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Chapter 14: Problem 77

A spaceship at rest in a certain reference frame \(S\) is given a speed increment of \(0.50 \mathrm{c}\). Relative to its new rest frame, it is then given a further \(0.50 \mathrm{c}\) increment. This process is continued until its speed with respect to its original frame \(S\) exceeds \(0.999 \mathrm{c}\). How many increments does this process require?

Short Answer

Expert verified
Five increments are required for the spaceship's speed to exceed \(0.999c\).

Step by step solution

01

Understand the Problem

We need to find out how many speed increments of \(0.50\, c\) are required until the spaceship's speed in the original reference frame exceeds \(0.999\, c\). Importantly, we must account for relativistic velocity addition since speeds are high relative to the speed of light \(c\).
02

Use Relativistic Velocity Addition Formula

In the relativistic context, when a body at velocity \(u\) is given a velocity increment \(v\), the new velocity \(u'\) is given by the formula:\[u = \frac{u + v}{1 + \frac{uv}{c^2}}\]Here, \(v = 0.50\, c\) is the increment, and initially \(u = 0\).
03

Calculate First Speed Increment

After the first increment:\[u_1 = \frac{0 + 0.50c}{1 + 0} = 0.50c\]The speed in the frame \(S\) after the first increment is \(0.50c\).
04

Compute Second Increment

For the second increment, using the velocity from Step 3:\[u_2 = \frac{0.50c + 0.50c}{1 + (0.50 \, c)(0.50 \, c)/c^2} = \frac{1.00c}{1.25} = 0.80c\]
05

Calculate Third Increment

For the third increment:\[u_3 = \frac{0.80c + 0.50c}{1 + (0.80 \, c)(0.50 \, c)/c^2} = \frac{1.30c}{1.40} = 0.9286c\]
06

Determine Fourth Increment

For the fourth increment:\[u_4 = \frac{0.9286c + 0.50c}{1 + (0.9286 \, c)(0.50 \, c)/c^2} = \frac{1.4286c}{1.4643} \approx 0.9756c\]
07

Calculate Fifth Increment

Compute for the fifth increment:\[u_5 = \frac{0.9756c + 0.50c}{1 + (0.9756 \, c)(0.50 \, c)/c^2} = \frac{1.4756c}{1.4878} \approx 0.9932c\]Now the speed has exceeded \(0.999c\).

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Key Concepts

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

Reference Frame
A reference frame is an essential concept in physics, especially in the study of motion. It serves as the backdrop against which the position and velocity of objects are measured. Imagine you're observing a spaceship from a stationary point on Earth; your observation point constitutes your reference frame.

  • In our exercise, reference frame "S" is the initial frame where the spaceship is at rest initially.
  • Each time an increment is applied, the speed of the spaceship is measured from this original frame "S".
This frame helps us understand changes in velocities due to subsequent increments, providing a baseline for computation. The concept of reference frames becomes crucial when dealing with high velocities, where relativistic effects come into play.

Why does this matter? Because velocities aren't absolute; they depend on the observer's frame. Understanding this helps grasp how the spaceship's speed surpasses a certain threshold relative to its starting point.
Speed of Light
The speed of light, denoted as "c", is one of the constants of nature, approximately equal to 299,792,458 meters per second. In physics, especially in relativistic contexts, it鈥檚 considered the ultimate speed limit for any object or signal.

  • In our context, the speed of the spaceship increments towards this limit.
  • As speeds approach "c", Newtonian physics can no longer be applied, necessitating the use of Einstein's theory of relativity.
This theory introduces changes in how velocities add, described by the relativistic velocity addition formula, which accounts for the effects as objects move closer to the speed of light.

Importantly, the speed of light being a constant governs how velocities combine. This is why even with continuous increments, the ship never quite reaches "c" but can approach it closely, such as exceeding 0.999c in our problem.
Velocity Increment
Velocity increment is a key factor in our exercise. It refers to the additional speed given to a moving object. In traditional, everyday scenarios, increments simply sum up 鈥� but not when close to the speed of light.

  • In this exercise, each increment is a significant portion of "c", specifically 0.50c.
  • The relativistic velocity addition formula is key to calculating the new velocity after each increment.
This formula takes into account the bounded nature of "c", allowing velocities to be added but never exceeding this natural limit. Each successive increment becomes less impactful in terms of increasing velocity when measured in the original reference frame, illustrating the principle of diminishing returns in relativistic speeds.

Understanding velocity increment through the lens of relativity guides us to the solution of how many increments it takes for the spaceship to exceed 0.999c, drawing from step-by-step increases and applying the relativistic formula iteratively.

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Most popular questions from this chapter

How much work must be done to increasc the speed of an electron (a) from \(0.18 c\) to \(0.19 \mathrm{c}\) and (b) from \(0.98 \mathrm{c}\) to \(0.99 \mathrm{c}\) ? Note that the speed increase is \(0.01 \mathrm{c}\) in both cascs.

The energy released in the explosion of \(1.00 \mathrm{~mol}\) of \(\mathrm{TNT}\) is \(3.40 \mathrm{MJ}\). The molar mass of TNT is \(0.227 \mathrm{~kg} / \mathrm {mol}\). What weight of TNT is needed for an explosive release of \(1.80 \times 10^{14} \mathrm{~J} ?\) (b) Can you carry that weight in a backpack, or is a truck or train required? (c) Suppose that in an cxplosion of a fission bomb, \(0.080 \%\) of the fissionable mass is converted to released energy. What weight of fissionable material is needed for an explosive release of \(1.80 \times 10^{14} \mathrm{~J} ?\) (d) Can you carry that weight in a backpack, or is a truck or train required?

When you cough, you expel air at high speed through the trachca and upper bronchi so that the air will remove cxcess mucus lining the pathway. You produce the high speed by this procedure: You brcathe in a large amount of air, trap it by closing the glottis (the narrow opening in the larynx ), increase the air pressure by contracting the lungs, partially collapse the trachca and upper bronchi to narrow the pathway, and then expel the air through the pathway by suddenly reopening the glottis. Assume that during the expulsion the volume flow rate is \(7.0 \times 10^{-7} \mathrm{~m}^{3} / \mathrm{s}\). What multiple of \(343 \mathrm{~m} / \mathrm{s}\) (the specd of sound \(v_{s}\) ) is the airspeed through the trachea if the trachea diameter (a) remains its normal value of \(14 \mathrm{~mm}\) and (b) contracts to \(5.2 \mathrm{~mm}\) ?

Space cruisers \(A\) and \(B\) are moving parallel to the positive direction of an \(x\) axis. Cruiser \(A\) is faster, with a relative speed of \(v=0.900 c,\) and has a proper length of \(L=200 \mathrm{~m}\). According to the pilot of \(A\), at the instant \((t=0)\) the tails of the cruisers are aligned, the noses are also. According to the pilot of \(B,\) how much later are the noses aligned?

The mass of an electron is \(9.10938188 \times 10^{31} \mathrm{~kg} .\) To eight significant figures, find the following for the given electron kinetic energy: (a) \(\gamma\) and (b) \(\beta\) for \(K=1.0000000 \mathrm {keV},\) (c) \(\gamma\) and \((\mathrm{d}) \beta\) for \(K=\) \(1.0000000 \mathrm{MeV},\) and then \((\mathrm{e}) \gamma\) and \((\mathrm{f}) \beta\) for \(K=1.0000000 \mathrm{GeV}\)
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