Problem 7 - University Physics with Modern Physics

Question

A spaceraft ties 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 shortest elapsed time?

Step-by-Step Solution

Answer

The clock on the spacecraft shows 1.10 hours less; it shows the shortest elapsed time.

Step 1: Calculate Total Time Taken by Earth Clock

According to the problem, the earth clock measures the round-trip time of the spacecraft as 365 days. Thus, the total time taken by the earth clock, in seconds, is:\(T_{\text{earth}} = 365 \times 24 \times 3600 = 31,536,000\) seconds.

Step 2: Determine Speed as a Fraction of Light Speed

The speed of the spacecraft is given as \(4.80 \times 10^{6} \text{ m/s}\). The speed of light is \(c = 3 \times 10^{8} \text{ m/s}\). Therefore, the speed as a fraction of the speed of light is:\(\frac{v}{c} = \frac{4.80 \times 10^{6}}{3 \times 10^{8}} = 0.016\).

Step 3: Calculate Time Dilation Factor (Lorentz Factor)

The time dilation factor, or Lorentz factor \(\gamma\), is calculated using the formula: \[\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}\] Substituting the value, we get: \[\gamma = \frac{1}{\sqrt{1- (0.016)^2}} \approx 1.000128\].

Step 4: Calculate Time Measured by Spacecraft Clock

The time measured by the spacecraft clock \(T_{\text{spacecraft}}\) can be found using the time dilation formula: \[T_{\text{spacecraft}} = \frac{T_{\text{earth}}}{\gamma}\] Substituting the values gives: \[T_{\text{spacecraft}} = \frac{31,536,000}{1.000128} \approx 31,532,036\] seconds.

Step 5: Determine Time Difference in Seconds

The difference in elapsed time \(\Delta T\) is given by: \[\Delta T = T_{\text{earth}} - T_{\text{spacecraft}}\] \[\Delta T = 31,536,000 - 31,532,036 = 3964 \] seconds.

Step 6: Convert Time Difference to Hours

Convert the time difference from seconds to hours by dividing by 3600 seconds per hour: \[\Delta T_{\text{hours}} = \frac{3964}{3600} \approx 1.10 \] hours.

Key Concepts

Lorentz factor

Time dilation is a fascinating concept in the realm of relativity, where the Lorentz factor \(\gamma\) provides a way to quantify the difference in elapsed time between two clocks moving relative to each other. This factor emerges from the theory of special relativity, introduced by Albert Einstein, and captures how time can slow down for objects moving close to the speed of light. The Lorentz factor is calculated using the formula \[\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}\]where \(v\) is the velocity of the moving object and \(c\) is the speed of light.

• The Lorentz factor becomes significant when the velocity of the object approaches the speed of light.
• In our spacecraft example, despite a high speed, the factor \(\gamma\) is only slightly larger than 1.
• This indicates that the time dilation effect here is relatively small because the speed of the spacecraft is much less than the speed of light.

Understanding the Lorentz factor helps us grasp why astronauts returning from speedy space journeys would experience just tiny shifts in their biological clocks compared to what happens on Earth.

relativity

Relativity fundamentally altered our understanding of space and time. Einstein's theory laid the groundwork for understanding how these two concepts are interwoven, especially through the phenomenon of time dilation.

• The principle of relativity holds that the laws of physics are the same for all observers, no matter how they are moving relative to one another.
• This means there's no absolute time that ticks at the same rate everywhere across the universe.
• Instead, time can stretch and contract based on the relative velocities of observers, showcased by the spacecraft example here.

When traveling at significant speeds, like that of our spacecraft, time appears to move slower compared to a stationary observer on Earth. In this case, the clock on the spacecraft ticks more slowly than the one on Earth due to the relative motion between them. This subtle difference accumulates into a measurable effect over long durations, like the year-long journey of our spacecraft. Relativity shows us that time isn't static but a dynamic aspect of our universe.

atomic clocks

Atomic clocks are crucial in demonstrating time dilation practically. These highly precise timepieces function by measuring the vibrations of atoms, often cesium or rubidium. These vibrations occur at consistent frequencies, allowing atomic clocks to count time with unparalleled accuracy. In our scenario, both the spacecraft and Earth have synchronized atomic clocks. They highlight how motion at high speeds affects timekeeping as described by relativity.

• The slight discrepancy in elapsed time between the two clocks provides compelling evidence of special relativity in action.
• Real-world experiments have confirmed these effects by comparing atomic clocks on jet planes or satellites.
• These clocks demonstrate the synchronization and minuscule slowing of time due to their relative motion.

Through these atomic timekeepers, time dilation isn't just a theoretical idea but an observable reality. Their use in scientific experiments and everyday technology (like GPS) underscores the significance of relativistic principles in modern life.