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A spaceship is traveling at two-thirds of the speed of light directly toward a stationary asteroid. If the spaceship turns on it headlights, what will be the speed of the light traveling from the spaceship to the asteroid as observed by a) someone on the spaceship? b) someone on the asteroid?

Short Answer

Expert verified
Answer: The observer on the spaceship and the observer on the asteroid both observe the speed of light to be the same universal constant, which is \(c = 3.0 \times 10^8\,m/s\).

Step by step solution

01

a) Speed of light observed by someone on the spaceship

Since the speed of light is a fundamental constant and is always \(3.0 \times 10^8\,m/s\) in any inertial frame, the observer on the spaceship will see the light beam traveling at the speed of light (\(c\)). Speed of light for spaceship observer = \(c = 3.0 \times 10^8\,m/s\).
02

b) Speed of light observed by someone on the asteroid

To find the speed of light observed by someone on the asteroid, we'll first find the velocity of the light according to the spaceship (i.e., the source frame of reference). Velocity of the spaceship, \(v = \frac{2}{3}c = \frac{2}{3}(3.0 \times 10^8\,m/s) = 2.0 \times 10^8\,m/s\). Next, we use the relativistic velocity addition formula to determine the velocity of light observed by the person on the asteroid, which is: \(u' = \frac{u+v}{1+\frac{uv}{c^2}}\), where \(u' = \) velocity of light observed by asteroid observer, \(u = c\) (speed of light), and \(v = 2.0 \times 10^8\,m/s\) (velocity of the spaceship). Now we plug in the values: \(u' = \frac{3.0 \times 10^8 + 2.0 \times 10^8}{1+\frac{(3.0 \times 10^8)(2.0 \times 10^8)}{(3.0 \times 10^8)^2}}\) Simplifying the equation: \(u' = \frac{5.0 \times 10^8}{1+\frac{2}{3}} = \frac{5.0 \times 10^8}{\frac{5}{3}}\) Thus, the speed of light observed by someone on the asteroid is also equal to the speed of light (\(c\)): \(u' = 3.0 \times 10^8\,m/s\).

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Key Concepts

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

Speed of Light
The speed of light is an essential concept in physics and a cornerstone of Einstein's theory of special relativity. It holds a constant value of approximately 299,792,458 meters per second, usually denoted as \(c\). One of the remarkable aspects of light's speed is that it remains the same across all inertial frames of reference, irrespective of the relative motion between the observer and the source of light.
This means that whether you're on a moving spaceship or standing on an asteroid, the speed of light will always measure \(c\). This crucial aspect is what distinguishes light from other speeds like those of sound or everyday objects.
It is important to remember that this constancy of light's speed leads to many non-intuitive consequences, such as time dilation and length contraction, which are central to the theory of special relativity.
Relativistic Velocity Addition
Relativistic velocity addition is a formula used in special relativity to calculate the resultant velocity of two objects moving relative to each other. Unlike classical velocity addition, where speeds are simply added, relativistic addition accounts for the effects of traveling close to the speed of light.
The formula is:
  • \( u' = \frac{u+v}{1+\frac{uv}{c^2}} \)
This formula ensures that no matter how fast two objects move in relation to one another, their combined speed never exceeds the speed of light, \(c\).
In the context of the problem, when we use the relativistic velocity addition, it ensures that even though it appears a moving spaceship and its light should have a combined speed greater than \(c\), the observed speed from the asteroid—thanks to relativistic corrections—remains \(3.0 \times 10^8 \) m/s, preserving the universal speed limit of light.
Inertial Frame
An inertial frame of reference is a hypothetical construct where a body not acted upon by forces moves at a constant velocity, which could be zero (rest) or a uniform motion. In simple terms, it's a perspective or viewpoint where the laws of physics take their simplest form.
The concept of inertial frame is essential for understanding observations in relativity. Each observer or object that is either at rest or moving at a constant speed without any acceleration can be considered to be in an inertial frame.
  • For example, the spaceship in the problem is moving at a constant velocity of two-thirds the speed of light. Therefore, it is in an inertial frame.
  • The stationary asteroid is also one, as it is not accelerating.
The constancy of the speed of light is observed similarly regardless of the inertial frame of the observer. This means that whether the observer is on the spaceship or on the asteroid, the speed of light they measure remains the same.

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Most popular questions from this chapter

Consider two clocks carried by observers in a reference frame moving at speed \(v\) in the positive \(x\) -direction relative to ours. Assume that the two reference frames have parallel axes, and that their origins coincide when clocks at that point in both frames read zero. Suppose the clocks are separated by distance \(l\) in the \(x^{\prime}-\) direction in their own reference frame; for instance, \(x^{\prime}=0\) for one clock and \(x^{\prime}=I\) for the other, with \(y^{\prime}=z^{\prime}=0\) for both. Determine the readings \(t^{\prime}\) on both clocks as functions of the time coordinate \(t\) in our reference frame.

Suppose NASA discovers a planet just like Earth orbiting a star just like the Sun. This planet is 35 light-years away from our Solar System. NASA quickly plans to send astronauts to this planet, but with the condition that the astronauts would not age more than 25 years during this journey. a) At what speed must the spaceship travel, in Earth's reference frame, so that the astronauts age 25 years during this journey? b) According to the astronauts, what will be the distance of their trip?

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At rest, a rocket has an overall length of \(L .\) A garage at rest (built for the rocket by the lowest bidder) is only \(L / 2\) in length. Luckily, the garage has both a front door and a back door, so that when the rocket flies at a speed of \(v=0.866 c\), the rocket fits entirely into the garage. However, according to the rocket pilot, the rocket has length \(L\) and the garage has length \(L / 4\). How does the rocket pilot observe that the rocket does not fit into the garage?

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