/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 How fast does a 250 -m spaceship... [FREE SOLUTION] | 91Ó°ÊÓ

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How fast does a 250 -m spaceship move relative to an observer who measures the ship's length to be \(150 \mathrm{m} ?\)

Short Answer

Expert verified
The spaceship moves at \(1.8 \times 10^8\) m/s relative to the observer.

Step by step solution

01

Understand the Concept

This problem is related to the relativistic effects of length contraction as described by Einstein's theory of special relativity. When an object is moving at a significant fraction of the speed of light, its length along the direction of motion appears contracted to an outside observer.
02

Write Down the Formula for Length Contraction

The formula for length contraction is given by:\[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where:- \( L \) is the contracted length (150 m), - \( L_0 \) is the proper length (250 m),- \( v \) is the velocity,- \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s).
03

Rearrange the Formula to Solve for Velocity

To find the velocity \( v \), rearrange the length contraction formula:\[ \sqrt{1 - \frac{v^2}{c^2}} = \frac{L}{L_0} \]Square both sides to eliminate the square root:\[ 1 - \frac{v^2}{c^2} = \left( \frac{L}{L_0} \right)^2 \]
04

Substitute Given Values and Solve

Substitute \( L = 150 \) m and \( L_0 = 250 \) m into the equation:\[ \left( \frac{150}{250} \right)^2 = \frac{v^2}{c^2} \]Simplify:\[ \left( \frac{3}{5} \right)^2 = \frac{v^2}{c^2} \]Calculate the left side:\[ \frac{9}{25} = \frac{v^2}{c^2} \]Now solve for \( v^2 \):\[ v^2 = \frac{9}{25}c^2 \]
05

Calculate the Velocity

To find \( v \), take the square root of both sides:\[ v = c \sqrt{\frac{9}{25}} \]Since \( c = 3 \times 10^8 \) m/s, this becomes:\[ v = 3 \times 10^8 \times \frac{3}{5} \]Calculate the product:\[ v = 1.8 \times 10^8 \text{ m/s} \]
06

Review the Result

The velocity of the spaceship relative to the observer is calculated to be \( 1.8 \times 10^8 \) m/s, which is a significant fraction of the speed of light, as expected in relativistic problems.

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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 profound theory proposed by Albert Einstein that revolutionized our understanding of space and time. Unlike Newtonian mechanics, where time and space are considered absolute and unchanged regardless of the observer, special relativity introduces a relative framework, where the laws of physics remain consistent across different inertial frames of reference.

A central idea in special relativity is that the speed of light, denoted by \( c \) and approximately equal to \( 3 \times 10^8 \) m/s, is constant for all observers, no matter their motion. This constancy leads to fascinating effects such as time dilation, where time slows down for fast-moving objects, and length contraction, where objects appear shorter in the direction of motion when moving close to the speed of light.

Length contraction plays a pivotal role in our exercise scenario, where a spaceship's perceived length changes with respect to an observer depending on the spaceship's velocity. As the spaceship travels at significant speeds, its length contracts along its direction of travel, embodying the principle of special relativity.
Einstein's Theory
Einstein's theory of special relativity, published in 1905, fundamentally changes our perception of time and space. It rests on two main postulates:
  • The laws of physics are identical in all inertial frames of reference, meaning there is no preferred viewpoint in the universe for observing the laws of physics.
  • The speed of light in a vacuum is constant, uninfluenced by the motion of the light source or the observer.


These postulates bring forth intriguing consequences such as the relativity of simultaneity, the equivalence of mass and energy as stated in the famous equation \( E = mc^2 \), and the observed phenomena of length contraction in the exercise. When observing the spaceship, an observer sees a contracted length due to its high speed, translating Einstein's abstract theory into a tangible effect.

The length contraction can be quantified using the formula \( L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \), where \( L_0 \) is the proper length and \( L \) is the contracted length. This equation shows how Einstein's theoretical ideas manifest in practical observations, such as determining the speed of the spaceship relative to an observer.
Velocity Calculation
Understanding velocity calculation in the context of special relativity adds an intriguing layer of complexity compared to classical physics. In our exercise, the key is finding the velocity at which the spaceship travels relative to the observer to result in the observed contraction of its length.

Starting with the length contraction formula \( L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \), we rearrange the equation to solve for \( v \), the spaceship's velocity. By isolating \( v \), we initially have the expression \( \sqrt{1 - \frac{v^2}{c^2}} = \frac{L}{L_0} \). Rearranging gives \( v^2 = \left( \frac{L}{L_0} \right)^2 c^2 \), and finally, we find \( v \) by taking the square root:

\[ v = c \cdot \sqrt{\frac{9}{25}} \l = 3 \times 10^8 \times \frac{3}{5} = 1.8 \times 10^8 \text{ m/s} \]

This velocity represents a significant fraction of the speed of light, reflecting the relativity-driven effects on the spaceship's length observed from the reference frame of a stationary observer. This exercise demonstrates not only how velocity affects the perceived length of objects in motion but also how critical understanding special relativity is to solving such problems in modern physics.

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Most popular questions from this chapter

A spaceship and an asteroid are moving in the same direction away from Earth with speeds of \(0.77 \mathrm{c}\) and \(0.41 \mathrm{c}\), respectively. What is the relative speed between the spaceship and the asteroid?

As a spaceship flies past with speed \(v,\) you observe that \(1.0000 \mathrm{s}\) elapses on the ship's clock in the same time that 1.0000 min elapses on Earth. How fast is the ship traveling, relative to the Earth? (Express your answer as a fraction of the speed of light.)

Predict/Explain You are in a spaceship, traveling directly away from the Moon with a speed of \(0.9 \mathrm{c}\). A light signal is sent in your direction from the surface of the Moon. (a) As the signal passes your ship, do you measure its speed to be greater than, less than, or equal to \(0.1 c ?\) (b) Choose the best explanation from among the following: I. The speed you measure will be greater than \(0.1 c ;\) in fact, it will be \(c,\) since all observers in inertial frames measure the same speed of light. II. You will measure a speed less than \(0.1 \mathrm{c}\) because of time dilation, which causes clocks to run slow. III. When you measure the speed you will find it to be \(0.1 c\) which is the difference between \(c\) and \(0.9 \mathrm{c}.\)

A street performer tosses a ball straight up into the air (event 1 ) and then catches it in his mouth (event 2 ). For each of the following observers, state whether the time they measure between these two events is the proper time or the dilated time: (a) the street performer; (b) a stationary observer on the other side of the street; \((c)\) a person sitting at home watching the performance on \(T V ;\) (d) a person observing the performance from a moving car.

A space probe with a rest mass of \(8.2 \times 10^{7} \mathrm{kg}\) and a speed of \(0.50 \mathrm{c}\) smashes into an asteroid at rest and becomes embedded within it. If the speed of the probe-asteroid system is \(0.26 \mathrm{c}\) after the collision, what is the rest mass of the asteroid?

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