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

To find the time dilation effect, we can use the time dilation formula from special relativity:

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

Where:

• $t'$ = dilated time (time measured by the observer)
• $t$ = proper time (time measured by the moving clock)
• $v$ = velocity of the spacecraft ($1.2 \times 10^7 \, \text{m s}^{-1}$)
• $c$ = speed of light ($3 \times 10^8 \, \text{m s}^{-1}$)

First, we calculate $\frac{v^2}{c^2}$:

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

Then we compute:

$t' = \frac{1 \, \text{year}}{\sqrt{1 - 0.016}} = \frac{1 \, \text{year}}{\sqrt{0.984}} \approx 1.008 \text{ years}$

Now converting the proper time of one year (365 days) to seconds:

$1 \, \text{year} = 31,536,000 \, \text{seconds}$

The dilated time amounting to about:

$t' \approx 31,536,259 \, \text{seconds}$

The difference in time recorded (dilated time - proper time):

$\Delta t = t' - t \approx 31,536,259 - 31,536,000 = 259 \, \text{seconds} \approx 4.32 \text{ minutes}$

Thus, the observed time will be longer than the time experienced on the spacecraft.