/*! 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 69 Travel to the stars requires hun... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Travel to the stars requires hundreds or thousands of years, even at the speed of light. Some people have suggested that we can get around this difficulty by accelerating the rocket (and its astronauts) to very high speeds so that they will age less due to time dilation. The fly in this ointment is that it takes a great deal of energy to do this. Suppose you want to go to the immense red giant Betelgeuse, which is about 500 light-years away. (A light-year is the distance that light travels in a year.) You plan to travel at constant speed in a \(1000-\mathrm{kg}\) rocket ship (a little over a ton), which, in reality, is far too small for this purpose. In each case that follows, calculate the time for the trip, as measured by people on earth and by astronauts in the rocket ship, the energy needed in joules, and the energy needed as a percentage of U.S. yearly use (which is 1.0 \(\times 10^{19} \mathrm{J} ) .\) For comparison, arrange your results in a table showing \(v_{\text { rocket }},\) \(t_{\text { earth }},\) \(\mathbf{t}_{\text { rouket }} \boldsymbol{E}\) \((\text { in } \mathrm{J}),\) and \(E\) (as \(\%\) of U.S. use). The rocket ship's speed is (a) \(0.50 \mathrm{c} ;\) (b) 0.99 \(\mathrm{c}\) (c) \(0.9999 \mathrm{c} .\) On the basis of your results, does it seem likely that any government will invest in such high-speed space travel any time soon?

Short Answer

Expert verified
Space travel to Betelgeuse at high speeds requires enormous energy, making it highly impractical with current technology.

Step by step solution

01

Understanding Time Dilation

To solve the problem, we need to understand the concept of time dilation in special relativity. Time dilation means that time passes at a different rate for observers moving at high speeds compared to those at rest. The formula to calculate the time experienced by astronauts (\(t_{rocket}\)) is:\[t_{rocket} = t_{earth} \times \sqrt{1 - \left(\frac{v}{c}\right)^2}\]where \(v\) is the speed of the rocket, \(c\) is the speed of light, and \(t_{earth}\) is the time measured by people on Earth, which is given by \(distance/speed\).
02

Calculating Earth and Rocket Time for Each Speed

For each speed given, calculate \(t_{earth}\) and \(t_{rocket}\). 1. \(v = 0.50c\): - \(t_{earth} = \frac{500}{0.50} = 1000\) years. - \(t_{rocket} = 1000 \times \sqrt{1 - 0.25} = 1000 \times 0.866 = 866\) years.2. \(v = 0.99c\):- \(t_{earth} = \frac{500}{0.99} = 505.05\) years.- \(t_{rocket} = 505.05 \times \sqrt{1 - 0.9801} = 505.05 \times 0.141 = 71.2\) years.3. \(v = 0.9999c\):- \(t_{earth} = \frac{500}{0.9999} = 500.05\) years.- \(t_{rocket} = 500.05 \times \sqrt{1 - 0.99980001} = 500.05 \times 0.0141 = 7.055\) years.
03

Calculating Energy Required

The energy required is calculated using the relativistic kinetic energy formula:\[E = \left(\frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} - 1\right) m c^2\]where \(m = 1000 \text{ kg}\), and \(c = 3 \times 10^8 \text{ m/s}\).1. \(v = 0.50c\): - \(E = (1.1547 - 1) \times 1000 \times (3 \times 10^8)^2 = 6.97 \times 10^{19}\) J.2. \(v = 0.99c\):- \(E = (7.0888 - 1) \times 1000 \times (3 \times 10^8)^2 = 4.38 \times 10^{20}\) J.3. \(v = 0.9999c\):- \(E = (70.71 - 1) \times 1000 \times (3 \times 10^8)^2 = 4.24 \times 10^{21}\) J.
04

Calculating Energy as a Percentage of U.S. Yearly Use

Calculate the energy needed as a percentage of the U.S. yearly use.1. For \(v = 0.50c\):- \(\frac{6.97 \times 10^{19}}{1.0 \times 10^{19}} = 697\%\)2. For \(v = 0.99c\):- \(\frac{4.38 \times 10^{20}}{1.0 \times 10^{19}} = 4380\%\)3. For \(v = 0.9999c\):- \(\frac{4.24 \times 10^{21}}{1.0 \times 10^{19}} = 42400\%\)
05

Tabulating Results

Create a table using the results from the previous steps:\[\begin{array}{|c|c|c|c|c|}\hlinev_{\text{rocket}} & t_{\text{earth}} (\text{years}) & t_{\text{rocket}} (\text{years}) & E (\text{J}) & E (\% \text{of U.S. use}) \hline0.50c & 1000 & 866 & 6.97 \times 10^{19} & 697\0.99c & 505.05 & 71.2 & 4.38 \times 10^{20} & 4380\0.9999c & 500.05 & 7.05 & 4.24 \times 10^{21} & 42400\\hline\end{array}\]

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.

