Question

A spaceship moves at a constant velocity of \(0.40 c\) relative to an Earth observer. The pilot of the spaceship is holding a rod, which he measures to be \(1.0 \mathrm{m}\) long. (a) The rod is held perpendicular to the direction of motion of the spaceship. How long is the rod according to the Earth observer? (b) After the pilot rotates the rod and holds it parallel to the direction of motion of the spaceship, how long is it according to the Earth observer?

Short answer

Answer: (a) 1.0 m (rod held perpendicular), (b) The length of the rod when held parallel can be calculated using the Lorentz contraction formula and the given values for L_0 and v.

Step-by-step solution

Determine the Lorentz factor

Calculate the Lorentz factor: \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\), using the given velocity \(v = 0.40c\).

Calculate length when the rod is perpendicular

Since the rod is held perpendicular to the direction of motion, there is no length contraction, and the length of the rod according to the Earth observer is the same as the proper length: \(L = L_0 = 1.0 \mathrm{m}\).

Calculate length when the rod is parallel

When the rod is held parallel to the direction of motion, we need to apply the Lorentz contraction formula: \(L = L_0 \cdot \sqrt{1 - \frac{v^2}{c^2}}\). Plug in the values for \(L_0\) and \(v\) to find the length of the rod according to the Earth observer. The answers to the problem are: (a) 1.0 m (rod held perpendicular), (b) the length of the rod when held parallel can be calculated using the formulas mentioned above.