Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

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

Key Concepts

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

Length Contraction
In special relativity, length contraction is a fascinating concept that describes how the length of an object traveling at high speeds appears shorter when viewed from a stationary reference frame. Imagine observing a spaceship traveling at near-light speed from Earth. The distance the spaceship crosses seems much shorter to the astronauts inside compared to those observing from Earth.

The formula for length contraction is:
  • Contracted length = Proper length \( \times \sqrt{1 - \frac{v^2}{c^2}} \)
Here:
  • Contracted length: The distance measured by the astronauts, appearing shorter due to their high-speed travel.
  • Proper length: The actual distance measured without any contraction; in our case, it's the 35 light-years measured from Earth.
  • \(v\): The speed of the spaceship.
  • \(c\): The speed of light.
The contraction occurs because, as an object moves closer to the speed of light, the effects of special relativity become more pronounced. This discrepancy in distance perception doesn't affect the passengers' sense of time on board, allowing them to experience a different duration of a journey in cosmic terms.
Speed of Light
The speed of light, denoted as \(c\), is one of the most critical aspects of physics, representing the fastest speed at which information or matter can travel through space. It is roughly \(3.0 \times 10^8\) meters per second. Most fascinating about the speed of light is how it remains constant, no matter the observer's motion or the light source's speed, a principle essential to the theory of special relativity.

A fundamental principle here is that nothing can travel faster than light. This rule underpins the calculations involving time dilation and length contraction, as it places a universal limit on speed. For instance, when calculating the required speed of NASA's spaceship, we use the time dilation formula, which is intertwined with the speed of light:
  • \(v = c \times \sqrt{1 - \left(\frac{\text{Proper time}}{\text{Dilated time}}\right)^2}\)
In NASA's hypothetical journey, the spaceship's speed calculation emphasizes that the required travel speed approaches the speed of light. As the spaceship gets closer to \(c\), relativistic effects, like time dilation and length contraction, become more evident. This immense speed challenge highlights why interstellar travel is still beyond reachable for current technology.
Special Relativity
Special relativity, introduced by Albert Einstein in 1905, is a revolutionary theory that transformed our understanding of space, time, and motion. It rests on two key postulates:
  • The laws of physics are the same in all inertial frames of reference.
  • The speed of light in a vacuum is the same for all observers, regardless of the motion of light sources or observers.
This theory results in some non-intuitive phenomena such as time dilation and length contraction, impacting our perception of time and space at high velocities.

Time dilation means that clocks moving at high speed will run slower to a stationary observer. In our NASA mission scenario, if the astronauts are supposed to age only 25 years while covering a 35 light-year trip, the spacecraft must travel at a velocity close to the speed of light to account for the time difference.

Special relativity also explains how objects moving at these speeds could appear contracted in the direction of motion, a concept known as length contraction. Although tricky to visualize, these phenomena are crucial for modern technology and scientific understanding, affecting global positioning systems, particle accelerators, and advanced physics research.

Einstein's insights continue to challenge and inspire, revealing the universe's surprising and counterintuitive nature.

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 HeNe laser onboard a spaceship moving toward a remote space station emits a beam of red light toward the space station. The wavelength of the beam, as measured by a wavelength meter on board the spaceship, is \(632.8 \mathrm{nm}\). If the astronauts on the space station see the beam as a blue beam of light with a measured wavelength of \(514.5 \mathrm{nm},\) what is the relative speed of the spaceship with respect to the space station? What is the shift parameter \(z\) in this case?

Although it deals with inertial reference frames, the special theory of relativity describes accelerating objects without difficulty. Of course, uniform acceleration no longer means \(d v / d t=g,\) where \(g\) is a constant, since that would have \(v\) exceeding \(c\) in a finite time. Rather, it means that the acceleration experienced by the moving body is constant: In each increment of the body's own proper time \(d \tau,\) the body acquires velocity increment \(d v=g d \tau\) as measured in the inertial frame in which the body is momentarily at rest. (As it accelerates, the body encounters a sequence of such frames, each moving with respect to the others.) Given this interpretation: a) Write a differential equation for the velocity \(v\) of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the "ground frame"). You can simplify your equation, remembering that squares and higher powers of differentials can be neglected. b) Solve this equation for \(v(t),\) where both \(v\) and \(t\) are measured in the ground frame. c) Verify that your solution behaves appropriately for small and large values of \(t\). d) Calculate the position of the body \(x(t),\) as measured in the ground frame. For convenience, assume that the body is at rest at ground-frame time \(t=0,\) at ground-frame position \(x=c^{2} / g\) e) Identify the trajectory of the body on a space-time diagram (Minkowski diagram, for Hermann Minkowski) with coordinates \(x\) and \(c t,\) as measured in the ground frame. f) For \(g=9.81 \mathrm{~m} / \mathrm{s}^{2},\) calculate how much time it takes the body to accelerate from rest to \(70.7 \%\) of \(c,\) measured in the ground frame, and how much ground-frame distance the body covers in this time.

An electron is accelerated from rest through a potential of \(1.0 \cdot 10^{6} \mathrm{~V}\). What is its final speed?

Suppose you are explaining the theory of relativity to a friend, and you have told him that nothing can go faster than \(300,000 \mathrm{~km} / \mathrm{s}\). He says that is obviously false: Suppose a spaceship traveling past you at \(200,000 \mathrm{~km} / \mathrm{s}\), which is perfectly possible according to what you are saying, fires a torpedo straight ahead whose speed is \(200,000 \mathrm{~km} / \mathrm{s}\) relative to the spaceship, which is also perfectly possible; then, he says, the torpedo's speed is \(400,000 \mathrm{~km} / \mathrm{s}\). How would you answer him?

Consider a particle moving with a speed less than \(0.5 c\). If the speed of the particle is doubled, by what factor will the momentum increase? a) less than 2 b) equal to 2 c) greater than 2

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.

Sign-up for free