Physics for Scientists and Engineers, Volume 1, Chapters 1-22
Physics for Scientists and Engineers, Volume 1, Chapters 1-22
8th Edition
ISBN: 9781439048382
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 39, Problem 39.71AP

An astronaut wishes to visit the Andromeda galaxy, making a one-way trip that will take 30.0 years in the spaceship’s frame of reference. Assume the galaxy is 2.00 million light-years away and his speed is constant. (a) How fast must he travel relative to Earth? (b) What will be the kinetic energy of his spacecraft, which has mass of 1.00 × 106 kg? (c) What is the cost of this energy if it is purchased at a typical consumer price for electric energy, 13.0¢ per kWh? The following approximation will prove useful:

1 1 + x = 1 x 2 for x < < 1

(a)

Expert Solution
Check Mark
To determine
The speed that an astronaut travel relative to Earth.

Answer to Problem 39.71AP

The speed that an astronaut travel relative to Earth is 11.12×1010 .

Explanation of Solution

Given info: The time for one way trip is 30.0years and distance of the galaxy is 2.00millionlightyears .

Formula to calculate the speed that an astronaut travel relative to Earth is,

v=dΔt

Here,

v is the speed that an astronaut travel relative to Earth.

d is the distance of the galaxy.

Δt is the relativistic time.

Formula to calculate the relativistic time for trip in frame of Earth reference is,

Δt=Δtp1(vc)2

Here,

Δtp is the time for one way trip.

c is the speed of light.

Substitute Δtp1(vc)2 for Δtp in above equation.

v=dΔtp1(vc)2

Substitute 30.0y for Δtp and 2.00millionlightyears for d .

v=2.00millionlightyears×106cy1millionlightyears30.0y1(vc)2vc=66.67×103(1(vc)2)(2.25×1010)(vc)2=(1(vc)2)(vc)2(1+2.25×1010)=1

Simplify the equation further,

vc=11+2.25×1010

Use the following approximation

11+x1x2

The ratio vc can be rewrite as

` vc=11+xvc1x2 (1)

Substitute 2.25×1010 for x .

vc12.25×10102vc11.12×1010

Conclusion:

Therefore, the speed that an astronaut travel relative to Earth is 11.12×1010 .

(b)

Expert Solution
Check Mark
To determine
The kinetic energy of the spacecraft.

Answer to Problem 39.71AP

The kinetic energy of the spacecraft is 6.00×1027J .

Explanation of Solution

Given info: The time for one way trip is 30.0years , distance of the galaxy is 2.00millionlightyears and mass of spacecraft is 1.00×106kg .

Formula to calculate the kinetic energy of the spacecraft is,

K=(γ1)mc2 (2)

Here,

γ is the relativistic factor.

Write the expression for the relativistic factor.

γ=11(vc)2 (3)

From equation (1), the ratio of vc can be given as,

vc1x2vc11+x1(vc)2111+x1(vc)2x1+x

Substitute x1+x for 1(vc)2 in equation (3).

γ1x1+x1+1x

Substitute 1+1x for γ in equation (2).

K=(1+1x1)mc2

Substitute 2.25×1010 for x , 1.00×106kg for m and 3×108m/s for c to find K .

K(1+12.25×10101)(1.00×106kg)(3×108m/s)26.00×1027J

Conclusion:

Therefore, the kinetic energy of the spacecraft is 6.00×1027J .

(c)

Expert Solution
Check Mark
To determine
The cost of the energy consume by the spacecraft.

Answer to Problem 39.71AP

The cost of the energy consume by the spacecraft is $2.17×1020 .

Explanation of Solution

Given info: The time for one way trip is 30.0years , distance of the galaxy is 2.00millionlightyears and mass of spacecraft is 1.00×106kg .

Write the expression for the total cost of the energy consume by the spacecraft.

cost=Energyconsume×consumerpricerate

Substitute 6.00×1027J for Energyconsume and 13.0perKWh for consumerpricerate .

cost=6.00×1027J×13.0×$1100perKWh×1KWh3.60×106J=$2.17×1020

Conclusion:

Therefore, the cost of the energy consume by the spacecraft is $2.17×1020 .

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Chapter 39 Solutions

Physics for Scientists and Engineers, Volume 1, Chapters 1-22

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