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\(\cdot\) Inside a spaceship flying past the earth at three-fourths the speed of light, a pendulum is swinging. (a) If each swing takes 1.50 s as measured by an astronat performing an experiment inside the spaceship, how long will the swing take as measured by a person at mission control on earth who is watching the experiment? (b) If each swing takes 1.50 s measured by a person at mission control on earth, how long will it take as measured by the astronaut in the spaceship?

Short Answer

Expert verified
(a) 2.27 s for Earth observer; (b) 0.99 s for astronaut.

Step by step solution

01

Understanding Time Dilation

According to Einstein's theory of relativity, time dilation is a difference in the elapsed time measured by two observers, due to a relative velocity between them. The time interval for an event, as measured in a stationary observer's frame (such as the astronaut inside the spaceship), is the proper time \(\Delta t_0\). The time interval as measured by an observer in another frame moving at a velocity \(v\) (the person at mission control) is \(\Delta t\). The relationship between these two times is given by the time dilation formula: \[\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}\] where \(c\) is the speed of light.
02

Calculating Time for Mission Control Observer (Part a)

For part (a), the spaceship travels at three-fourths the speed of light, so \(v = 0.75c\). The proper time \(\Delta t_0\) is 1.50 s. We substitute into the time dilation formula: \[\Delta t = \frac{1.50}{\sqrt{1 - \left(\frac{0.75c}{c}\right)^2}} = \frac{1.50}{\sqrt{1 - 0.5625}}\] \[= \frac{1.50}{\sqrt{0.4375}} = \frac{1.50}{0.6614} \approx 2.27 \text{ seconds}\] So, the swing takes approximately 2.27 seconds for the observer on Earth.
03

Calculating Proper Time for Astronaut (Part b)

For part (b), the situation is reversed. The time measured by mission control is 1.50 s, and we need to find the proper time \(\Delta t_0\) experienced by the astronaut. Rearranging the time dilation formula gives: \[\Delta t_0 = \Delta t \cdot \sqrt{1 - \left(\frac{v}{c}\right)^2}\] \[= 1.50 \cdot \sqrt{1 - 0.5625} = 1.50 \cdot 0.6614 \approx 0.99 \text{ seconds}\] Thus, the swing takes approximately 0.99 seconds for the astronaut on the spaceship.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Einstein's theory of relativity
One of the most revolutionary ideas in physics comes from Albert Einstein, who developed the theory of relativity. This theory changed how we understand space and time. It is not just one thing; it includes the special relativity and general relativity theories. Special relativity applies to objects moving at constant speeds, especially those close to the speed of light.

A key insight from special relativity is that as objects move at high speeds, time can appear to "stretch," resulting in what we call "time dilation." This means time doesn't flow at the same rate for everyone everywhere. People moving at different speeds will experience time differently. It's an important concept, especially when we look at things like spaceships moving at speeds close to the speed of light. Relativity tells us that the laws of physics are the same for all observers regardless of their velocity. This makes time dilation quite important in understanding space travel.
proper time
"Proper time" refers to the time measured by an observer who is at rest relative to the entire system being measured. In simpler terms, it's the time an observer would measure if they were right there with the clock, watching it tick.

For example, if you're in a spaceship watching a pendulum swing, the time you measure for each swing is the proper time. This is different from the time measured by someone observing from a distance, like the mission control on Earth. The proper time is always the shortest time interval. It's a fundamental piece of special relativity and is critical to calculating how long processes take when there's a significant relative velocity involved.
relative velocity
Relative velocity is a measure of how fast one object is moving in relation to another. In the context of Einstein's theory of relativity, it plays a crucial role because it influences observations of time and distance.

Consider the spaceship from our example. It's moving at three-fourths the speed of light relative to the Earth. So, depending on where you are—inside the spaceship or on Earth—you're going to see time and movement differently. This is because your frame of reference affects these perceptions. The faster the relative velocity between the observers, the more pronounced the effects on time and space become. It's this relative speed that causes the fascinating effects predicted by relativity, like time dilation.
speed of light
The speed of light is an incredible constant in the universe. It's the fastest speed anything can move through space, approximately 299,792 kilometers per second. In the context of relativity, it's the ultimate speed limit and has profound implications for time and space.

Einstein's equations show us that as objects move quicker and closer to the speed of light, strange things begin to happen. One of these is time dilation, where time seems to slow down from an outside observer's perspective. This speed isn't just a number; it's crucial in the equations of relativity that explain why time feels different in different frames of reference. It's why, as you approach the speed of light, the effects of relativity become more noticeable, stretching out time and altering distances.

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Most popular questions from this chapter

\(\bullet\) (a) At what speed does the momentum of a particle differ by 1.0\(\%\) from the value obtained with the nonrelativistic expression \(m v ?\) (b) Is the correct relativistic value greater or less than that obtained from the nonrelativistic expression?

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A meterstick moves past you at great speed. Its motion relative to you is parallel to its long axis. If you measure the length of the moving meterstick to be \(1.00 \mathrm{ft}(1 \mathrm{ft}=0.3048 \mathrm{m})-\) for example, by comparing it with a l-foot ruler that is at rest relative to you, at what speed is the meterstick moving relative to you?

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