Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 33: Problem 46

An advanced civilization has developed a spaceship that goes, with respect to the galaxy, only \(50 \mathrm{km} / \mathrm{s}\) slower than light. (a) According to the ship's crew, how long does it take to cross the galaxy's 100,000 -ly diameter? (b) What's the galactic diameter measured in the ship's reference frame?

Short Answer

Expert verified
Part (a) The crew finds it takes them about \(2.82843 * 10^{15}\) seconds or about 90,000 years to cross the galaxy. (b) In terms of the ship's reference frame, the diameter of the galaxy is about 35,355.34 light years.

Step by step solution

01

Determine Relative Speed

First, calculate the ship's speed with respect to light's speed. According to the problem, it goes \(50 \mathrm{km/s}\) slower than light. Given that the speed of light is approximately \(3*10^5 \mathrm{km/s}\), calculate the relative speed (v) using the equation \(v = c - v_{ship}\), where \(v_{ship}\) is \(50 \mathrm{km/s}\).
02

Calculate Lorentz factor

Second, compute the Lorentz factor, which is a function of the speed of the spaceship relative to the speed of light. Use the equation \(\gamma = 1/\sqrt{1-(v/c)^2}\).
03

Compute Time for Crew

Using the Lorentz factor, calculate the time dilation experienced by the ship's crew. The time experienced by the crew (\(t_{crew}\)) can be computed using the equation \(t_{crew} = \gamma * t_{galaxy}\), where \(t_{galaxy}\) is the time it would take to cross the galaxy in the galaxy's frame (given as 100,000 light years in the problem). A light-year is the distance light travels in one year, so the speed of light can be used to convert this time into seconds.
04

Compute Galactic Diameter for Crew

Now, calculate the length contraction. From the crew's reference frame, the galaxy's diameter (\(D_{crew}\)) can be calculated using the equation \(D_{crew} = D_{galaxy}/\gamma\), where \(D_{galaxy}\) is the given diameter of the galaxy (100,000 light years).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Find the kinetic energy of an electron moving at (a) \(0.0010 c\) (b) \(0.60 c,\) and \((\mathrm{c}) 0.99 c .\) Use suitable approximations where possible.

What's special about the special theory of relativity?

A spaceship travels at \(0.80 c\) from Earth to a star 10 light years distant, as measured in the Earth-star reference frame. Let event A be the ship's departure from Earth and event B its arrival at the star. (a) Find the distance and time between the two events in the Earth-star frame. (b) Repeat for the ship's frame. (Hint: The distance in the ship frame is the distance an observer has to move with respect to that frame to be at both events-not the same as the Lorentz-contracted distance between Earth and star.) (c) Compute the square of the spacetime interval in both frames to show explicitly that it's invariant.

How fast would you have to move relative to a meter stick for it to be 99 cm long in your reference frame?

You've been named captain of NASA's first interstellar mission since the Voyager robotic spacecraft. You board your spaceship, accelerate quickly to \(0.8 c,\) and cruise at constant speed toward Proxima Centauri, the closest star to our Sun. Proxima Centauri is 4 light-years distant as measured in the two stars' common rest frame. On the way, you conduct various medical experiments to determine the effects of a long space voyage on the human body. Taking your pulse, you find a. it's significantly slower than when you're on Earth. b. it's the same as when you're on Earth. c. it's significantly faster than when you're on Earth.
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Nuclear Physics

Read Explanation

Force

Read Explanation

Atoms and Radioactivity

Read Explanation

Electromagnetism

Read Explanation

Modern Physics

Read Explanation
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.

Sign-up for free