Question

A spacecraft of proper length Lo = 48.0 m is moving in the +x direction at a constant Lorentz factor y = 2.20 relative to an inertial frame, the Earth frame. The spacecraft is aligned with the direction of motion. The tip of the spacecraft crosses the origin (x = 0) at tA = 0 in the Earth frame and t'A = 0 in the frame of the moving spacecraft. a. What is the speed parameter of the spaceship B? b. What is the measured length of the spacecraft in the Earth frame? What is the measured length of the spacecraft in the spaceship frame? C. At what time in the Earth frame does the tail of the spaceship cross the Earth frame origin tB? d. At what time in the spaceship frame does the tail of the spaceship cross the Earth frame origin t'B?

Answer 1

(a)

y = 1 - 에 C

$\gamma =\frac {1}{\sqrt {1-\beta^2}}$

where

$\beta =\frac {v}{c}$

$2.20 =\frac {1}{\sqrt {1-\beta^2}}$

$\mathbf{\beta = 0.890}$

(b)

$L_{e} = L_0/\gamma=\frac {48.0}{2.20}\: \text{m}= 21.82\: \text{m}$

$L_{s} = L_0=48.0\: \text{m}$

(c)

$t_{B}=\frac {L_e}{v}=\frac {21.82}{0.890\times 3\times 10^8}\: s = 8.17\times 10^{-8}\: s$

(d)

$t_{s}=\frac {L_s}{v}=\frac {48.0}{0.890\times 3\times 10^8}\: s = 1.79\times 10^{-7}\: s$