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Chapter 20: Problem 22

An airplane whose rest length is \(42.4 \mathrm{~m}\) is moving with respect to the Earth at a constant speed of \(522 \mathrm{~m} / \mathrm{s}\). (a) By what fraction of its rest length will it appear to be shortened to an observer on Earth? (b) How long would it take by Earth clocks for the airplane's clock to fall behind by \(1 \mu \mathrm{s}\) ? (Assume that only special relativity applies.)

Short Answer

Expert verified
For (a), the length of the airplane will appear to be shortened by \(0.000455\) or \(0.0455 \% \) of the length it is at rest to an observer on Earth. For (b), it would take approximately \(1.000001 \mu s\) by Earth clocks for the airplane's clock to fall behind by \(1 \mu s\).

Step by step solution

01

Identify Given Values

Identify the given values: Rest length of the plane \(L_0=42.4m\), Speed of the plane \(v=522m/s\), Time delay \(\Delta t=1 \mu s\). Also recall the speed of light \(c=3.0x10^{8}m/s\).
02

Calculate Length Contraction

Using the formula for the length contraction \(L=L_0/\gamma\) where \(\gamma=1/\sqrt{1-(v/c)^2}\), substitute the given values to find the length \(L\) and then calculate the percentage of rest length that it appears shortened, which is \((1 - L/L_0) x 100\% \)
03

Calculate Time Dilation

Using the formula for the time dilation \(\Delta T=\gamma*\Delta t\) (where \(\Delta T\) is the time according to an Earth observer and \(\Delta t\) is a proper time on the moving airplane, which is given), substitute the values to determine the time \(\Delta T\). Finally, find how much time the airplane's clock falls behind the earth clock, which is \(\Delta T - \Delta t\).

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