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Chapter 26: Problem 5

An astronaut wears a new Rolex watch on a journey at a speed of $2.0 \times 10^{8} \mathrm{m} / \mathrm{s}$ with respect to Earth. According to mission control in Houston, the trip lasts \(12.0 \mathrm{h}\). How long is the trip as measured on the Rolex?

Short Answer

Expert verified
Answer: The trip's duration as measured on the astronaut's Rolex watch is approximately 57,915 seconds or approximately 16.1 hours.

Step by step solution

01

Identify the given values

We are given the speed of the astronaut's journey, \(v = 2.0 \times 10^8 \, \mathrm{m/s}\), with respect to Earth, and the trip lasts a time duration \(t_{earth} = 12.0 \, \mathrm{h}\) according to mission control in Houston. Our task is to determine the time duration as measured by the astronaut's Rolex watch, \(t_{astronaut}\).
02

Convert the Earth time duration to seconds

Before proceeding with the time dilation formula, we need to convert the given Earth time duration from hours to seconds. This will make our calculations consistent in terms of units: \[t_{earth} = 12.0 \, \mathrm{h} \times \frac{3600 \, \mathrm{s}}{1 \, \mathrm{h}} = 43200 \, \mathrm{s}\]
03

Apply the time dilation formula

Next, we will use the time dilation formula derived from the Lorentz transformation to relate the time experienced by the astronaut to the time experienced by mission control on Earth. The formula is as follows: \[t_{astronaut} = \frac{t_{earth}}{\sqrt{1 - \frac{v^2}{c^2}}}\] In this equation, \(v\) is the velocity of the astronaut with respect to Earth, \(t_{earth}\) is the time duration experienced on Earth, and \(c\) represents the speed of light, which is approximately \(3.0 \times 10^8 \, \mathrm{m/s}\).
04

Calculate the time experienced by the astronaut

Now we will plug in the given values and solve for the time experienced by the astronaut, \(t_{astronaut}\): \[t_{astronaut} = \frac{43200 \, \mathrm{s}}{\sqrt{1 - \frac{(2.0 \times 10^8 \, \mathrm{m/s})^2}{{(3.0 \times 10^8 \, \mathrm{m/s})}^2}}}\]
05

Compute the final result

By calculating the value inside the square root and solving the equation above, we can find the value of \(t_{astronaut}\): \[t_{astronaut} \approx \frac{43200 \, \mathrm{s}}{\sqrt{0.5556}} \approx \frac{43200 \, \mathrm{s}}{0.7454} \approx 57915 \, \mathrm{s}\] So, the trip's duration as measured on the Rolex watch is approximately \(57915 \, \mathrm{s}\) or approximately \(16.1 \, \mathrm{h}\).

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Most popular questions from this chapter

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? \(\left(1 \mathrm{u}=931.494 \mathrm{MeV} / \mathrm{c}^{2} .\right)\)
A spaceship is traveling away from Earth at \(0.87 c .\) The astronauts report home by radio every \(12 \mathrm{h}\) (by their own clocks). At what interval are the reports sent to Earth, according to Earth clocks?
An engineer in a train moving toward the station with a velocity \(v=0.60 c\) lights a signal flare as he reaches a marker \(1.0 \mathrm{km}\) from the station (according to a scale laid out on the ground). By how much time, on the stationmaster's clock, does the arrival of the optical signal precede the arrival of the train?
Kurt is measuring the speed of light in an evacuated chamber aboard a spaceship traveling with a constant velocity of \(0.60 c\) with respect to Earth. The light is moving in the direction of motion of the spaceship. Siu- Ling is on Earth watching the experiment. With what speed does the light in the vacuum chamber travel, according to Siu-Ling's observations?
Show that each of these statements implies that \(v<c\) which means that \(v\) can be considered a nonrelativistic speed: (a) $\gamma-1<<1 \text { [Eq. }(26-14)] ;\( (b) \)K<<m c^{2}\( [Eq. \)(26-15)] ;(\text { c }) p<m c[\text { Eq. }(26-16)] ;\( (d) \)K \approx p^{2} /(2 m)$.
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