Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 26, Problem 67P

(a)

To determine

The speed of the ship travelling according to an Earth observer.

(a)

Expert Solution
Check Mark

Answer to Problem 67P

The speed of the ship travelling according to an Earth observer is 0.625c_.

Explanation of Solution

The time elapse between the two events is 50.0ns, the length of the ship is 12.0m.

Let us use the symbols with primes for the astronaut’s frame and symbols without primes for the Earth’s frame. Earth measures the proper time between the events since they occur at the same location in that frame so that Δt0=Δt which is 50.0ns. The astronaut’s measures the proper length of the ship since it is at rest in that frame so that L=L0 which is 12.0m.

Write the expression for the contracted length in the Earth’s frame.

    L=L0γ                                                                                                                       (I)

Here, L is the contracted length in the Earth frame, L0 is the proper length of the ship, γ is the Lorentz factor.

Write the expression for the speed of the ship measured in the Earth’s frame.

    v=LΔt                                                                                                                        (II)

Here, v is the speed of the ship measured in the Earth’s frame, Δt is the change in time.

Use equation (I) in equation (II).

    v=L0γΔt                                                                                                                   (III)

Rearrange the above equation,

    vγ=L0Δt                                                                                                                  (IV)

Write the expression for the Lorentz factor.

    γ=11v2/c2                                                                                                        (V)

Here, v is the speed of the spaceship, c is the speed of light.

Conclusion:

Substitute 12.0m for L0, and 50.0ns for Δt in equation (IV).

    vγ=12.0m/s50.0ns=12.0m/s50.0ns×1s1×109ns=2.40×108m/s×1c3×108m/s=0.800c                                                                            (VI)

Use equation (V) and (VI) in equation (IV)and solve for to v.

    v(11v2/c2)=0.800cv2=(0.800c)2(1v2/c2)v=0.6401.640c=0.625c

Therefore, the speed of the ship travelling according to an Earth observer is 0.625c_.

(b)

To determine

The elapsed time between light flashes in the astronaut’s frame of reference.

(b)

Expert Solution
Check Mark

Answer to Problem 67P

The elapsed time between light flashes in the astronaut’s frame of reference is 64.0ns_.

Explanation of Solution

The clock measures the proper time interval Δt0 in the reference frame in which the clock is at rest. In any other frame, the time interval is longer.

Write the expression for the time dilation.

    Δt=γΔt0                                                                                                              (VII)

Here, Δt is the time elapsed in the astronauts frame.

Use equation (V) in equation (VII),

    Δt=Δt1v2c2                                                                                                       (VIII)

Conclusion:

Substitute 50.0ns for Δt and 0.625c for v in equation (VIII) to find Δt.

    Δt=50.0ns1(0.625c)2c2=64.0ns

Therefore, the elapsed time between light flashes in the astronaut’s frame of reference is 64.0ns_.

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