A student models a spacecraft journey that takes one year - AQA - A-Level Physics - Question 4 - 2017 - Paper 7

To find the time dilation, we need to use the formula for time dilation in special relativity:

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

Where:

• $t$ is the proper time (time for the observer), which is 1 year or 31,536,000 seconds.
• $v$ is the velocity of the spacecraft, $1.2 \times 10^7 \, \text{m s}^{-1}$.
• $c$ is the speed of light, approximately $3 \times 10^8 \, \text{m s}^{-1}$.

Calculating $\frac{v^2}{c^2}$ gives:

$\frac{(1.2 \times 10^7)^2}{(3 \times 10^8)^2} = \frac{1.44 \times 10^{14}}{9 \times 10^{16}} \approx 0.016 \text{ (approx.)}$

Now, substituting into the time dilation formula:

$t' \approx \frac{31,536,000}{\sqrt{1 - 0.016}} \approx \frac{31,536,000}{0.998} \approx 31,578,684.98 \, \text{seconds}$

The time difference, or dilation, is:

$\Delta t = t' - t = 31,578,684.98 - 31,536,000 \approx 25,259.98 \, \text{seconds} \approx 7 \, \text{hours} \; 4 \, \text{minutes}$

This means the clock on the spacecraft will record approximately 7 hours more than the observer's clock, supporting the student's prediction.