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Chapter 26: Problem 19

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

Expert verified
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

01

Determine the Lorentz factor

Calculate the Lorentz factor: \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\), using the given velocity \(v = 0.40c\).
02

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: \(L = L_0 = 1.0 \mathrm{m}\).
03

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.

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Most popular questions from this chapter

Electron A is moving west with speed \(\frac{3}{5} c\) relative to the lab. Electron \(\mathrm{B}\) is also moving west with speed \(\frac{4}{5} c\) relative to the lab. What is the speed of electron \(\mathrm{B}\) in a frame of reference in which electron \(\mathrm{A}\) is at rest?
A spaceship travels at constant velocity from Earth to a point 710 ly away as measured in Earth's rest frame. The ship's speed relative to Earth is $0.9999 c .$ A passenger is 20 yr old when departing from Earth. (a) How old is the passenger when the ship reaches its destination, as measured by the ship's clock? (b) If the spaceship sends a radio signal back to Earth as soon as it reaches its destination, in what year, by Earth's calendar, does the signal reach Earth? The spaceship left Earth in the year 2000.
Two atomic clocks are synchronized. One is put aboard a spaceship that leaves Earth at \(t=0\) at a speed of \(0.750 c .\) (a) When the spaceship has been traveling for \(48.0 \mathrm{h}\) (according to the atomic clock on board), it sends a radio signal back to Earth. When would the signal be received on Earth, according to the atomic clock on Earth? (b) When the Earth clock says that the spaceship has been gone for \(48.0 \mathrm{h},\) it sends a radio signal to the spaceship. At what time (according to the spaceship's clock) does the spaceship receive the signal?
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A cosmic-ray proton entering the atmosphere from space has a kinetic energy of \(2.0 \times 10^{20} \mathrm{eV} .\) (a) What is its kinetic energy in joules? (b) If all of the kinetic energy of the proton could be harnessed to lift an object of mass \(1.0 \mathrm{kg}\) near Earth's surface, how far could the object be lifted? (c) What is the speed of the proton? Hint: \(v\) is so close to \(c\) that most calculators do not keep enough significant figures to do the required calculation, so use the binomial approximation: $$\text { if } \gamma \gg 1, \quad \sqrt{1-\frac{1}{\gamma^{2}}} \approx 1-\frac{1}{2 \gamma^{2}}$$
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