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Chapter 26: Problem 29

Rocket ship Able travels at \(0.400 c\) relative to an Earth observer. According to the same observer, rocket ship Able overtakes a slower moving rocket ship Baker that moves in the same direction. The captain of Baker sees Able pass her ship at \(0.114 c .\) Determine the speed of Baker relative to the Earth observer.

Short Answer

Expert verified
Answer: The velocity of rocket ship Baker relative to the Earth observer is approximately \(0.273c\).

Step by step solution

01

Write down the problem information

We have the following information: - Velocity of rocket ship Able relative to Earth Observer, \(v_A = 0.400c\) - Velocity of rocket ship Able relative to rocket ship Baker, \(v_{AB} = 0.114c\) We need to find the velocity of rocket ship Baker relative to Earth Observer, \(v_B\).
02

Write down the relativistic velocity addition formula

The relativistic velocity addition formula is given by \(v_{AB} = \cfrac{v_A - v_B}{1 - \cfrac{v_A v_B}{c^2}}\)
03

Rearrange the formula to isolate \(v_B\)

To isolate \(v_B\), we can first multiply both sides by the denominator: \(v_{AB}(1 - \cfrac{v_A v_B}{c^2}) = v_A - v_B\) Next, expand the left side and simplify: \((v_{AB} - \cfrac{v_{AB} v_A v_B}{c^2}) = v_A - v_B\) Now, move the terms with \(v_B\) to the left side and all other terms to the right side: \(v_B + \cfrac{v_{AB} v_A v_B}{c^2} = v_A - v_{AB}\) factor \(v_B\) from the left side: \(v_B(1 + \cfrac{v_{AB} v_A}{c^2}) = v_A - v_{AB}\) Finally, divide by \((1 + \cfrac{v_{AB} v_A}{c^2})\) to isolate \(v_B\): \(v_B = \cfrac{v_A - v_{AB}}{1 + \cfrac{v_{AB} v_A}{c^2}}\)
04

Substitute the given values and solve for \(v_B\)

Substitute \(v_A = 0.400c\) and \(v_{AB} = 0.114c\) into the formula for \(v_B\) and solve: \(v_B = \cfrac{0.400c - 0.114c}{1 + \cfrac{0.114c \cdot 0.400c}{c^2}}\) \(v_B = \cfrac{0.286c}{1 + \cfrac{0.0456c^2}{c^2}}\) \(v_B = \cfrac{0.286c}{1 + 0.0456}\) \(v_B = \cfrac{0.286c}{1.0456}\) \(v_B \approx 0.273c\)
05

Write down the final answer

The velocity of rocket ship Baker relative to the Earth observer is approximately \(0.273c\).

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Most popular questions from this chapter

A spaceship traveling at speed \(0.13 c\) away from Earth sends a radio transmission to Earth. (a) According to Galilean relativity, at what speed would the transmission travel relative to Earth? (b) Using Einstein's postulates, at what speed does the transmission travel relative to Earth?
Event A happens at the spacetime coordinates $(x, y, z, t)=(2 \mathrm{m}, 3 \mathrm{m}, 0,0.1 \mathrm{s})$ and event B happens at the spacetime coordinates \((x, y, z, t)=\left(0.4 \times 10^{8} \mathrm{m}\right.\) $3 \mathrm{m}, 0,0.2 \mathrm{s}) .$ (a) Is it possible that event A caused event B? (b) If event B occurred at $\left(0.2 \times 10^{8} \mathrm{m}, 3 \mathrm{m}, 0,0.2 \mathrm{s}\right)$ instead, would it then be possible that event A caused event B? [Hint: How fast would a signal need to travel to get from event \(\mathrm{A}\) to the location of \(\mathrm{B}\) before event \(\mathrm{B}\) occurred?]
Find the conversion between the mass unit \(\mathrm{MeV} / c^{2}\) and the SI unit of mass.
A spaceship travels toward Earth at a speed of \(0.97 c .\) The occupants of the ship are standing with their torsos parallel to the direction of travel. According to Earth observers, they are about \(0.50 \mathrm{m}\) tall and $0.50 \mathrm{m}$ wide. What are the occupants' (a) height and (b) width according to others on the spaceship?
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.]
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