Chapter 15, Problem 16A

(a)

To determine

To rank: The speeds from greatest to least.

Answer to Problem 16A

A= B= C.

Explanation of Solution

Introduction: According to the Special Theory of Relativity, the speed of light is equal in all inertial frames irrespective of the state of motion of the emitting body.

When spaceship emits brief flashes of light, then the observer to the right, in front of the approaching spaceship would see the flashes coming with equal speed because he is standing right in front of it. And, the speed of light is constant in all inertial frames i.e. $c=3\times {10}^{8}m/s$ .

Conclusion: Thus, the observer will see the flashes equally for all three spaceships.

(b)

To determine

To rank: The frequency from greatest to least.

Answer to Problem 16A

C> A> B.

Explanation of Solution

Introduction: According to the Special Theory of Relativity, the speed of light is equal in all inertial frames irrespective of the state of motion of the emitting body.

After observing the circles of flashes already emitted by the spaceship, it is obvious that the spaceship in the circles C is approaching faster. Thus, the frequency of observing flashes will be greatest in C.

However, in circle A, flashes would appear at a lesser frequency than C due to distance of spaceship in the flash circle.

In circle B, spaceship is at the center of the concentering circles which are at equal distance to each other. Thus, the frequency of observing flashes from circle B would be least.

Conclusion: Thus, frequency of observing flashes by observer would be in the order C> A> B.

(c)

To determine

To rank: The speeds of the spaceship as seen by an Earth observer from greatest to least.

Answer to Problem 16A

C> A> B.

Explanation of Solution

Introduction: According to the Special Theory of Relativity, the speed of light is equal in all inertial frames irrespective of the state of motion of the emitting body.

Spaceships are emitting the flashes in the interval of 1-second. As seen in the circles, spaceship in circle C is gaining its speed approaching Earth. When speed is calculated by an observer on Earth, the sum would be the actual speed of the spaceship and the distance light covered in reaching the Earth.

As seen in circle A, spaceship is emitting single a second later but is faster in approaching Earth as compared to B.

Conclusion: Thus, the observer on Earth will observe the spaceships approaching him in the ranking C> A> B.