/*! 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 34 You’re standing on an asteroid... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You’re standing on an asteroid when you see your best friend rocketing by in her new spaceship. As she goes by, you notice that the front and rear of her ship coincide exactly with the 400-m-diameter of another nearby asteroid that is stationary with respect to you. However, you happen to know that your friend’s spaceship measured 500 m long in the showroom. What is your friend’s speed relative to you?

Short Answer

Expert verified
The friend's relative speed is approximately 180,000 km/s.

Step by step solution

01

Identify the given information

We know the proper length \( L_0 \) of the spaceship is 500 m, the contracted length \( L \) is 400 m, and the speed of light \( c \) is \( 3*10^8 \) m/s.
02

Set up the length contraction formula

The length contraction formula is \(L = L_0\sqrt{1 - v^2/c^2}\). Replace \( L \), \( L_0 \) and \( c \) with their given values. We get \( 400 = 500\sqrt{1 - v^2/(3*10^8)^2} \).
03

Solve for \( v \)

Solving for \( v \) involves some algebraic manipulations. Divide both sides by 500, then square both sides to get rid of the square root, subtract 1 from both sides, then multiply by \( c^2 \) and finally take the square root.
04

Calculation

Here are the calculations. After dividing by 500 and squaring we get \( 0.64 = 1 - v^2/(3*10^8)^2 \). After subtracting 1 from both sides, multiplying by \( c^2 \) and then taking square root we get \( v = sqrt(0.36 * (3*10^8)^2) \).
05

Final Solution

After doing the final calculation, we get \( v \) approximately to be \( 0.6 * 3*10^8 \) m/s or 180,000 km/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Length Contraction
Length contraction is a fascinating concept from the theory of relativity by Albert Einstein. It tells us that objects moving at high speeds appear shorter along their direction of motion, from the viewpoint of a stationary observer. Imagine your friend zooming past you in a spaceship. To you, outside and stationary, the spaceship looks shorter than it actually is.
This phenomenon is more noticeable at speeds close to the speed of light. The formula governing this is:
  • \(L = L_0\sqrt{1 - v^2/c^2}\)
  • where \(L\) is the contracted length, \(L_0\) is the proper length, and \(v\) is the velocity of the moving object.

The spaceship in your exercise is measured at 400 meters as it passes. However, its true uncontracted length, when stationary, is 500 meters. This discrepancy embodies length contraction in action. Understanding this helps you know how speed affects the observation of moving objects.
Proper Length
Proper length, symbolized as \(L_0\), is the actual length of an object measured in the frame of reference where the object is at rest. In simpler terms, it's the length measurement in the object's own rest frame.
In the given problem, your friend's spaceship has a proper length of 500 meters. This is the length that would be measured if the spaceship were stationary in front of you or if you were moving along with it.
Think of proper length as the 'true' or 'original' length of an object, free from the influence of high speeds. In everyday speeds, the concept doesn't come into play significantly. But at relativistic velocities, such as near the speed of light, the difference between proper and observed length becomes pronounced. This reinforces the idea that space measurements can change depending on motion and reference frames.
Speed of Light
The speed of light, denoted as \(c\), is a universal constant and critical to the theory of relativity. It is approximately \(3 \times 10^8\) meters per second. Interestingly, it is not just the speed at which light travels, but also sets the ultimate speed limit in the universe.
In relativity, this constant is essential in formulas like length contraction and time dilation. Using \(c\) ensures that such equations account for high-speed travel influences on space and time.
  • The speed of light creates a barrier where classical intuition about speed and distance changes.
It explains why, in the problem you're solving, length contraction occurs at specific velocities: those significant enough compared to the speed of light. By solving for your friend's speed, you take into account this profound limit, reflecting the constant speed of light's role in weaving the fabric of space and time.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A baseball pitcher can throw a ball with a speed of 40 m/s. He is in the back of a pickup truck that is driving away from you. He throws the ball in your direction, and it floats toward you at a lazy 10 m/s. What is the speed of the truck?

An astronaut in a spacesuit on a spacewalk is a completely sealed system, with no matter lost or gained. In principle, the energy the astronaut uses will result in a decrease in mass. Suppose on a gentle spacewalk his metabolic power plus the power of the suit is 250 W. What is the change in mass, in kg, on a 1 hour spacewalk

A very fast-moving train car passes you, moving to the right at 0.50c. You measure its length to be 12 m. Your friend David flies past you to the right at a speed relative to you of 0.80c. How long does David measure the train car to be?

A rocket traveling at 0.500c sets out for the nearest star, Alpha Centauri, which is 4.25 ly away from earth. It will return to earth immediately after reaching Alpha Centauri. What distance will the rocket travel and how long will the journey last according to (a) stay-at-home earthlings and (b) the rocket crew?

An astronaut travels to a star system 4.5 ly away at a speed of 0.90c. Assume that the time needed to accelerate and decelerate is negligible. a. How long does the journey take according to Mission Control on earth? b. How long does the journey take according to the astronaut? c. How much time elapses between the launch and the arrival of the first radio message from the astronaut saying that she has arrived?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.