A spacecraft is travelling at a constant speed relative to a nearby planet - Scottish Highers Physics - Question 8 - 2019

To determine the speed of the spacecraft relative to the observer, we can use the concept of length contraction from special relativity.

The formula for length contraction is given by: $L = L_0 \sqrt{1 - \frac{v^2}{c^2}}$ where:

• $L_0$ is the proper length (length measured in the spacecraft): 275 m
• $L$ is the contracted length (length measured by the observer on the planet): 125 m
• $v$ is the speed of the spacecraft
• $c$ is the speed of light (~$3.00 \times 10^8$ m/s)

We can rearrange the formula to solve for $v$:

1. Substitute the known values into the equation: $125 = 275 \sqrt{1 - \frac{v^2}{(3.00 \times 10^8)^2}}$

2. Divide both sides by 275: $\frac{125}{275} = \sqrt{1 - \frac{v^2}{(3.00 \times 10^8)^2}}$ $$ 3. Square both sides: $\left( \frac{125}{275} \right)^2 = 1 - \frac{v^2}{(3.00 \times 10^8)^2}$

3. Simplify the left side: $\frac{15625}{75625} = 1 - \frac{v^2}{(9.00 \times 10^{16})}$

4. Rearranging yields: $\frac{v^2}{(9.00 \times 10^{16})} = 1 - \frac{15625}{75625}$

5. This simplifies further: $\frac{v^2}{(9.00 \times 10^{16})} = \frac{60000}{75625}$

6. Multiply both sides by $9.00 \times 10^{16}$: $v^2 = \frac{60000}{75625} \cdot (9.00 \times 10^{16})$

7. Finally, take the square root: $v = \sqrt{\frac{60000}{75625} \cdot (9.00 \times 10^{16})} \approx 2.67 \times 10^8 \text{ ms}^{-1}$

Thus, the speed of the spacecraft relative to the observer on the nearby planet is approximately 2.67 x 10^8 ms^-1, corresponding to option C.