Problem — Contraction of a spaceship

Problem

Contraction. A spaceship, at rest, is $L = 100$ m long. An observer on the ground measures it to be $L' = 60$ m long. At what speed is the spaceship travelling? Draw the situation comparing the lengths in the two frames.

Solution

Length contraction. The proper length is $L = 100$ m (spaceship at rest); the observer on the ground measures the contracted length $L' = L/\gamma$. So:

$\frac{L'}{L} = \frac{1}{\gamma} = \sqrt{1 - \frac{v^2}{c^2}}$

Numerical substitution. $\dfrac{L'}{L} = \dfrac{60}{100} = 0{,}60$:

$\sqrt{1 - \frac{v^2}{c^2}} = 0{,}60 \quad\Rightarrow\quad 1 - \frac{v^2}{c^2} = 0{,}36 \quad\Rightarrow\quad \frac{v^2}{c^2} = 0{,}64$

Speed.

$\frac{v}{c} = \sqrt{0{,}64} = 0{,}80$

$\ev{v = 0{,}80\,c}$

For the spaceship to appear contracted by 40%, it must travel at $0{,}80\,c$.

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Topics: Relatività ristretta Concepts: Contrazione delle lunghezze Skills: Impostazione simbolica