Problem

The length of a spaceship is measured to be exactly one-third of its proper length. What is the speed of the spaceship relative to the observer?

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Data

• Proper length of the spaceship ($L_p$) = reference length
• Measured length ($L$) = $\frac{1}{3} L_p$
• Speed of light ($c$) = constant

To Find: Speed of the spaceship relative to the observer ($v$).

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Prerequisite Concepts

Relativistic length contraction is given by:

$$L = L_p \sqrt{1 - \frac{v^2}{c^2}},$$

where:

• $L$ is the contracted length observed,
• $L_p$ is the proper length,
• $v$ is the speed of the moving object,
• $c$ is the speed of light.

Rearranging:

$$\sqrt{1 - \frac{v^2}{c^2}} = \frac{L}{L_p}.$$

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Solution

1. From the given data:

$$L = \frac{1}{3} L_p.$$

Substituting into the contraction equation:

$$\sqrt{1 - \frac{v^2}{c^2}} = \frac{L}{L_p} = \frac{1}{3}.$$

2. Squaring both sides:

$$1 - \frac{v^2}{c^2} = \left(\frac{1}{3}\right)^2,$$ $$1 - \frac{v^2}{c^2} = \frac{1}{9}.$$

3. Rearranging:

$$\frac{v^2}{c^2} = 1 - \frac{1}{9}.$$

Simplify:

$$\frac{v^2}{c^2} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}.$$

4. Solving for $v$:

$$v^2 = \frac{8}{9} c^2.$$ $$v = \sqrt{\frac{8}{9}} c.$$

5. Simplifying:

$$v = \frac{\sqrt{8}}{3} c.$$

Approximation:

$$v \approx 0.9428 c.$$

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Answer

The speed of the spaceship relative to the observer is approximately $v = 0.9428c$.