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Chapter 26: Problem 10

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 \(.(1-v^{2} / c^{2})^{-1 / 2} \approx 1+v^{2} /(2 c^{2}) .\)]

Short Answer

Expert verified
Answer: Approximately \(3.56 \times 10^{-10}\, \text{seconds}\) more time elapses between ticks of the clock when observed by an observer at rest on Earth compared to when observed in the clock's rest frame.

Step by step solution

01

Time dilation formula

The time dilation formula in special relativity states that the observed time interval, \(t\), in a stationary frame is related to the proper time interval, \(t_0\), in the moving frame by: \(t = \dfrac{t_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}\) where \(v\) is the relative velocity between the two frames and \(c\) is the speed of light.
02

Apply the binomial approximation

We will use the given binomial approximation to simplify the time dilation formula: \((1 - \frac{v^2}{c^2})^{-1/2} \approx 1 + \frac{v^2}{2c^2}\) Now, the formula for the time dilation becomes: \(t \approx t_0 (1 + \frac{v^2}{2c^2})\)
03

Plug in the given values

We are given that \(t_0 = 1 \, \text{second}\), and \(v = 8.0 \, \text{km/s} = 8000 \, \text{m/s}\) (convert kilometers to meters). Also, given that the speed of light, \(c \approx 3 \times 10^8 \, \text{m/s}\), we can apply these values to the binomial simplification of the time dilation formula: \(t \approx 1 \, \text{second} \left( 1 + \dfrac{(8000 \, \text{m/s})^2}{2 (3 \times 10^8 \, \text{m/s})^2} \right)\)
04

Evaluate the time dilation

Calculate the time dilation with the given values: \(t \approx 1 \, \text{second} \left( 1 + \dfrac{64 \times 10^6}{2 \times 9 \times 10^{16}} \right)\) \(t \approx 1 \, \text{second} \left( 1 + \dfrac{32}{9 \times 10^{10}} \right)\) Now, evaluate the expression inside the parentheses: \(1 + \dfrac{32}{9 \times 10^{10}} = 1 + 3.56 \times 10^{-10}\) Therefore, the observed time between ticks for an observer at rest on Earth is: \(t \approx 1 \, \text{second} \left( 1 + 3.56 \times 10^{-10} \right) \approx 1.000000356 \, \text{second}\)
05

Find the difference in time

We can now find the difference in time between the clock ticks as measured by an observer at rest on Earth and the clock ticks in the rest frame of the clock: \(\Delta t = t - t_0 \approx 1.000000356 \, \text{seconds} - 1 \, \text{second} \approx 3.56 \times 10^{-10} \, \text{seconds}\) So, about \(3.56 \times 10^{-10}\, \text{seconds}\) more than a second elapses between ticks of the clock as measured by an observer at rest on Earth.

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Most popular questions from this chapter

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.
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