Problem 9 - College Physics

Question

A spacecraft flies away from the earth with a speed of \(4.80 \times 10^{6} \mathrm{m} / \mathrm{s}\) relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days \((1\) year) later, as measured by the clock that remained on earth. What is the difference in the elapsed times on the two clocks, measured in hours? Which clock, the one in the spacecraft or the one on earth, shows the smaller elapsed time?

Step-by-Step Solution

Answer

The clock on the spacecraft shows 1.5 hours less than the clock on Earth.

Step 1: Analyze the Problem

We need to find the time difference between two clocks: one on Earth and one on a spacecraft moving at a high speed. This requires understanding the effects of time dilation from special relativity.

Step 2: Use Time Dilation Formula

The time dilation formula is given by \( \Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( \Delta t \) is the time interval measured by the stationary observer (on Earth), \( \Delta t_0 \) is the proper time interval experienced by the moving observer (spacecraft), \( v \) is the velocity of the spacecraft, and \( c \) is the speed of light (approximately \(3 \times 10^8 \text{ m/s}\)).

Step 3: Calculate the Lorentz Factor

Calculate \( \sqrt{1 - \frac{v^2}{c^2}} \) where \( v = 4.80 \times 10^6 \text{ m/s} \). First compute \( \frac{v^2}{c^2} = \left(\frac{4.80 \times 10^6}{3 \times 10^8}\right)^2 = 2.56 \times 10^{-4} \). Therefore, \( \sqrt{1 - 2.56 \times 10^{-4}} \approx 0.999872 \).

Step 4: Calculate Proper Time on the Spacecraft

Given \( \Delta t = 365 \text{ days} \times 24 \text{ hours/day} = 8760 \text{ hours} \), compute \( \Delta t_0 = \Delta t \times 0.999872 \). This gives \( \Delta t_0 \approx 8758.5 \text{ hours} \).

Step 5: Determine the Time Difference

The time difference is \( 8760 \text{ hours} - 8758.5 \text{ hours} = 1.5 \text{ hours} \).

Step 6: Conclude Which Clock Shows Smaller Elapsed Time

The clock on the spacecraft shows a smaller elapsed time (8758.5 hours) compared to the clock on Earth (8760 hours) due to time dilation.

Key Concepts

Special Relativity

Special relativity is a theory proposed by Albert Einstein, which revolutionized the understanding of space and time. One of its fundamental principles is that the laws of physics are the same for all observers, regardless of their relative motion. This means that phenomena such as the speed of light remain constant for all observers, regardless of their speed or direction of motion.

In the context of this exercise, special relativity helps explain why time seems to move at different rates for observers in different frames of reference. It allows us to predict the amount of time that passes in a high-speed spacecraft compared to a stationary observer on Earth, resulting in the effect known as time dilation.

Special relativity fundamentally shifts how we perceive the flow of time when dealing with high velocities close to the speed of light. This is crucial when dealing with precise time measurement, such as with atomic clocks, where even minute differences in elapsed time can be significant.

Lorentz Factor

The Lorentz factor is a crucial concept in special relativity and is used to calculate time dilation and length contraction effects. Represented by the variable \( \gamma \), the Lorentz factor can be expressed with the formula:
\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]

Here, \( v \) is the velocity of the moving object, and \( c \) is the speed of light. In our specific scenario, the spacecraft traveling at \( 4.80 \times 10^{6} \text{ m/s} \) results in a very small value for \( \frac{v^2}{c^2} \), approximately \( 2.56 \times 10^{-4} \).

This translates to a Lorentz factor that's slightly greater than one (about 0.999872 in our exercise). The Lorentz factor quantifies how much time dilation occurs, meaning it's how much time slows down for fast-moving observers relative to stationary ones.

The closer the speed of an object gets to the speed of light, the larger the Lorentz factor becomes, thereby increasing the effects of time dilation.

Proper Time

Proper time denotes the time interval measured by an observer who is at rest relative to the event being timed. It is the shortest time interval possible between two events and represents the time registered by an observer who sees the events occur in the same location, like the clock in the spacecraft.

In the exercise, proper time (\( \Delta t_0 \)) is the time on the atomic clock aboard the spacecraft, which experiences the flight's journey away from and back to Earth, compared to the observer on Earth who measures the journey’s time as \( \Delta t = 365 \) days. Special relativity uses proper time to illustrate why moving observers measure different elapsed times than stationary observers.

The spacecraft's atomic clock shows this proper time as it is co-moving with the craft, thus experiencing all the motion's effects directly. Therefore, calculating proper time lets us compare time intervals between moving and stationary frames.

Atomic Clock Synchronization

Atomic clock synchronization involves aligning the timing of clocks to ensure precise measurements. An atomic clock uses the vibrations of atoms to keep time with extraordinary accuracy. In the exercise, both the Earth's clock and the spacecraft's clock were synchronized before the journey.

By synchronizing the clocks before the spacecraft departs, the exercise can adequately measure the effects of time dilation and observe how differently the clocks keep time upon the spacecraft's return. This synchronization allows an accurate comparison of the time elapsed in different reference frames.

The differences in time readings after the journey underscore the reliability of Einstein’s theories. As the atomic clocks are very precise, even the minimal time discrepancies (such as 1.5 hours found in our exercise) are easily measurable. Such comparisons are vital in scientific experiments and technologies reliant on precise timing, such as GPS systems.