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Magazine Chapter 20: Problem 19
A spaceship makes the 300 -million-km round-trip journey to the Sun in just \(20 \mathrm{~min}\), as measured by clocks on Earth. How much time elapses on the spaceship's clock? (a) \(20 \mathrm{~min}\) (b) \(17 \mathrm{~min}\) (c) \(14 \mathrm{~min}\); (d) \(11 \mathrm{~min}\).
Short Answer
Expert verified
(b) 17 minutes
Step by step solution
01
Understanding the Problem
The problem is about time dilation, a concept in special relativity. We need to find out how much time elapses on the spaceship's clock when it travels a round-trip distance of 300 million km to the Sun in 20 minutes, as measured by Earth.
02
Determine the Speed
Calculate the speed of the spaceship using the formula \(v = \frac{d}{t}\), where \(d = 300 \text{ million km} = 3 \times 10^8 \text{ km}\) and \(t = 20 \text{ min} = \frac{1}{3} \text{ hr} = 1200 \text{ seconds}\). Therefore, \(v = \frac{3 \times 10^8 \text{ km}}{1200 \text{ s}} = 2.5 \times 10^5 \text{ km/s}\).
03
Calculate the Lorentz Factor
To find the time experienced on the spaceship, calculate the Lorentz factor (gamma, \(\gamma\)) using \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\), where \(c = 3 \times 10^5 \text{ km/s}\). First, \(\frac{v^2}{c^2} = \left(\frac{2.5 \times 10^5}{3 \times 10^5}\right)^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}\). Thus, \(\gamma = \frac{1}{\sqrt{1 - \frac{25}{36}}} = \frac{1}{\sqrt{\frac{11}{36}}} = \frac{6}{\sqrt{11}} \approx 1.13\).
04
Calculate the Time on the Spaceship Clock
Use the relationship between the time interval measured on Earth and on the spaceship: \(t' = \frac{t}{\gamma}\), where \(t = 20 \text{ min}\) and \(\gamma \approx 1.13\). Therefore, \(t' = \frac{20 \text{ min}}{1.13} \approx 17.7 \text{ min}\). Consequently, the closest answer option is 17 minutes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Special Relativity
Special Relativity is a fundamental concept in physics developed by Albert Einstein. This theory introduces the idea that the laws of physics are the same for all observers, regardless of their motion. A key aspect of Special Relativity is the realization that time and space are interconnected in a four-dimensional space-time.
- Time is not an absolute entity but changes depending on the observer's relative speed.
- At very high speeds, approaching the speed of light, time as measured by moving clocks (time dilation) becomes significantly different from time measured by stationary clocks.
- Special Relativity has profound implications for understanding the universe, particularly when dealing with high-velocity objects like spaceships.
Lorentz Factor
The Lorentz Factor, denoted by the Greek letter gamma (), is a crucial element in calculating aspects of special relativity, including time dilation. It's essentially a measure of how much time, length, and relativistic mass change for an object moving close to the speed of light.
- The Lorentz Factor formula is given by: = \(\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\), where:
- \(v\) is the velocity of the moving object,
- \(c\) is the speed of light.
- As the speed of the object approaches the speed of light, \(\gamma\) increases, indicating significant time dilation.
- In the exercise, the spaceship's velocity brings us to a Lorentz Factor of approximately 1.13, suggesting a noticeable but not extreme dilation effect.
Spaceship Time Calculation
The essence of time dilation in a practical scenario is exemplified through spaceship time calculation. In the given problem, the spaceship travels to the Sun and back at relativistic speeds.
- To find how much time passes in different frames (Earth vs. spaceship), you need to apply the Lorentz Factor.
- The calculation involves computing how much slower time progresses on the spaceship compared to the time observed on Earth.
- Using the formula \(t' = \frac{t}{\gamma}\):
- \(t\) is the time as measured on Earth, 20 minutes in this case.
- \(\gamma\) is approximately 1.13, as calculated previously.
- The result, approximately 17.7 minutes, indicates how time for the passengers onboard the spaceship passes slower than the time passed on Earth.
Physics Problem Solving
Physics problem-solving involves breaking down complex situations into understandable bits, applying correct formulas, and understanding concepts at the core of the problem.
- Identify what you are solving, such as calculating time dilation or velocity.
- Use known formulas like the Lorentz Factor and time dilation equations to compute the desired quantities.
- Verify the calculations and ensure they are physically sensible. For example, in this scenario, understanding that 17.7 minutes aligns reasonably with physically achievable motions.
- Engage with the problem conceptually to know when relativistic effects become significant.
