Question

A spacecraft has speed 𝑣_𝑠=𝑓𝑐 with respect to the earth, where 𝑐 is the speed of light in vacuum. An observer in the spacecraft measures the time of one complete rotation of the earth to be 48 hours. The value of 𝑓 is (rounded off to two decimal places).

Answer 1

Correct Answer: 0.85 - 0.88

Given: 
Period of Earth's rotation in Earth frame = \( T_0 = 24\ \text{hours} \)
Period measured in spacecraft frame = \( T = 48\ \text{hours} \)
Spacecraft speed = \( v_s = f\,c \)

Time dilation relation:
\[ T = \gamma T_0 \] \[ \gamma = \frac{T}{T_0} = \frac{48}{24} = 2 \] Relativistic factor:
\[ \gamma = \frac{1}{\sqrt{1 - f^2}} \] Thus, \[ 2 = \frac{1}{\sqrt{1 - f^2}} \] \[ \sqrt{1 - f^2} = \frac{1}{2} \] \[ 1 - f^2 = \frac{1}{4} \] \[ f^2 = \frac{3}{4} \] \[ f = \sqrt{\frac{3}{4}} = 0.866 \] Final Answer:
\[ f \approx 0.87 \]