Two physics students are in an airport building on their way to visit CERN - Scottish Highers Physics - Question 4 - 2016

The speed of light is constant in all inertial frames of reference. Therefore, regardless of the speed of the plane moving at 0.8c, the speed of the emitted light as measured by the observer is:

$c = 3 \times 10^8 \, \text{m/s}$

This remains the same for all observers.

To calculate the length of the plane as measured by the stationary observer, we use the principle of length contraction, which is given by the formula:

$L = L_0 \sqrt{1 - \frac{v^2}{c^2}}$

Where:

• $L_0 = 71 \, \text{m}$ (proper length of the plane)
• $v = 0.8c$
• $c = 3 \times 10^8 \, \text{m/s}$

Substituting the values:

$L = 71 \cdot \sqrt{1 - (0.8)^2} \quad (c=1)$ $L = 71 \cdot \sqrt{1 - 0.64}$ $L = 71 \cdot \sqrt{0.36}$ $L = 71 \cdot 0.6 = 42.6 \, \text{m}$

Therefore, the length of the plane travelling at 0.8c as measured by the stationary observer is 42.6 m.

The statement about the clocks on board the plane running slower when the plane is travelling at relativistic speeds is correct. From the perspective of the stationary observer, time dilation occurs. This means that time passes more slowly on board the plane compared to the stationary observer's reference frame. Therefore, the clocks on the plane will indeed run slower as viewed by the stationary observer.