Question

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.150 s . The first officer on the craft measures the searchlight to be on for 18.0 ms . Part A Which of these two measured times is the proper time? 0.150 s 18.0 ms Part B What is the speed of the spacecraft relative to the earth, expressed as a fraction of the speed of light, c?

Answer 1

The proper time interval is 18.0 ms. The speed of the spacecraft relative to the earth is approximately 0.866c, or 86.6% of the speed of light.

Step-by-step explanation:

The proper time interval is the time measured in the frame of reference where the event takes place. In this case, the first officer on the spacecraft measures the searchlight to be on for 18.0 ms. Therefore, the proper time is 18.0 ms.

To determine the speed of the spacecraft relative to the earth, we can use the concept of time dilation. The time measured in one frame of reference will be dilated or stretched relative to another frame of reference moving at a different velocity. Given the time measured by the first officer and the duration observed from your frame of reference (0.150 s), we can calculate the speed of the spacecraft relative to the earth.

Using the formula for time dilation, t' = t / √(1 - (v² / c²)), where t' is the time measured in the moving frame of reference, t is the proper time, v is the velocity of the moving frame of reference, and c is the speed of light, we can solve for v. Plugging in the known values, we can find that the speed of the spacecraft relative to the earth is approximately 0.866c, or 86.6% of the speed of light.

Answer 2

The proper time is the time measured on the spacecraft, which is 18.0 ms. To find the speed of the spacecraft relative to Earth, we would employ time dilation formulas from special relativity.

Step-by-step explanation:

The spaceflight scenario described involves concepts of time dilation in the theory of special relativity.

Part A: Proper Time

The proper time is the time interval measured by an observer who sees the events at the same location in their frame of reference. In this scenario, the proper time is the time measured on the spacecraft, which is 18.0 ms.

Part B: Speed of the Spacecraft

We can use the time dilation formula Δt = Δt0 / √(1 - v²/c²) to determine the speed of the spacecraft. Here, Δt is the time interval measured on Earth (0.150 s), and Δt0 is the proper time (18.0 ms or 0.018 s). Solving for v gives us the speed of the spacecraft relative to Earth:

1. First, express Δt and Δt0 in seconds.
2. Then, substitute Δt and Δt0 into the time dilation formula and solve for v².
3. Take the square root of v² to get v, and express it as a fraction of the speed of light, c.