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A spaceship of proper length \(300 \mathrm{~m}\) takes \(0.75 \mu \mathrm{s}\) to pass an Earth observer. Determine the speed of this spaceship as measured by the Earth observer.

Short Answer

Expert verified
After substituting the given values into the final equation and solving, we find that the speed of the spaceship as measured by the Earth observer is approximately \(2.25 \times 10^8 m/s\).

Step by step solution

01

Understand the problem

The proper length \(L_0\) of the spaceship is given as 300 m. The time \(t\) for the ship to pass the observer is 0.75 microseconds (\(0.75 \times 10^{-6}\) seconds). We need to find the velocity (\(v\)) of the spaceship relative to the observer on Earth.
02

Apply the Length Contraction Formula

In special relativity, the length \(L\) of an object moving with velocity \(v\) relative to an observer is given by \(L = L_0/\gamma\), where \(L_0\) is the proper length and \(\gamma = 1/\sqrt{1-(v/c)^2}\) is the Lorentz factor. Here, the spaceship passes the observer, meaning the observer sees the length \(L\) of the spaceship passing in time \(t\). We can write this as \(L = v \times t\). Rearranging and substituting for \(L\) from the length contraction formula gives us \(v = L_0/(\gamma t)\).
03

Solve the equation for velocity

As we want to find the velocity of the spaceship, we can substitute for \(\gamma\) in the formula for \(v\). This gives us \(v = L_0 t /\sqrt{1+(L_0 t/c)^2}\). We know that the speed of light \(c\) is \(3 \times 10^8 m/s\), so we can substitute for \(L_0\), \(t\), and \(c\) to find the velocity \(v\).

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Key Concepts

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

Length Contraction
Length contraction is a fundamental concept in Einstein's theory of relativity. It describes how an object appears to be shorter when it moves at high speeds relative to an observer. This effect is only noticeable at speeds close to the speed of light. For example, a spaceship traveling close to the speed of light will look shorter to an observer on Earth than it does to passengers inside the spaceship.

Mathematically, length contraction is described by the formula:
  • \[ L = \frac{L_0}{\gamma} \]
Where:
  • \( L \) is the length observed by an outsider,
  • \( L_0 \) is the proper length (the length of the object in its rest frame), and
  • \( \gamma \) is the Lorentz factor.
This means that in the spaceship problem from the exercise, the spaceship's proper length is 300 meters. But, for an observer on Earth, it appears contracted. Understanding this concept is crucial because it helps us explain real-world phenomena like the size changes and time dilation experienced in high-speed travel.
Proper Length
The proper length is the length of an object measured by an observer who is at rest relative to the object. This is the object's actual length when it is not moving relative to the observer. In the context of our exercise, the proper length of the spaceship is given as 300 meters. This is the length that passengers within the spaceship or any observers moving with it would measure.

Proper length is an important parameter in relativistic physics as it forms the baseline measurement for calculating length contraction. When using the length contraction formula, the proper length is what we're starting with to find the contracted length viewed by an observer in motion relative to the object.

The proper length (\( L_0 \)) is the longest length an observer can measure since length appears shorter during motion relative to another observer when viewed from a different frame of reference. This essential concept allows us to understand why objects never exceed the speed of light as their length contracts as they speed up.
Lorentz Factor
The Lorentz factor, denoted by \( \gamma \), is central to the ideas of special relativity. It quantifies how much time, length, and relativistic mass scale with velocity, especially as objects approach the speed of light.

Here's how it's defined:
  • \[ \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \]
Where:
  • \( v \) is the velocity of the object, and
  • \( c \) is the speed of light in a vacuum (\( 3 \times 10^8 \, \text{meters/second} \)).
As the object's speed nears the speed of light, \( \gamma \) increases dramatically, which affects how time and length appear to an observer. This factor is responsible for both time dilation and length contraction.

In our exercise, the Lorentz factor helps us determine how much the spaceship shrinks from the perspective of the Earth observer. As the speed approaches the speed of light, \( \gamma \) becomes large, which results in significant length contraction and differing observations of time and lengths between different frames of reference. The Lorentz factor is what enables the equations of special relativity, making it a cornerstone in understanding high-speed physics.

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Most popular questions from this chapter

An interstellar space probe is launched from Earth. After a brief period of acceleration, it moves with a constant velocity, \(70.0 \%\) of the speed of light. Its nuclear-powered batteries supply the energy to keep its data transmitter active continuously. The batteries have a lifetime of \(15.0\) years as measured in a rest frame. (a) How long do the batteries on the space probe last as measured by mission control on Earth? (b) How far is the probe from Earth when its batteries fail as measured by mission control? (c) How far is the probe from Earth as measured by its built-in trip odometer when its batteries fail? (d) For what total time after launch are data received from the probe by mission control? Note that radio waves travel at the speed of light and fill the space between the probe and Earth at the time the battery fails.

An astronaut wishes to visit the Andromeda galaxy, making a one-way trip that will take \(30.0\) years in the spaceship's frame of reference. Assume the galaxy is \(2.00\) million light-years away and his speed is constant. (a) How fast must he travel relative to Earth? (b) What will be the kinetic energy of his spacecraft, which has mass of \(1.00 \times\) \(10^{6} \mathrm{~kg} ?\) (c) What is the cost of this energy if it is purchased at a typical consumer price for electric energy, \(13.0\) cents per kWh? The following approximation will prove useful: $$ \frac{1}{\sqrt{1+x}}=1-\frac{x}{2} \quad \text { for } x<1 $$

A meterstick moving at \(0.900 c\) relative to Earth approaches an observer at rest with respect to Earth. (a) What is the meterstick's length, according to the observer? (b) Qualitatively, how would the answer change if the observer were moving relative to Earth in the opposite direction?

An atomic clock moves at \(1000 \mathrm{~km} / \mathrm{h}\) for \(1.00 \mathrm{~h}\) as measured by an observer using an identical clock on Earth. How many nanoseconds slow will the moving clock be at the end of the \(1.00-\mathrm{h}\) interval? Hint: The following approximation is helpful: \(\sqrt{1-x} \approx(1-x / 2)\) for \(x<1\).

A cube of steel has a volume of \(1.00 \mathrm{~cm}^{3}\) and a mass of \(8.00 \mathrm{~g}\) when at rest on Earth. If this cube is now given a speed \(v=0.900 c\), what is its density as measured by a stationary observer? Note that relativistic density is \(E_{R} / c^{2} V\).

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