Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 35: Problem 73

Two stationary space stations are separated by a distance of \(100 .\) light- years, as measured by someone on one of the space stations. A spaceship traveling at \(0.950 c\) relative to the space stations passes by one of the space stations heading directly toward the other one. How long will it take to reach the other space station, as measured by someone on the spaceship? How much time will have passed for a traveler on the spaceship as it travels from one space station to the other, as measured by someone on one of the space stations? Round the answers to the nearest year.

Short Answer

Expert verified
Answer: It will take approximately 105 years for the spaceship to reach the other space station, as measured by someone on the spaceship. For a traveler on the spaceship, 337 years will have passed as measured by someone on one of the space stations during the journey.

Step by step solution

01

Identify the given values

We're given: - Distance between the space stations (D): 100 light-years - Speed of the spaceship (v): 0.950 c
02

Calculate the time it takes to reach the other space station, as measured by someone on the spaceship

We can use the formula for time as: \(time = \frac{distance}{speed}\) which can be rewritten as: \(t_{spaceship} = \frac{D}{v}\) Then, \(t_{spaceship} = \frac{100}{0.950 c}\) \(t_{spaceship} = \frac{100}{0.950} \,\text{years}\) \(t_{spaceship} \approx 105 \,\text{years}\)
03

Calculate time dilation factor

We can find the time dilation factor using the formula: \(\gamma = \frac{1}{\sqrt{1- \frac{v^2}{c^2}}}\) where \(c\) is the speed of light. \(\gamma = \frac{1}{\sqrt{1- \frac{(0.950c)^2}{c^2}}}\) \(\gamma = \frac{1}{\sqrt{1- 0.9025}}\) \(\gamma \approx 3.203\)
04

Calculate the time that passes for a traveler on the spaceship, as measured by someone on the space station

We can use the time dilation factor to find the time that passes for a traveler on the spaceship (\(t_{station}\)) as: \(t_{station} = \gamma \times t_{spaceship}\) \(t_{station} = 3.203 \times 105\,\text{years}\) \(t_{station} \approx 337\,\text{years}\) So, it will take approximately 105 years for the spaceship to reach the other space station, as measured by someone on the spaceship, and 337 years will have passed for a traveler on the spaceship as it travels from one space station to the other, as measured by someone on one of the space stations.

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

An astronaut on a spaceship traveling at a speed of \(0.50 c\) is holding a meter stick parallel to the direction of motion. a) What is the length of the meter stick as measured by another astronaut on the spaceship? b) If an observer on Earth could observe the meter stick, what would be the length of the meter stick as measured by that observer?

Which quantity is invariant-that is, has the same value-in all reference frames? a) time interval, \(\Delta t\) d) space-time interval, b) space interval, \(\Delta x\) \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\) c) velocity, \(v\)

In the age of interstellar travel, an expedition is mounted to an interesting star 2000.0 light-years from Earth. To make it possible to get volunteers for the expedition, the planners guarantee that the round trip to the star will take no more than \(10.000 \%\) of a normal human lifetime. (At that time the normal human lifetime is 400.00 years.) What is the minimum speed the ship carrying the expedition must travel?

Find the value of \(g\), the gravitational acceleration at Earth's surface, in light-years per year, to three significant figures.

Robert, standing at the rear end of a railroad car of length \(100 . \mathrm{m},\) shoots an arrow toward the front end of the car. He measures the velocity of the arrow as \(0.300 c\). Jenny, who was standing on the platform, saw all of this as the train passed her with a velocity of \(0.750 c .\) Determine the following as observed by Jenny: a) the length of the car b) the velocity of the arrow c) the time taken by arrow to cover the length of the car d) the distance covered by the arrow
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Astrophysics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Modern Physics

Read Explanation

Scientific Method Physics

Read Explanation

Translational Dynamics

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