Question

Imagine a spaceship traveling at a constant speed through outer space. The length of the ship, as measured by a passenger aboard the ship, is 28.2 m. An observer on Earth, however, sees the ship as contracted by 16.1 cm along the direction of motion. What is the speed of the spaceship with respect to the Earth

Answer 1

$3.20×10^7\:\text{m/s}$

Explanation:

Let

$L = 28.2\:\text{m}$

$L' = 28.2\:\text{m} - 0.161\:\text{m} = 28.039\:\text{m}$

The Lorentz length contraction formula is given by

$L' = L\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)}$

where L is the length measured by the moving observer and L' is the length measured by the stationary Earth-based observer. We can rewrite the above equation as

$\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)} = \dfrac{L'}{L}$

Taking the square of the equation, we get

$1 - \left(\dfrac{v^2}{c^2}\right) = \left(\dfrac{L'}{L}\right)^2$

or

$1 - \left(\dfrac{L'}{L}\right)^2 = \left(\dfrac{v}{c}\right)^2$

Solving for v, we get

$v = c\sqrt{1 - \left(\dfrac{L'}{L}\right)^2}$

$\:\:\:\:=(3×10^8\:\text{m/s})\sqrt{1 - \left(\dfrac{28.039\:\text{m}}{28.2\:\text{m}}\right)^2}$

$\:\:\:\:=3.20×10^7\:\text{m/s} = 0.107c$