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Chapter 26: Problem 82

An astronaut has spent a long time in the Space Shuttle traveling at $7.860 \mathrm{km} / \mathrm{s} .$ When he returns to Earth, he is 1.0 s younger than his twin brother. How long was he on the shuttle? [Hint: Use an approximation from Appendix A.5 and beware of calculator round off errors.]

Short Answer

Expert verified
Answer: The astronaut spent approximately 41,712 seconds, or about 11 hours and 35 minutes, on the Space Shuttle.

Step by step solution

01

Write down the time dilation formula.

The formula for time dilation due to the effects of special relativity is given by: \(t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}}\) Here, \(t'\) is the dilated time experienced by the astronaut, \(t\) is the time experienced on Earth, \(v\) is the velocity of the astronaut, and \(c\) is the speed of light.
02

Calculate the difference in time experienced by the twins.

The problem states that the astronaut is 1.0 s younger than his twin brother on Earth. Therefore, we can write the difference in time as: \(\Delta t = t - t' = 1.0\,\text{s}\) We need to find the value of \(t\), which is the time spent by the astronaut on the shuttle.
03

Rearrange the time dilation formula to find t.

We need to express t in terms of \(t'\), \(v\), and \(c\). First, rewrite the formula as: \(t = t' \sqrt{1 - \frac{v^2}{c^2}}\) Then, substitute the value of \(\Delta t\) into the equation: \(t = (t + 1.0\,\text{s}) \sqrt{1 - \frac{v^2}{c^2}}\)
04

Insert the given values and solve for t.

We are given the astronaut's velocity, \(v = 7.86 \,\text{km/s}\), which we need to convert to meters per second: \(v = 7.86 \times 10^3 \,\text{m/s}\) We also need to use the speed of light, \(c = 3 \times 10^8 \,\text{m/s}\). Now, we can substitute these values into the rearranged time dilation formula: \(t = (t + 1.0\,\text{s}) \sqrt{1 - \frac{(7.86 \times 10^3\, \text{m/s})^2}{(3 \times 10^8 \,\text{m/s})^2}}\) Solve for \(t\): \(t \approx 41712\,\text{s}\) The astronaut has spent about 41,712 seconds on the shuttle, which is about 11 hours and 35 minutes.

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Most popular questions from this chapter

A clock moves at a constant velocity of \(8.0 \mathrm{km} / \mathrm{s}\) with respect to Earth. If the clock ticks at intervals of one second in its rest frame, how much more than a second elapses between ticks of the clock as measured by an observer at rest on Earth? [Hint: Use the binomial approximation \(.(1-v^{2} / c^{2})^{-1 / 2} \approx 1+v^{2} /(2 c^{2}) .\)]
An observer in the laboratory finds that an electron's total energy is $5.0 \mathrm{mc}^{2} .$ What is the magnitude of the electron's momentum (as a multiple of \(m c\) ), as observed in the laboratory?
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Radon decays as follows: $^{222} \mathrm{Rn} \rightarrow^{218} \mathrm{Po}+\alpha .$ The mass of the radon-222 nucleus is 221.97039 u, the mass of the polonium- 218 nucleus is \(217.96289 \mathrm{u},\) and the mass of the alpha particle is 4.00151 u. How much energy is released in the decay? \(\left(1 \mathrm{u}=931.494 \mathrm{MeV} / \mathrm{c}^{2} .\right)\)
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