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:

$\frac{1}{\sqrt{1+x}}=1-\frac{x}{2}\text{for}x<<1$

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 $1-1.12\times {10}^{-10}$.

Explanation of Solution

Given info: The time for one way trip is $30.0\text{years}$ and distance of the galaxy is $2.00\text{million}\text{light}\text{years}$.

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

$v=\frac{d}{\Delta t}$

Here,

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

$d$ is the distance of the galaxy.

$\Delta t$ is the relativistic time.

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

$\Delta t=\frac{\Delta t_{\text{p}}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$

Here,

$\Delta t_{\text{p}}$ is the time for one way trip.

$c$ is the speed of light.

Substitute $\frac{\Delta t_{\text{p}}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$ for $\Delta t_{\text{p}}$ in above equation.

$v=\frac{d}{\frac{\Delta t_{\text{p}}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}}$

Substitute $30.0\text{y}$ for $\Delta t_{\text{p}}$ and $2.00\text{million}\text{light}\text{years}$ for $d$.

$\begin{array}{l}v=\frac{2.00\text{million}\text{light}\text{years}\times \frac{{10}^{6}c\cdot \text{y}}{1\text{million}\text{light}\text{years}}}{\frac{30.0\text{y}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}}\\ \frac{v}{c}=66.67\times {10}^3\left(\sqrt{1-{\left(\frac{v}{c}\right)}^2}\right)\\ \left(2.25\times {10}^{-10}\right){\left(\frac{v}{c}\right)}^2=\left(1-{\left(\frac{v}{c}\right)}^2\right)\\ {\left(\frac{v}{c}\right)}^2\left(1+2.25\times {10}^{-10}\right)=1\end{array}$

Simplify the equation further,

$\frac{v}{c}=\frac{1}{\sqrt{1+2.25\times {10}^{-10}}}$

Use the following approximation

$\frac{1}{\sqrt{1+x}}\approx 1-\frac{x}{2}$

The ratio $\frac{v}{c}$ can be rewrite as

` $\begin{array}{l}\frac{v}{c}=\frac{1}{\sqrt{1+x}}\\ \frac{v}{c}\approx 1-\frac{x}{2}\end{array}$ (1)

Substitute $2.25\times {10}^{-10}$ for $x$.

$\begin{array}{l}\frac{v}{c}\approx 1-\frac{2.25\times {10}^{-10}}{2}\\ \frac{v}{c}\approx 1-1.12\times {10}^{-10}\end{array}$

Conclusion:

Therefore, the speed that an astronaut travel relative to Earth is $1-1.12\times {10}^{-10}$.

To determine

The kinetic energy of the spacecraft.

Answer to Problem 39.71AP

The kinetic energy of the spacecraft is $6.00\times {10}^{27}\text{J}$.

Explanation of Solution

Given info: The time for one way trip is $30.0\text{years}$, distance of the galaxy is $2.00\text{million}\text{light}\text{years}$ and mass of spacecraft is $1.00\times {10}^6\text{kg}$.

Formula to calculate the kinetic energy of the spacecraft is,

$K=\left(\gamma -1\right)m{c}^2$ (2)

Here,

$\gamma$ is the relativistic factor.

Write the expression for the relativistic factor.

$\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$ (3)

From equation (1), the ratio of $\frac{v}{c}$ can be given as,

$\begin{array}{c}\frac{v}{c}\approx 1-\frac{x}{2}\\ \frac{v}{c}\approx \frac{1}{\sqrt{1+x}}\\ 1-{\left(\frac{v}{c}\right)}^2\approx 1-\frac{1}{1+x}\\ 1-{\left(\frac{v}{c}\right)}^2\approx \frac{x}{1+x}\end{array}$

Substitute $\frac{x}{1+x}$ for $1-{\left(\frac{v}{c}\right)}^2$ in equation (3).

$\begin{array}{c}\gamma \approx \frac{1}{\sqrt{\frac{x}{1+x}}}\\ \approx \sqrt{1+\frac{1}{x}}\end{array}$

Substitute $\sqrt{1+\frac{1}{x}}$ for $\gamma$ in equation (2).

$K=\left(\sqrt{1+\frac{1}{x}}-1\right)m{c}^2$

Substitute $2.25\times {10}^{-10}$ for $x$, $1.00\times {10}^6\text{kg}$ for $m$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ to find $K$.

$\begin{array}{c}K\approx \left(\sqrt{1+\frac{1}{2.25\times {10}^{-10}}}-1\right)\left(1.00\times {10}^6\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ \approx 6.00\times {10}^{27}\text{J}\end{array}$

Conclusion:

Therefore, the kinetic energy of the spacecraft is $6.00\times {10}^{27}\text{J}$.

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 $\$2.17\times {10}^{20}$.

Explanation of Solution

Given info: The time for one way trip is $30.0\text{years}$, distance of the galaxy is $2.00\text{million}\text{light}\text{years}$ and mass of spacecraft is $1.00\times {10}^6\text{kg}$.

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

$\text{cost}=\text{Energy}\text{consume}\times \text{consumer}\text{price}\text{rate}$

Substitute $6.00\times {10}^{27}\text{J}$ for $\text{Energy}\text{consume}$ and $13.0\not\subset \text{per}\text{KWh}$ for $\text{consumer}\text{price}\text{rate}$.

$\begin{array}{c}\text{cost}=6.00\times {10}^{27}\text{J}\times 13.0\not\subset \times \frac{\$1}{100\not\subset }\text{per}\text{KWh}\times \frac{1\text{KWh}}{3.60\times {10}^6\text{J}}\\ =\$2.17\times {10}^{20}\end{array}$

Conclusion:

Therefore, the cost of the energy consume by the spacecraft is $\$2.17\times {10}^{20}$.