Problem 9 - Fundamentals of Physics

Question

A spaceship of rest length \(130 \mathrm {~m}\) races past a timing station at a speed of \(0.740 \mathrm{c}\). (a) What is the length of the spaceship as measured by the timing station? (b) What time interval will the station clock record between the passage of the front and back cnds of the ship?

Step-by-Step Solution

Answer

(a) 87.44 m; (b) 3.94 x 10^-7 s.

Step 1: Identify the Problem

For part (a), we need to find the length of a spaceship as measured by an observer stationary relative to a timing station. This involves the concept of length contraction in special relativity. For part (b), we need to determine the time interval the timing station records for the spaceship's passage, considering its speed and contracted length.

Step 2: Understand the Length Contraction Formula

The length contraction formula in special relativity is given by: \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \] where \( L \) is the contracted length, \( L_0 \) is the rest length, \( v \) is the velocity of the spaceship, and \( c \) is the speed of light.

Step 3: Solve for Contracted Length (Part a)

Substitute the given values into the formula. The rest length \( L_0 = 130 \, \text{m} \), the velocity \( v = 0.740 \, c \).\[L = 130 \, \text{m} \times \sqrt{1 - (0.740)^2}\]Calculate the value:\[L = 130 \, \text{m} \times \sqrt{1 - 0.5476} = 130 \, \text{m} \times \sqrt{0.4524}\]\[L \approx 130 \, \text{m} \times 0.6726 \approx 87.44 \, \text{m}\]

Step 4: Calculate Time Interval (Part b)

The time interval \( \Delta t \) is calculated by dividing the contracted length \( L \) by the velocity \( v \), given as a fraction of the speed of light: \[\Delta t = \frac{L}{v} = \frac{87.44 \, \text{m}}{0.740 \, c}\]First, express the speed of light \( c \) in m/s:\[\Delta t = \frac{87.44 \, \text{m}}{0.740 \times 3 \times 10^8 \, \text{m/s}}\]Perform the calculation:\[\Delta t \approx \frac{87.44}{222 \times 10^6} \approx 3.94 \times 10^{-7} \, \text{s}\]

Step 5: Answer the Questions

Part (a): The length of the spaceship as measured by the timing station is approximately 87.44 m. Part (b): The time interval recorded by the station clock is approximately 3.94 x 10^-7 s.

Key Concepts

Length Contraction

Length contraction is one of the surprising outcomes of Einstein’s theory of special relativity. Imagine you have a spaceship racing past a timing station at a high speed. To an observer at the station, the spaceship appears shorter than its actual length. This phenomenon is known as length contraction.

• It occurs only along the direction of motion.
• The faster an object moves, the more pronounced the contraction.
• It becomes significant at speeds approaching the speed of light.

The formula for length contraction is:\[L = L_0 \sqrt{1 - \frac{v^2}{c^2}}\]Here, - \(L\) is the length observed by the stationary observer.- \(L_0\) is the rest length, or the length of the object in its own rest frame.- \(v\) represents the object's velocity.- \(c\) is the speed of light.In our exercise, the spaceship's rest length (\(L_0\)) is 130 m, and it's traveling at 0.740c. The calculation showed that its contracted length (\(L\)) is approximately 87.44 m, meaning it appears shorter to someone at the timing station.

Rest Length

The rest length, sometimes referred to as proper length, is a fundamental concept in relativity. It is the length of an object as measured in its own rest frame, where it is not moving.

Think of the rest length as the 'true' length of the object. When an object is at rest with respect to the observer, its length is not contracted. This measurement is crucial for understanding how motion affects observations in relativistic contexts.

• Rest length is the maximum length the object can have when measured by an observer who is at rest relative to the object.
• It remains unchanged regardless of how an object is moving relative to other observers.

In the exercise, the spaceship's rest length is given as 130 m. This is the length it would have if it were not moving relative to the timing station. It's a constant value and unaffected by the spaceship's speed.

Time Dilation

Time dilation is another intriguing result of special relativity, which affects how time is perceived differently for observers in different frames of motion.

In the scenario involving the spaceship, the timing station measures the time interval for the spaceship's back and front end passage. Due to its high speed (0.740c), a time dilation effect occurs.

• Time will seem to move slower for the astronaut in the spaceship.
• The faster the spaceship's relative speed, the greater the time dilation effect.
• For the observer at the station, the moving clock (spaceship) ticks slower than a stationary clock.

The time interval calculated from the spaceship's contracted length and its speed (\(\Delta t = \frac{L}{v}\)) is approximately \(3.94 \times 10^{-7}\) s. This is how long the station records the event, showing the effects of relativistic speeds.