Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 35: Problem 31

Suppose NASA discovers a planet just like Earth orbiting a star just like the Sun. This planet is 35 light-years away from our Solar System. NASA quickly plans to send astronauts to this planet, but with the condition that the astronauts would not age more than 25 years during this journey. a) At what speed must the spaceship travel, in Earth's reference frame, so that the astronauts age 25 years during this journey? b) According to the astronauts, what will be the distance of their trip?

Short Answer

Expert verified
The spaceship's speed is approximately 2 * sqrt(6) * 10^8 m/s / 7 in the Earth's reference frame. b) What is the distance of the trip according to the astronauts? The distance of the trip according to the astronauts is approximately (35 light-years) * (5/7).

Step by step solution

01

a) Finding the spaceship's speed

We will use the time dilation formula for this step. The time dilation formula can be given as: Dilated time = Proper time / sqrt(1 - v^2/c^2) where - Dilated time is the time experienced in the Earth's frame (35 years), - Proper time is the time experienced by the astronauts (25 years), - v is the spaceship's speed we need to determine, - c is the speed of light (approx. 3.0 x 10^8 m/s). Let's rearrange the equation to solve for v: v = c * sqrt(1 - (Proper time / Dilated time)^2) Substituting the given values and calculating the speed, we get: v = (3.0 x 10^8 m/s) * sqrt(1 - (25 years / 35 years)^2)
02

a) Calculating the spaceship's speed

Calculate the spaceship's speed: v ≈ (3.0 x 10^8 m/s) * sqrt(1 - (5/7)^2) v ≈ (3.0 x 10^8 m/s) * sqrt(1 - 25/49) v ≈ (3.0 x 10^8 m/s) * sqrt(24/49) v ≈ (3.0 x 10^8 m/s) * (2 * sqrt(6) / 7) v ≈ 2 * sqrt(6) * 10^8 m/s / 7 So, the spaceship must travel at approximately 2 * sqrt(6) * 10^8 m/s / 7 in the Earth's reference frame.
03

b) Finding the distance according to the astronauts

We will use the length contraction formula for this step. The formula can be given as: Contracted length = Proper length * sqrt(1 - v^2/c^2) where - Contracted length is the distance according to the astronauts, - Proper length is the distance measured in the Earth's frame (35 light-years), - v is the calculated spaceship's speed, and - c is the speed of light. Let's rearrange the equation to insert the values: Contracted length = (35 light-years) * sqrt(1 - (2 * sqrt(6) * 10^8 m/s / 7)^2 / (3.0 x 10^8 m/s)^2)
04

b) Calculating the distance according to the astronauts

Calculate the contracted length: Contracted length ≈ (35 light-years) * sqrt(1 - (24/49)) Contracted length ≈ (35 light-years) * sqrt(25/49) Contracted length ≈ (35 light-years) * (5/7) So, according to the astronauts, the distance of their trip would be approximately (35 light-years) * (5/7).

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

Michelson and Morley used an interferometer to show that the speed of light is constant, regardless of Earth's motion through any perceived luminiferous aether. An analogy can be understood from the different times it takes for a rowboat to travel two different round-trip paths in a river that flows at a constant velocity \((u)\) downstream. Let one path be for a distance \(D\) directly across the river, then back again; and let the other path be the same distance \(D\) directly upstream, then back again. Assume that the rowboat travels at constant speed, \(v\) (with respect to the water), for both trips. Neglect the time it takes for the rowboat to turn around. Find the ratio of the cross-stream time divided by the upstream-downstream time, as a function of the given constants.

A particle of rest mass \(m_{0}\) travels at a speed \(v=0.20 c\) How fast must the particle travel in order for its momentum to increase to twice its original momentum? a) \(0.40 c\) c) \(0.38 c\) e) \(0.99 c\) b) \(0.10 c\) d) \(0.42 c\)

Sam sees two events as simultaneous: (i) Event \(A\) occurs at the point (0,0,0) at the instant 0: 00: 00 universal time; (ii) Event \(B\) occurs at the point \((500, \mathrm{~m}, 0,0)\) at the same moment. Tim, moving past Sam with a velocity of \(0.999 c \hat{x}\), also observes the two events. a) Which event occurred first in Tim's reference frame? b) How long after the first event does the second event happen in Tim's reference frame?

In the twin paradox example, Alice boards a spaceship that flies to a space station 3.25 light-years away and then returns with a speed of \(0.65 c\). a) Calculate the total distance Alice traveled during the trip, as measured by Alice. b) With the aforementioned total distance, calculate the total time duration for the trip, as measured by Alice.

At what speed will the length of a meter stick look \(90.0 \mathrm{~cm} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Astrophysics

Read Explanation

Force

Read Explanation

Engineering Physics

Read Explanation

Medical Physics

Read Explanation

Fundamentals of 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