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Chapter 26: Problem 28

A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of \(0.60 c\) and \(0.80 c .\) What is the relative speed of the two ships as measured by a passenger on either one of the spaceships?

Short Answer

Expert verified
Answer: The relative speed of the two ships as measured by a passenger on either one of the spaceships is approximately \(0.9459c\).

Step by step solution

01

Identify the given velocities

The given velocities of the two spaceships are \(0.60c\) and \(0.80c.\) We will use these values in the formula for relative velocity in special relativity.
02

Use the relativistic velocity addition formula

To find the relative velocity of the spaceships, use the formula \(v_{rel} = \frac{v_1 + v_2}{1 + \frac{v_1 \cdot v_2}{c^2}},\) where \(v_1\) and \(v_2\) are the velocities of the spaceships. Substitute the given values: \(v_{rel} = \frac{0.60c + 0.80c}{1 + \frac{(0.60c) (0.80c)}{c^2}}.\)
03

Simplify the expression

Now, simplify the expression: \(v_{rel} = \frac{1.40c}{1 + \frac{0.48c^2}{c^2}}.\)
04

Cancel common terms

Cancel out the \(c^2\) terms in the denominator: \(v_{rel} = \frac{1.40c}{1 + 0.48}.\)
05

Calculate the relative velocity

Finally, calculate the relative velocity by dividing the numerator by the denominator: \(v_{rel} = \frac{1.40c}{1.48} = 0.9459c.\) The relative speed of the two ships as measured by a passenger on either one of the spaceships is approximately \(0.9459c.\)

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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 \(A ?\)
A constant force, acting for \(3.6 \times 10^{4} \mathrm{s}(10 \mathrm{h}),\) brings a spaceship of mass \(2200 \mathrm{kg}\) from rest to speed \(0.70 c\) (a) What is the magnitude of the force? (b) What is the initial acceleration of the spaceship? Comment on the magnitude of the answer.
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?
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