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Chapter 37: Problem 51

The starships of the Solar Federation are marked with the symbol of the federation, a circle, while starships of the Denebian Empire are marked with the empire's symbol, an ellipse whose major axis is 1.40 times longer than its minor axis \((a=1.40 b\) in Fig. P37.51). How fast, relative to an observer, does an empire ship have to travel for its marking to be confused with the marking of a federation ship?

Short Answer

Expert verified
The Denebian ship must travel at approximately 70% the speed of light for its marking to be confused with a circle.

Step by step solution

01

Understand the Problem

We are given two different symbols for two entities: a circle for the Solar Federation and an ellipse (with its major axis 1.40 times its minor axis) for the Denebian Empire. We need to find the speed at which the Denebian ship must travel such that its marking appears circular, making it indistinguishable from the federation's symbol.
02

Recall the Concept of Length Contraction

According to the theory of relativity, lengths contract along the direction of motion. The contracted length can be determined by the formula: \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where \( L \) is the contracted length, \( L_0 \) is the proper length (length at rest), \( v \) is the velocity of the object, and \( c \) is the speed of light.
03

Establish Relations for Ellipse Dimensions

The ellipse is defined by its semi-major axis \( a = 1.40b \) and semi-minor axis \( b \). For the ellipse to look like a circle when contracted, its major axis \( a' \) must equal \( b \). That is, the contracted major axis must be equal to its semi-minor axis: \[ a' = a \sqrt{1 - \frac{v^2}{c^2}} = b \]
04

Set Up the Equation to Solve for Speed

Using the relationship from our length contraction concept, substitute the modified length of the ellipse's major axis:\[ 1.40b \sqrt{1 - \frac{v^2}{c^2}} = b \]Divide both sides by \( b \):\[ 1.40 \sqrt{1 - \frac{v^2}{c^2}} = 1 \]
05

Isolate the Contraction Factor

Solve for the length contraction factor by isolating the square root:\[ \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{1.40} \]
06

Square Both Sides to Eliminate the Square Root

Square both sides to solve for \( v^2 \):\[ 1 - \frac{v^2}{c^2} = \left( \frac{1}{1.40} \right)^2 \] \[ \frac{v^2}{c^2} = 1 - \left( \frac{1}{1.40} \right)^2 \]
07

Solve for Velocity

Calculate \( \left( \frac{1}{1.40} \right)^2 \) and solve for \( v \):\[ \left( \frac{1}{1.40} \right)^2 = \left( \frac{10}{14} \right)^2 = \frac{100}{196} \approx 0.5102 \]Substitute:\[ \frac{v^2}{c^2} = 1 - 0.5102 = 0.4898 \]Then:\[ v^2 = 0.4898 c^2 \]\[ v = c \sqrt{0.4898} \approx 0.7c \]
08

Interpret the Solution

The Denebian ship has to travel at approximately 70% the speed of light for its elliptical symbol to appear circular to the observer.

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Key Concepts

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

Relativity Theory
Relativity theory, particularly Einstein鈥檚 theory of special relativity, introduces the concept of length contraction. This theory suggests that objects physically contract along the direction of motion as they approach the speed of light.
This contraction is only noticeable at velocities approaching that of light. The fundamental formula describing length contraction is:
  • \(L = L_0 \sqrt{1 - \frac{v^2}{c^2}}\)
where:
  • \(L\) is the contracted length (observed length).
  • \(L_0\) is the proper length (length of the object at rest).
  • \(v\) is the velocity of the moving object.
  • \(c\) is the speed of light.
This mathematical expression implies that the faster an object moves, the shorter it appears to an observer in the line of motion. The contracted form of the object can create fascinating optical illusions, such as transforming an ellipse into what appears to be a circle. Understanding these principles provides a key insight into how velocities close to the speed of light affect the perception of objects in motion.
Ellipse Geometry
The ellipse is a fundamental shape in geometry, characterized by two axes: the semi-major axis and the semi-minor axis. In the given scenario, the Denebian Empire's symbol is shaped as an ellipse with its major axis being 1.40 times longer than its minor axis. An ellipse can be defined mathematically as:
  • The equation in a standard form: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
  • Here, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.
An important feature of ellipse geometry crucial to understanding this exercise is how one can manipulate its perceived geometry through relative motion. When viewed at high speed, under the influence of length contraction, its major axis shortens. This shortening brings the major and minor axes closer, changing the ellipse's perceived shape significantly. Thus, at a specific speed, the ellipse's marking on the Denebian ship would visually match the circular marking of the Solar Federation.
Starship Identification
Starship identification within this exercise involves distinguishing between two factions by the geometric shape of their symbols. The Solar Federation uses a circle, while the Denebian Empire employs an ellipse鈥攁 longer shape with the major axis being 1.40 times the minor axis. The challenge lies in the situation where the Denebian starship moves at such a high velocity that its elliptical marking appears circular due to length contraction. At around 70% the speed of light, the contracted major axis length of the ellipse matches the minor axis length, making the two shapes indistinguishable to an observer. In simpler terms, high-speed motion masks the true identifying symbol, making it possible for the Denebian Empire鈥檚 starship to mimic, visually, the symbols of the Solar Federation. This scenario highlights how sophisticated concepts like relativity can impact something seemingly straightforward like starship identification in advanced space adventures.

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Most popular questions from this chapter

A spaceship flies past Mars with a speed of 0.985\(c\) relative to the surface of the planet. When the spaceship is directly overhead, a signal light on the Martian surface blinks on and then off. An observer on Mars measures that the signal light was on for 75.0\(\mu\) s. (a) Does the observer on Mars or the pilot on the spaceship measure the proper time? (b) What is the duration of the light pulse measured by the pilot of the spaceship?

A rocket ship flies past the earth at 85.0\(\%\) of the speed of light. Inside, an astronaut who is undergoing a physical examination is having his height measured while he is lying down parallel to the direction the rocket ship is moving. (a) If his height is measured to be 2.00 \(\mathrm{m}\) by his doctor inside the ship, what height would a person watching this from earth measure for his height? (b) If the earth-based person had measured \(2.00 \mathrm{m},\) what would the doctor in the spaceship have measured for the astronaut's height? Is this a reasonable height? (c) Suppose the astronaut in part (a) gets upafter the examination and stands with his body perpendicular to the direction of motion. What would the doctor in the rocket and the observer on earth measure for his height now?

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot fies past you in her spaceracer at a constant speed of 0.800\(c\) relative to you. At the instant the spaceracer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled \(1.20 \times 10^{8} \mathrm{m}\) past you, what does the race pilot read on her timer? (b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her? (c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

Electromagnetic radiation from a star is observed with an earth-based telescope. The star is moving away from the earth with a speed of 0.600\(c .\) If the radiation has a frequency of \(8.64 \times 10^{14} \mathrm{Hz}\) in the rest frame of the star, what is the frequency measured by an observer on earth?

A spaceship moving at constant speed \(u\) relative to us broadcasts a radio signal at constant frequency \(f_{0 .}\) As the spaceship approaches us, we receive a higher frequency \(f ;\) after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_{0}\) , and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_{0} ?\) (Hint: In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals 1\(/ T .\) Use the time dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_{0}=345 \mathrm{MHz}\) as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758\(c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f-f_{0} ?(\mathrm{c})\) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?
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