Problem 9 - Fundamentals of Physics

Question

A spaceship of rest length \(130 \mathrm{~m}\) races past a timing station at a speed of \(0.740 c .\) (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 ends of the ship?

Step-by-Step Solution

Answer

The length is approximately 87.44 m and the time interval is approximately 3.94 x 10^{-7} seconds.

Step 1: Understand the Problem

We have a spaceship with a rest length of \(130\ m\) and it is traveling at a speed \(0.740c\) relative to a timing station. We are to find the length of this spaceship as seen by the timing station and the time it takes for the entire spaceship to pass by the timing station.

Step 2: Apply the Length Contraction Formula

The length observed by a stationary observer is given by the length contraction formula: \( L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \), where \( L_0 = 130\ m \), \( v = 0.740 c \), \( c \) is the speed of light, and \( L \) is the contracted length. Substitute the known values:\[ L = 130 \sqrt{1 - (0.740)^2} \]

Step 3: Calculate the Contracted Length

First, calculate \( (0.740)^2 = 0.5476 \). Then compute \( 1 - 0.5476 = 0.4524 \). Take the square root to get \( \sqrt{0.4524} \approx 0.6726 \). Multiply by the rest length:\[ L = 130 \times 0.6726 \approx 87.44 \ m \]Thus, the length measured by the timing station is approximately \( 87.44 \ m \).

Step 4: Calculate the Time Interval for the Ship to Pass

The time interval measured by the timing station is the time it takes for the entire spaceship to pass. This is given by \( \Delta t = \frac{L}{v} \), where \( L \) is the length observed by the station (87.44 m) and \( v = 0.740c \). To find \( \Delta t \), use the formula:\[ \Delta t = \frac{87.44}{0.740 \times 3 \times 10^8} \]Calculate \( 0.740 \times 3 \times 10^8 = 2.22 \times 10^8 \). Then compute:\[ \Delta t = \frac{87.44}{2.22 \times 10^8} \approx 3.94 \times 10^{-7} \ s \]Thus, the time interval is approximately \( 3.94 \times 10^{-7} \) seconds.

Key Concepts

Special Relativity

Special relativity is a revolutionary theory proposed by Albert Einstein in 1905. It explained how the laws of physics are the same for all observers, no matter their relative speed. This theory introduces two important ideas: the constancy of the speed of light and how time and space are interwoven.

• One consequence of special relativity is that the speed of light, approximately \(3 \times 10^8 \ \mathrm{m/s}\), is the same for all observers.
• Another is that time and space are not absolute; their measurements can change depending on the observer's movement.

This theory impacts how we understand phenomena such as time dilation and length contraction. In the context of the exercise, these effects are crucial to calculating how the length of the spaceship changes as it moves at high speeds.

Spaceship Velocity

The velocity of a spaceship can drastically influence the measurements from an observer's perspective due to relativistic effects. In the problem, the spaceship travels at \(0.740c\) relative to a timing station, where \(c\) is the speed of light.

• When objects move at a sizable fraction of the speed of light, like the spaceship in this exercise, relativistic effects such as length contraction occur.
• This means the spaceship's length as seen by a stationary observer would appear shorter than its actual rest length.

Calculating these effects involves understanding how velocity, a time rate of change of position, interacts with the fundamental principles of relativity.

Time Dilation

Time dilation is a key concept in special relativity, describing how a moving clock ticks slower compared to a stationary one. As the spaceship moves quickly past the timing station, relativistic effects significantly impact the time interval measured.

• In the exercise, time dilation is reflected in how observers perceive the duration of the spaceship passing by.
• The time interval is stretched when viewed from a stationary frame relative to the moving spaceship.

The formula \(\Delta t = \frac{L}{v}\) calculates the time it takes for the spaceship to pass a timing station. While length contraction affects the observed length \(L\), time dilation helps determine how long an event appears to take. Understanding both time dilation and length contraction is vital in comprehending how time and space behave under high velocities.

Speed of Light

The speed of light is a fundamental constant in physics, denoted as \(c\). It plays a critical role in special relativity and affects every calculation involving relativistic speeds.

• In our exercise, the spaceship's velocity is a large fraction of the speed of light, specifically \(0.740c\).
• This highlights the need for relativistic equations when measuring things like length and time.

In numerous physics problems, the speed of light not only sets a universal speed limit but also ensures that all observers measure it as the same, regardless of their relative speeds. Its constancy unifies various phenomena, such as length contraction and time dilation, under the umbrella of relativity.