A spacecraft passes the Earth at a speed of 0.4c - Scottish Highers Physics - Question 7 - 2022

To determine the time interval measured by the observer on Earth, we employ the concept of time dilation from the theory of relativity. The formula for time dilation is:

$t_0 = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}}$

Where:

• (t_0) is the proper time (time measured in the spacecraft),
• (t) is the dilated time (time measured by the observer on Earth),
• (v) is the speed of the spacecraft (0.4c), and
• (c) is the speed of light.

First, we compute the time dilation factor:

$\sqrt{1 - \frac{(0.4c)^2}{c^2}} = \sqrt{1 - 0.16} = \sqrt{0.84} \approx 0.9165$

Thus, substituting our values:

$t = \frac{2.5 s}{0.9165} \approx 2.73 s$