/s}$ .

The rocket releases a missile, such that the relative speed of the missile with respect to the rocket is denoted as $v_{\text{B}}$ .

Formula used:

The expression for the speed of the missile relative to the observer is

  $v=\frac{v_{\text{A}}+v_{\text{B}}}{\left(1+\frac{v_{\text{A}}{v}_{\text{B}}}{c^{\text{2}}}\right)}$

Calculation:

Consider the velocity of the laser beam (equivalent to the speed of light) is $v_{\text{B}}=3.00\times {10}^8\text{m/s}$ .

The speed of the laser beam relative to the unmoving observer is

  $\begin{array}{l}v=\frac{v_{\text{A}}+v_{\text{B}}}{\left(1+\frac{v_{\text{A}}{v}_{\text{B}}}{c^{\text{2}}}\right)}\\ v=\frac{\left(11km/s\times \frac{1000\text{m}}{1\text{km}}\right)+3\times {10}^8\text{m/s}}{\left\{1+\frac{\left(11km/s\times \frac{1000\text{m}}{1\text{km}}\right)\left(3\times {10}^8\text{m/s}\right)}{{\left(3\times {10}^8\text{m/s}\right)}^{\text{2}}}\right\}}\\ v=\frac{11000m/s+3\times {10}^8\text{m/s}}{\left\{1+\frac{\left(11000m/s\right)\left(3\times {10}^8\text{m/s}\right)}{{\left(3\times {10}^8\text{m/s}\right)}^{\text{2}}}\right\}}\\ v=3\times {10}^8\text{m/s}\end{array}$

Consider the rocket moves with the speed, $v_{\text{A}}=\frac{c}{2}$ .

Consider the speed of the missile relative to the rocket is $v_{\text{B}}=\frac{c}{2}$ .

The speed of the missile relative to the unmoving observer as follows:

  $\begin{array}{l}v=\frac{v_{\text{A}}+v_{\text{B}}}{\left(1+\frac{v_{\text{A}}{v}_{\text{B}}}{c^{\text{2}}}\right)}\\ v=\frac{\left(\frac{c}{2}+\frac{c}{2}\right)}{\left\{1+\frac{\left(\frac{c}{2}\times \frac{c}{2}\right)}{c^{\text{2}}}\right\}}\\ v=\frac{c}{\left\{1+\frac{1}{4}\right\}}\\ v=\frac{4}{5}c\end{array}$

  $\begin{array}{l}v=\frac{4}{5}\times \left(3\times {10}^8\text{m/s}\right)\\ v=2.40\times {10}^8\text{m/s}\end{array}$

Conclusion:

Therefore, the speed of the laser beam relative to the unmoving observer is $3\times {10}^8\text{m/s}$ .The speed of the missile relative to the unmoving observer is $2.40\times {10}^8\text{m/s}$ .