Problem 6 - University Physics with Modern Physics

Question

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot flies past you in her spaceracer at a constant speed of 0.800 c relative to you. At the instant the space-racer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled \(1.20 \times 10^{8} \mathrm{in}\) past you, what does the race pilot read on her timer? (b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her? (c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

Step-by-Step Solution

Answer

(a) 0.833 seconds; (b) 2.00 x 10^8 meters; (c) 0.5 seconds.

Step 1: Understanding the Problem

This exercise involves the relative motion of two observers in space, with speeds approaching the speed of light. We're dealing with special relativity due to high velocities, given as 0.800c (where c is the speed of light). We need to calculate time and distance measurements from two different frames of reference.

Step 2: Calculations for Part (a)

The velocity of the race pilot relative to the space utility vehicle is 0.800c. We need to find the time on the race pilot's clock when the space utility vehicle measures that she has traveled 1.20 x 10^8 meters.Using the formula for time dilation: \[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \( t = \frac{d}{v} \), \( v = 0.800c \), and \( d = 1.20 \times 10^8 \) meters.Calculate: \[ t = \frac{1.20 \times 10^8}{0.800 \times 3.00 \times 10^8} \approx 0.5 \, \text{seconds} \]Thus, \[ t' = \frac{0.5}{\sqrt{1 - 0.6400}} \approx \frac{0.5}{0.6} \approx 0.833 \text{ seconds} \]

Step 3: Calculations for Part (b)

Now, we calculate the distance the race pilot measures from her perspective when her timer reads 0.833 seconds. To her, the space utility moves in the opposite direction but at the same speed (0.800c). Using the formula \[ d' = v \cdot t' \],we find:\[ d' = 0.800 \times 3.00 \times 10^8 \times 0.833 \approx 2.00 \times 10^8 \, \text{meters}\]

Step 4: Calculations for Part (c)

Finally, find the time you will read on your timer when the race pilot reads 0.833 seconds on her timer. We have to account for time dilation from the race pilot's frame:Since the time is already given for the race pilot as 0.833 seconds, due to the symmetry of time dilation,\[ t = 0.833 \times \sqrt{1 - 0.6400} \approx 0.5 \, \text{seconds}\] for your measurement.

Key Concepts

Time Dilation

Time dilation is a fascinating concept from Einstein's theory of special relativity. It describes how time experienced by an observer moving at a significant fraction of the speed of light differs from time experienced by a stationary observer. This discrepancy occurs because, at high velocities, time seems to slow down.

In the exercise scenario, the race pilot is moving at 0.800 times the speed of light or 0.800c. When you, as the pilot of the space utility vehicle, measure the movement of the spaceracer, the time on the race pilot's clock will be less than what your clock measures.

To find the time on the race pilot's clock, you use the formula for time dilation, which is: \[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \]where:

• \(t'\) is the time observed in the space racer’s frame,
• \(t\) is the time observed in your frame,
• \(v\) is the velocity relative to you,
• \(c\) is the speed of light.

Calculating step-by-step helps visualize how relative motion affects time. It's an excellent demonstration of how everyday intuitions about time can change dramatically at cosmic speeds.

Relative Motion

Relative motion is key in understanding scenarios involving two moving objects. In this case, both the space utility vehicle and the space racer are moving, but the focus is on the space racer's speed relative to your vehicle.

Each observer measures time and distance differently based on their relative motion. Due to the symmetry in their speeds (both consider the other moving at 0.800c away from them), their observations are tied directly to their respective frames of reference.

From your perspective, when the race pilot has moved a certain distance, time passes differently due to your high-speed difference. Interestingly, from the race pilot’s perspective, while she reads 0.833 seconds on her timer, she perceives different distances for your movement. This difference encapsulates the core idea where each observer sees the other in motion, yet differently from themselves.

Speed of Light

In special relativity, the speed of light (\(c\)) is a universal constant, approximately \(3.00 \times 10^8\) meters per second. It sets the ultimate speed limit in the universe and greatly influences the nature of time and space.

Since neither object in the exercise can exceed this speed, their relative motion calculations are based on fractions of light speed, hence the notation \(0.800c\).

Understanding light's role here involves recognizing how it shapes the measurements from different frames of reference. This constant speed ensures that regardless of the observers' motions, both the space racer and the utility vehicle perceive the limits of their speed relative to light unchanged. It highlights the revolutionary impact of relativity theories on physics, demonstrating the principle of invariant speed regardless of relative movement, a centerpiece in modern physics discussions.