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Chapter 37: Problem 50

The distance to a particular star, as measured in the earth's frame of reference, is 7.11 light-years ( 1 light-year is the distance that light travels in \(1 \mathrm{y}\) ). A spaceship leaves the earth and takes \(3.35 \mathrm{y}\) to arrive at the star, as measured by passengers on the ship. (a) How long does the trip take, according to observers on earth? (b) What distance for the trip do passengers on the spacecraft measure?

Short Answer

Expert verified
The trip takes approximately 4.82 years as observed from earth and the passengers on the spaceship measure a distance of approximately 3.35 light-years for the trip.

Step by step solution

01

Calculate Earth Time

The spaceship's time-travel is \(3.35 \, \mathrm{y}\) and the distance in light years is \(7.11 \, \mathrm{light-years}\). This indicates that the ship traveled at a speed of \(\frac{7.11}{3.35} = 2.12 \, \text{light-years per year}\) as measured from the ship. Since the speed of light is 1 light-year per year, we can say that the velocity \(v\) is \(2.12c\). Now, to find the time as observed on Earth, we'll need to make use of the time dilation principle from the theory of relativity, which states that time observed in a moving frame of reference (ship's frame) is shorter than the time observed in a stationary frame (Earth's frame). The formula for time dilation is \(t=\frac{t_0}{\sqrt{1-\left(\frac{v}{c}\right)^2}}\). Substitute \(t_0 = 3.35 \, \mathrm{y}\) and \(v = 2.12c\) into this formula to get the time observed on earth.
02

Calculate Spacecraft Distance

For finding the distance observed by passengers, it's known that the length observed in a moving frame (length in the spaceship) is shorter than the length observed in a stationary frame (length observed on earth), a result of length contraction in relativity. The formula for this phenomenon is \(L = L_0 \sqrt{1-\left(\frac{v}{c}\right)^2}\). \(L_0 = 7.11 \, \mathrm{light-years}\) is the distance observed on earth. Substituting \(v = 2.12c\) into this formula gives the distance observed in the spaceship.

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Key Concepts

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

Time Dilation
Time dilation is a fascinating concept from Einstein鈥檚 theory of special relativity. It explains how time can appear to pass at different rates in different frames of reference.
For example, if you are traveling at a very high speed in a spaceship, time for you (as observed from the ship) appears to move slower compared to an observer on Earth.
The formula for time dilation is \( t = \frac{t_0}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \), where:
  • \( t \) is the time observed in the stationary frame (Earth).
  • \( t_0 \) is the time observed in the moving frame (spaceship).
  • \( v \) is the velocity of the moving frame.
  • \( c \) is the speed of light.
When the speed \( v \) approaches that of light \( c \), time dilation becomes more pronounced.
Length Contraction
Length contraction is another intriguing result from special relativity, revealing that objects traveling at high speeds appear shorter in the direction of motion to a stationary observer. This effect is a direct consequence of the same relativistic principles that cause time dilation.
The formula to calculate the contracted length is \( L = L_0 \sqrt{1 - \left(\frac{v}{c}\right)^2} \), where:
  • \( L \) is the length measured in the moving frame (spaceship).
  • \( L_0 \) is the original length measured in the stationary frame (Earth).
  • \( v \) is the velocity of the moving object.
  • \( c \) is the speed of light.
As the velocity \( v \) approaches the speed of light \( c \), the length \( L \) becomes significantly shorter.
Speed of Light
The speed of light, denoted as \( c \), is a fundamental constant in the universe and is approximately \( 299,792,458 \) meters per second, or \( 1 \) light-year per year when considering astronomical distances. According to the theory of relativity, nothing can travel faster than the speed of light.
This principle shapes the very framework of how time and space interact.
When objects move close to the speed of light, unusual effects like time dilation and length contraction become significant. The significance of the speed of light is its role in nearly all modern physics equations related to relativity.
Frame of Reference
A frame of reference in physics is a set of coordinates or a viewpoint used to measure or observe physical phenomena. A crucial aspect of relativity is understanding that the observations can differ depending on the observer's frame of reference.
For instance, an observer on Earth perceives time and distance differently compared to an observer traveling in a high-speed spaceship.
In the realm of special relativity:
  • The stationary frame of reference often makes measurements from an "at rest" perspective, like Earth's frame.
  • The moving frame of reference measures from within the moving object, such as a spaceship.
Every scenario is relative to the speed and direction of the observer's frame, making relativity the universal rule for such high-speed interactions.

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Most popular questions from this chapter

A proton (rest mass \(1.67 \times 10^{-27} \mathrm{~kg}\) ) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the proton's speed?

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