Question

A spaceship is moving at a constant velocity of \(0.70 c\) relative to an Earth observer. The Earth observer measures the length of the spaceship to be $40.0 \mathrm{m}$. How long is the spaceship as measured by its pilot?

Short answer

Answer: The length of the spaceship as measured by the pilot is approximately 56.19 meters.

Step-by-step solution

Understand the length contraction formula

The length contraction formula is given by: \(L = L_{0} \sqrt{1 - \frac{v^2}{c^2}}\) where \(L\) is the contracted length measured by the observer, \(L_0\) is the proper length measured in the rest frame of the object (in this case, the pilot), \(v\) is the relative velocity of the object, and \(c\) is the speed of light.

Rearrange the formula to solve for the proper length

We need to find \(L_0\), the proper length as measured by the pilot. Rearrange the formula to solve for \(L_0\): \(L_0 = \frac{L}{\sqrt{1 - \frac{v^2}{c^2}}}\)

Substitute the given values and solve for the proper length

Plug in the given values: \(L = 40.0\,\text{m}\) and \(v = 0.70c\): \(L_0 = \frac{40.0\,\text{m}}{\sqrt{1 - \frac{(0.70c)^2}{c^2}}}\) Simplify the expression in the square root: \(L_0 = \frac{40.0\,\text{m}}{\sqrt{1 - (0.70^2)}}\)

Calculate the proper length

Now, calculate the proper length: \(L_0 = \frac{40.0\,\text{m}}{\sqrt{1 - 0.49}}\) \(L_0 = \frac{40.0\,\text{m}}{\sqrt{0.51}}\) \(L_0 \approx 56.19\,\text{m}\) So, the length of the spaceship as measured by the pilot is approximately 56.19 meters.