Question

A rocket of length 18.0 m is moving at speed \(0.9c\) (where \(c\) is the speed of light) parallel to its own length, relative to the earth. The length of the rocket measured in meters by an observer on earth (rounded off to two decimal places) is

Hint:

Hint:
Use \(L = L_0\sqrt{1-v^2/c^2}\) with \(L_0=18.0\) m and \(v=0.9c\).

Answer 1

Correct Answer: 7.85

Step 1: Set up the Lorentz transformation.
Let the rocket's rest frame be $S'$, moving at speed $v = 0.9c$ along the earth frame $S$. The two ends of the rocket sit at $x'_1 = 0$ and $x'_2 = L_0 = 18.0$ m in $S'$ at all times, since the rocket does not move in its own frame. The Lorentz transformation gives, for the earth frame coordinate of each end at earth time $t$:
\[ x' = \gamma(x - vt) \]

Step 2: Measure both ends at the same earth time.
A length measured on earth means finding both end positions $x_1, x_2$ at the same instant $t$ in the earth frame. Applying the transformation to both ends at this common $t$ and subtracting:
\[ x'_2 - x'_1 = \gamma(x_2 - x_1) \quad \Rightarrow \quad L_0 = \gamma L \quad \Rightarrow \quad L = \frac{L_0}{\gamma} \]

Step 3: Work out the Lorentz factor and the length.
\[ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0.81}} = \frac{1}{\sqrt{0.19}} = \frac{1}{0.4359} = 2.294 \]
\[ L = \frac{18.0}{2.294} = 7.846 \text{ m} \]

Final Answer:
Deriving it from the coordinate transformation instead of quoting the contraction formula gives the same 7.85 m. \[ \boxed{L = 7.85 \text{ m}} \]