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Chapter 26: Problem 65

A starship takes 3.0 days to travel between two distant space stations according to its own clocks. Instruments on one of the space stations indicate that the trip took 4.0 days. How fast did the starship travel, relative to that space station?

Short Answer

Expert verified
Answer: The relative speed between the starship and the space station was approximately 1.29 * 10^8 m/s.

Step by step solution

01

Identify the given information

We are given: 1. Time experienced by the starship: \(t_{starship} = 3.0 \, days\) 2. Time experienced by the space station: \(t_{station} = 4.0 \, days\) Our goal is to find the relative speed, \(v\), between the starship and the space station.
02

Convert time to seconds

First, we need to convert both given times to seconds, as follows: 1 day = 24 hours = 1440 minutes = 86400 seconds \(t_{starship} = 3.0 \, days * 86400 \, s/day = 259200 \, s\) \(t_{station} = 4.0 \, days * 86400 \, s/day = 345600 \, s\)
03

Apply the time dilation formula

According to the time dilation formula in special relativity, we have: \(t_{station} = \frac{t_{starship}}{\sqrt{1 - \frac{v^2}{c^2}}}\) where \(t_{station}\) is the time experienced by the space station, \(t_{starship}\) is the time experienced by the starship, \(v\) is the relative speed between the starship and the space station, and \(c\) is the speed of light (approximately \(3.0 * 10^8 \, m/s\)). We need to solve this equation for the relative speed \(v\).
04

Rearrange the time dilation formula

First, divide both sides of the equation by \(t_{starship}\): \(\frac{t_{station}}{t_{starship}} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) Square both sides: \(\left(\frac{t_{station}}{t_{starship}}\right)^2 = \frac{1}{1 - \frac{v^2}{c^2}}\) Next, isolate the term containing the relative speed \(v\): \(\frac{v^2}{c^2} = 1 - \frac{1}{\left(\frac{t_{station}}{t_{starship}}\right)^2}\) Finally, solve for the relative speed \(v\): \(v = c * \sqrt{1 - \frac{1}{\left(\frac{t_{station}}{t_{starship}}\right)^2}}\)
05

Calculate the relative speed

Now, plug in the values of \(t_{starship}\), \(t_{station}\), and \(c\) to find the relative speed \(v\): \(v = (3.0 * 10^8 \, m/s) * \sqrt{1 - \frac{1}{\left(\frac{345600 \, s}{259200 \, s}\right)^2}}\) \(v \approx 1.29 * 10^8 \, m/s\) Therefore, the starship traveled at a relative speed of approximately \(1.29 * 10^8 \, m/s\) with respect to the space station.

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Most popular questions from this chapter

Particle \(A\) is moving with a constant velocity \(v_{\mathrm{AE}}=+0.90 c\) relative to an Earth observer. Particle \(B\) moves in the opposite direction with a constant velocity \(v_{\mathrm{BE}}=-0.90 c\) relative to the same Earth observer. What is the velocity of particle \(B\) as seen by particle \(A ?\)
An object of mass \(0.12 \mathrm{kg}\) is moving at $1.80 \times 10^{8} \mathrm{m} / \mathrm{s}$ What is its kinetic energy in joules?
An astronaut in a rocket moving at \(0.50 c\) toward the Sun finds himself halfway between Earth and the Sun. According to the astronaut, how far is he from Earth? In the frame of the Sun, the distance from Earth to the Sun is \(1.50 \times 10^{11} \mathrm{m}\).
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 \(.(1-v^{2} / c^{2})^{-1 / 2} \approx 1+v^{2} /(2 c^{2}) .\)]
Muons are created by cosmic-ray collisions at an elevation \(h\) (as measured in Earth's frame of reference) above Earth's surface and travel downward with a constant speed of \(0.990 c .\) During a time interval of \(1.5 \mu \mathrm{s}\) in the rest frame of the muons, half of the muons present at the beginning of the interval decay. If one fourth of the original muons reach Earth before decaying, about how big is the height \(h ?\)
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