Answer to Question #266049 in Mechanics | Relativity for Dinesh

Question

A spacecraft with a proper length of 300 m takes 0.75 ms to pass an Earth-based observer. Determine the speed of the spacecraft as measured by the Earth observer.

Answer 1

$l_0=300 \ (m)$

$t=0.75\cdot10^{-6}\ (s)$

$l=l_0\sqrt{1-v^2/c^2}\to l^2=l_0^2(1-v^2/c^2)$

$l=vt\to l^2=\frac{v^2}{c^2}(ct)^2$ . After equating these two expressions we will get

$l_0^2(1-v^2/c^2)=\frac{v^2}{c^2}(ct)^2\to l_0^2=\frac{v^2}{c^2}(l_0^2+(ct)^2)$

$300^2=\frac{v^2}{c^2}(300^2+(c\cdot0.00000075)^2)\to v=0.8c$ . Answer