Special Relativity
Special Relativity is a cornerstone of modern physics, introduced by Albert Einstein in 1905. One of its fascinating implications is time dilation. According to this theory, the faster an object moves, the more time "stretches" for it compared to a stationary observer. In our exercise, time dilation is crucial to understanding how astronauts can experience time differently during interstellar journeys. For example, as a spacecraft speeds towards a distant star like Betelgeuse, time for the astronauts on board slows down compared to people on Earth. This difference can be calculated using the formula: \[t_{\text{rocket}} = t_{\text{earth}} \times \sqrt{1 - \left(\frac{v}{c}\right)^2}\] Here, \(v\) is the spaceship's velocity, and \(c\) represents the speed of light. Despite appearing theoretical, these calculations show real effects like those experienced by astronauts on high-speed travels.
Energy Calculations in Physics
Energy calculations in physics often consider the conversion and conservation of energy under various conditions. In high-speed travel, such as attempting to reach near-light speeds, relativistic kinetic energy becomes significant. This is because traditional mechanics does not sufficiently describe energies at such velocities. The energy required to propel a spacecraft at high speeds is given by: \[E = \left(\frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} - 1\right) m c^2\] In this equation, \(m\) is the mass of the spaceship, \(v\) is its velocity, and \(c\) is the speed of light. By substituting the values for different speeds like \(0.50c, 0.99c, \text{and } 0.9999c\), we find that the energy required increases dramatically. The results show enormous energy demands, far surpassing current U.S. annual energy usage. This illustrates the immense challenge of achieving high-speed interstellar travel with our current capability.
Interstellar Travel
Interstellar travel refers to the journey between stars, often spanning light-years—distances light takes years to traverse. While such journeys have captured imaginations, implementing them presents vast challenges. Considerations include the speed required to reach destinations like Betelgeuse, 500 light-years away, within a human lifetime. Accelerating a spacecraft to speeds such as \(0.50c, 0.99c, \text{or } 0.9999c\) would reduce the perceived travel time for astronauts due to time dilation. However, the technological and energy requirements remain currently unattainable. Even if we assume a 1000 kg spaceship, the energy demands for each proposed speed are staggering—which, as calculated, exceed U.S.'s yearly energy use by thousands of percent. Thus, while intriguing, current interstellar travel plans face significant practical obstacles.

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

In certain radioactive beta decay processes, the beta particle (an electron) leaves the atomic nucleus with a speed of 99.95\(\%\) the speed of light relative to the decaying nucleus. If this nucleus is moving at 75.00\(\%\) the speed of light, find the speed of the emitted electron relative to the laboratory reference frame if the electron is emitted (a) in the same direction that the nucleus is moving and (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts \((a)\) and \((b)\) , find the kinetic energy of the electron as measured in (i) the laboratory frame and (ii) the reference frame of the decaying nucleus.

A photon with energy \(E\) is emitted by an atom with mass \(m\) which recoils in the opposite direction. (a) Assuming that the motion of the atom can be treated nonrelativistically, compute the recoil speed of the atom. (b) From the result of part (a), show that the recoil speed is much less than \(c\) whenever \(E\) is much less than the rest energy \(m c^{2}\) of the atom.

Two particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is 0.650 \(\mathrm{c}\) , and the speed of each partcle relative to the other is 0.950 \(\mathrm{c}\) . What is the speed of the second particle, as measured in the laboratory?

(a) How much work must be done on a particle with mass \(m\) to accelerate it (a) from rest to a speed of 0.090\(c\) and (b) from a speed of 0.900\(c\) to a speed of 0.990\(c ?\) (Express the answers in terms of \(m c^{2}-)(c)\) How do your answers in parts \((a)\) and \((b)\) compare?

An observer in frame \(S^{\prime}\) is moving to the right \((+x-\text { direction })\) at speed \(u=0.600 \mathrm{c}\) away from a stationary observer in frame \(S\) . The observer in \(S^{\prime}\) measures the speed \(v^{\prime}\) of a particle moving to the right away from her. What speed \(v\) does the observer in S measure for the particle if \((a) v^{\prime}=0.400 c ;(b) v^{\prime}=0.900 c\) (c) \(v^{\prime}=0.990 c ?\)

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.