A spacecraft is travelling at a speed of 0.200c relative to the Earth - Scottish Highers Physics - Question 7 - 2018

To calculate the time duration as measured by an observer on Earth, we can use the time dilation formula from the theory of relativity:

$t' = t \sqrt{1 - \frac{v^2}{c^2}}$

Where:

• $t'$ is the time interval measured by the observer on Earth.
• $t$ is the time interval in the spacecraft's reference frame (20.0 s).
• $v$ is the speed of the spacecraft (0.200c).
• $c$ is the speed of light.

First, we calculate $v^2/c^2$:

$\frac{v^2}{c^2} = (0.200)^2 = 0.040$

Now, we substitute this into the formula:

$t' = 20.0 \sqrt{1 - 0.040} = 20.0 \sqrt{0.960}$

Calculating the square root:

$\sqrt{0.960} \approx 0.980$

Now, substituting back:

$t' \approx 20.0 \times 0.980 \approx 19.6 s$