Ch 36: Special Relativity

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 36: Special Relativity

Problem 52b

Chapter 36, Problem 52b

The quantity dE/dv, the rate of increase of energy with speed, is the amount of additional energy a moving object needs per 1 m/s increase in speed. A 25,000 kg rocket is traveling at 0.90c. How much additional energy is needed to increase its speed by 1 m/s?

Verified step by step guidance

Step 1: Recognize that the problem involves relativistic energy, as the rocket is traveling at a speed close to the speed of light (0.90c). The total relativistic energy of an object is given by the formula:

E_{total} = γ mc^2

, where

γ = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

is the Lorentz factor,

m

is the mass of the rocket,

v

is its velocity, and

c

is the speed of light.

Step 2: To find the rate of increase of energy with speed, calculate the derivative of the total energy with respect to velocity,

\frac{dE}{dv}

. Start by differentiating the relativistic energy formula:

\frac{d}{dv} E_{total} = \frac{d}{dv} (γ mc^2)

. Use the chain rule to differentiate

γ

with respect to

v

Step 3: The derivative of the Lorentz factor

γ

is:

\frac{d}{dv} γ = \frac{v^{2}}{c^{2}} * \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

Step 4: Substitute the given values into the derivative expression. The mass of the rocket is

25000

kg, the initial velocity is

0.90 c

, and the speed of light is

c = 3.00 \times 10^8

m/s. Plug these values into the derivative formula to compute

\frac{dE}{dv}

Step 5: Once

\frac{dE}{dv}

is calculated, multiply it by the change in velocity, which is

1

m/s, to find the additional energy required to increase the rocket's speed by 1 m/s. This will give the final result.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Relativistic Energy

In the context of special relativity, the energy of an object increases as its speed approaches the speed of light (c). The relativistic energy is given by the equation E = γmc², where γ (gamma) is the Lorentz factor, which accounts for the effects of relativity. As an object's speed increases, its mass effectively increases, requiring more energy to continue accelerating.

Lorentz Factor

The Lorentz factor, denoted as γ (gamma), is a crucial component in relativistic physics that describes how much time, length, and relativistic mass increase as an object approaches the speed of light. It is calculated using the formula γ = 1 / √(1 - v²/c²), where v is the object's velocity. This factor becomes significant at high speeds, affecting the energy required for further acceleration.

Kinetic Energy in Relativity

In classical mechanics, kinetic energy is given by KE = 1/2 mv². However, in relativistic physics, the kinetic energy must be adjusted to account for the increase in mass and energy as speed approaches c. The relativistic kinetic energy is expressed as KE = (γ - 1)mc², highlighting that the energy required to increase speed becomes significantly larger as the object's velocity nears the speed of light.

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Related Practice

Textbook Question

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Textbook Question

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The Stanford Linear Accelerator (SLAC) accelerates electrons to v = 0.99999997c in a 3.2-km-long tube. If they travel the length of the tube at full speed (they don’t, because they are accelerating), how long is the tube in the electrons’ reference frame?

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