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Suppose that you are traveling on board a spacecraft that is moving with respect to the earth at a speed of \(0.975 c .\) You are breathing at a rate of 8.0 breaths per minute. As monitored on earth, what is your breathing rate?

Short Answer

Expert verified
The breathing rate as observed from Earth is approximately 1.76 breaths per minute.

Step by step solution

01

Identify Given Variables

We have a spacecraft moving at a speed \( v = 0.975c \), where \( c \) is the speed of light. The breathing rate on the spacecraft is given as 8.0 breaths per minute.
02

Understand Time Dilation Concept

According to the theory of relativity, time dilation occurs when an object moves at a significant fraction of the speed of light relative to an observer. The time experienced in the spacecraft, referred to as the proper time \( T_0 \), will appear to move slower for an observer on Earth.
03

Apply Time Dilation Formula

The time dilation formula is given by:\[ T = \frac{T_0}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \]where \( T \) is the observed time on Earth, \( T_0 \) is the proper time, and \( v \) is the velocity of the spacecraft. Substitute \( T_0 = 1 \) minute (for 1 minute of breathing) and \( v = 0.975c \).
04

Calculate Dilated Time

Substitute the values into the formula:\[ T = \frac{1}{\sqrt{1 - (0.975)^2}} \]This calculates to approximately \( T \approx 4.55 \) minutes. This is the time it takes for the observer on Earth to see 1 minute of breathing in the spacecraft.
05

Calculate Breathing Rate as Observed from Earth

The breathing rate in the spacecraft is 8.0 breaths per minute. As 1 minute on the spacecraft corresponds to about 4.55 minutes on Earth, the breathing rate as observed from Earth is:\[ \text{Breaths per minute from Earth} = \frac{8.0 \text{ breaths}}{4.55 \text{ minutes}} \approx 1.76 \text{ breaths per minute} \]

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Key Concepts

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

Relativity
Relativity is one of the cornerstone principles of modern physics, introduced by Albert Einstein. It fundamentally altered how we understand time and space. In the context of our exercise, relativity explains why observers in different frames (like a spacecraft moving near light-speed and an observer on Earth) perceive time differently. Einstein's theory of relativity consists of two parts: the special theory of relativity and the general theory of relativity. The special theory, which concerns our problem, deals with how the laws of physics are the same for all non-accelerating observers and reveals how light speed remains constant in all inertial frames. **Key Takeaways** - Moving clocks tick more slowly than ones at rest from an outside observer's perspective. - Distance and time can change based on the velocity of the observer. - The implications of relativity are broad and include things like the time dilation effect we calculate in the given problem.
Velocity
Velocity is more than just speed; it’s the speed of something in a given direction. In the context of relativity, velocity becomes crucial since it can greatly influence how time is experienced. In our problem, the spacecraft's velocity is given as a significant fraction of the speed of light, specifically 0.975c, where "c" is the constant speed of light. When an object, like our spacecraft, moves at this tremendous velocity, the relativistic effects come into play. One such effect is time dilation, where time seems to flow at a different rate for observers in different frames due to their relative velocities. **Key Takeaways** - Velocity is the combination of speed and direction. - High velocities, especially near-light speed, cause noticeable relativistic effects. - Knowing the velocity is essential for calculating how time will seem to differ between moving objects and stationary observers, which is the focus of the exercise.
Proper Time
Proper time refers to the time measured by an observer at rest with respect to the event being observed. In simpler terms, it's the "true" time experienced by something moving along with the observer. In our exercise, the proper time is the time as experienced by the astronaut inside the moving spacecraft.In relativity, this proper time is represented by the symbol \( T_0 \). It's the time interval experienced by an observer who is not experiencing any relative motion concerning the event being timed. For instance, if you're on the spacecraft and measuring your breathing rate, that rate experienced by you represents the proper time.**Key Takeaways**- Proper time is the time interval for an observer moving with the event they are timing.- It differs from the time measured by a stationary observer when the event is moving at significant speeds.- Understanding proper time is crucial for resolving problems involving time dilation, like how Earth observers perceive the astronaut’s breathing rate in our problem.

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Most popular questions from this chapter

As observed on earth, a certain type of bacterium is known to double in number every 24.0 hours. Two cultures of these bacteria are prepared, in number every 24.0 hours. Two cultures of these bacteria are prepared, each consisting initially of one bacterium. One culture is left on earth and the other placed on a rocket that travels at a speed of \(0.866 c\) relative to the earth. At a time when the earthbound culture has grown to 256 bacteria, how many bacteria are in the culture on the rocket, according to an earthbased observer?

An unstable high-energy particle is created in the laboratory, and it moves at a speed of \(0.990 \mathrm{c}\). Relative to a stationary reference frame fixed to the laboratory, the particle travels a distance of \(1.05 \times 10^{-3} \mathrm{m}\) before disintegrating. What are (a) the proper distance and (b) the distance measured by a hypothetical person traveling with the particle? Determine the particle's (c) proper lifetime and (d) its dilated lifetime.

A spacecraft approaching the earth launches an exploration vehicle. After the launch, an observer on earth sees the spacecraft approaching at a speed of \(0.50 c\) and the exploration vehicle approaching at a speed of \(0.70 \mathrm{c}\). What is the speed of the exploration vehicle relative to the spaceship?

Two kilograms of water are changed (a) from ice at \(0^{\circ} \mathrm{C}\) into liquid water at \(0^{\circ} \mathrm{C}\) and \((\mathbf{b})\) from liquid water at \(100^{\circ} \mathrm{C}\) into steam at \(100^{\circ} \mathrm{C}\). For each situation, determine the change in mass of the water.

An unstable particle is at rest and suddenly breaks up into two fragments. No external forces act on the particle or its fragments. One of the fragments has a velocity of \(+0.800 \mathrm{c}\) and a mass of \(1.67 \times 10^{-27} \mathrm{kg}\), and the other has a mass of \(5.01 \times 10^{-27} \mathrm{kg} .\) What is the velocity of the more massive fragment? (Hint: This problem is similar to Example 6 in Chapter \(7 .\) )

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