Question

A low-flying earth satellite travels at about 8000 m/s. What is the factor \(\gamma\) for this speed? As observed from the ground, by how much would a clock traveling at this speed differ from a ground-based clock after one hour (as measured by the latter)? What is the percent difference?

Short answer

\(\gamma \approx 1.000000003555\); Time difference: 0.0000128 s; Percent difference: \(3.56 \times 10^{-7}\%\).

Step-by-step solution

Understanding the Problem

We need to calculate the Lorentz factor \(\gamma\) for a satellite moving at 8000 m/s and determine the time difference between a clock on the satellite and a ground-based clock. Finally, we'll find the percentage difference in time measurements.

Calculating the Lorentz Factor \(\gamma\)

The Lorentz factor is given by the equation \(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\), where \(v\) is the speed of the object and \(c\) is the speed of light (approximately \(3 \times 10^8\) m/s). Substituting the given speed \(v = 8000\) m/s:\[\gamma = \frac{1}{\sqrt{1 - \left(\frac{8000}{3 \times 10^8}\right)^2}} \approx 1.000000003555\]

Calculating Time Dilation

Time dilation can be calculated using the formula \(\Delta t' = \gamma \Delta t\), where \(\Delta t'\) is the time interval for the moving clock and \(\Delta t\) is the time interval for the stationary clock. For a one-hour interval \(\Delta t = 3600\) seconds:\[\Delta t' = 1.000000003555 \times 3600 \approx 3600.0000128\] seconds.

Finding the Time Difference

The difference in time measured between the clocks is \(\Delta t' - \Delta t\), which calculates to:\[\Delta t' - \Delta t = 3600.0000128 - 3600 = 0.0000128\] seconds.

Calculating the Percent Difference

The percent difference in time measurements is calculated by \(\left(\frac{\text{Time Difference}}{\Delta t}\right) \times 100\%\):\[\left(\frac{0.0000128}{3600}\right) \times 100\% \approx 3.56 \times 10^{-7}\%\]

Key concepts

Time Dilation

Time dilation is a fascinating concept that arises from the effects of special relativity. It describes how time can run at different rates for observers in different frames of reference. This phenomenon occurs when one observer is moving at a significant fraction of the speed of light relative to another. In the context of our satellite problem, time dilation makes the time on the satellite pass slightly slower than on Earth.

To calculate this, we use the time dilation formula:

• \[\Delta t' = \gamma \Delta t\]
• Where \(\Delta t\) is the time interval for a stationary clock on Earth, and \(\Delta t'\) is the time interval for the moving satellite clock.

For our satellite moving at 8000 m/s, a one-hour time interval on a ground clock appears slightly longer when observed from the satellite. The calculated difference may be small, only about 0.0000128 seconds, but it highlights time dilation's reality and importance in physics. Understanding time dilation helps in applications like GPS satellites, where precise timekeeping is essential.

Special Relativity

Special relativity, a theory proposed by Albert Einstein, revolutionized our understanding of space and time. It provides a framework for how objects behave at high velocities, close to the speed of light. The theory introduces two main postulates:

• The laws of physics are the same in all inertial frames of reference.
• The speed of light in a vacuum is constant and does not change regardless of the motion of the light source or observer.

From these, consequences like time dilation and length contraction arise. Special relativity greatly impacts how we view the universe. It's important in technologies that require high-speed motion, including satellite communication systems.
The Lorentz factor \(\gamma\) is a pivotal element in special relativity, modifying time intervals to account for high speeds. In our exercise, calculating \(\gamma\) for the satellite gives insight into how time progresses differently in various reference frames. This factor converges to 1 for low speeds but becomes significant as velocities approach the speed of light. Understanding this concept opens the door to grasping deeper physics at work in high-speed phenomena.

Satellite Motion

Satellites orbit the earth due to the perfect balance of velocity and gravitational pull, often moving at speeds such as 8000 m/s, as in our exercise example. Although this speed is relatively low compared to the speed of light, special relativity still plays a role.
In satellite motion, precision timing is crucial, especially in systems like GPS, where synchronization errors can lead to significant navigation inaccuracies. Because of time dilation, clocks on satellites tick slightly slower than those on Earth. Engineers use the Lorentz factor to adjust satellite clocks, ensuring high-precision measurements.

• Satellites are constantly in motion, subject to relativistic effects such as time dilation.
• The need for precise understanding of these effects is critical for accurate satellite-based technology.
• Satellite motion illustrates how special relativity applies to real-world engineering challenges.

Incorporating these effects lets us harness technology that is crucial for modern navigation and communication. So, although satellite speeds aren't near the speed of light, relativistic physics remains important, showing the theory's broad reach.