Chapter 26, Problem 43AP

An astronaut wishes to visit the Andromeda galaxy, making a one-way trip that will take 30.0 years in the space-ship’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 cents per kWh? The following approximation will prove useful:

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

To determine

The speed of the observer relative to Earth.

Answer to Problem 43AP

The speed of the observer relative to Earth is $1-1.13\times {10}^{-10}$.

Explanation of Solution

Given Info:

The distance is $2.00\times {10}^6\text{ly}$.

The time taken to travel is $30.0\text{yr}$.

Formula to calculate the time is,

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

• $\Delta t_p$ is the time measured by the observer in spacecraft to cover the distance
• $\Delta t$ is time measured by the observer at Earth
• $c$ is speed of light
• $v$ is the speed of spacecraft

Formula to calculate the speed is,

$\begin{array}{c}v=\frac{d}{\Delta t}\\ =\frac{d}{\frac{\Delta t_p}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}}\end{array}$

• $d$ is the distance covered

Substitute $2.00\times {10}^6\text{ly}$ for $d$ and $30.0\text{yr}$ for $\Delta t$ to find $v$.

$\begin{array}{c}v=\frac{d}{\Delta t}\\ =\frac{\left(2.00\times {10}^6\text{ly}\right)\left(\frac{1c}{\text{ly}}\right)}{\frac{30.0\text{yr}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}}\\ \Rightarrow \\ \left(1.50\times {10}^{-5}\right)\left(\frac{v}{c}\right)=\sqrt{1-{\left(\frac{v}{c}\right)}^2}\\ \Rightarrow \\ \frac{v}{c}\approx 1-\frac{2.25\times {10}^{-10}}{2}\\ =1-1.13\times {10}^{-10}\end{array}$

Thus, the speed of the observer relative to Earth is $1-1.13\times {10}^{-10}$.

Conclusion:

The speed of the observer relative to Earth is $1-1.13\times {10}^{-10}$.

To determine

The kinetic energy of the spacecraft.

Answer to Problem 43AP

The kinetic energy of the spacecraft is $5.99\times {10}^{27}\text{J}$.

Explanation of Solution

Given Info:

The mass of spacecraft is $1.00\times {10}^6\text{kg}$.

The speed of light is $3.00\times {10}^8\text{m/s}$.

Formula to kinetic energy is,

$\begin{array}{c}K.E=\left(\gamma -1\right)m{c}^2\\ =\left(\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}-1\right)m{c}^2\end{array}$

• $m$ is the mass of spacecraft
• $K.E$ is the kinetic energy

Substitute $1-1.13\times {10}^{-10}$ for $v/c$, $1.00\times {10}^6\text{kg}$ for $m$ and $3.00\times {10}^8\text{m/s}$ for $c$ to find $K.E$.

$\begin{array}{c}K.E=\left(\frac{1}{\sqrt{1-{\left(1-1.13\times {10}^{-10}\right)}^2}}-1\right)\left(1.00\times {10}^6\text{kg}\right){\left(3.00\times {10}^8\text{m/s}\right)}^2\\ =5.99\times {10}^{27}\text{J}\end{array}$

Thus, the kinetic energy of the spacecraft is $5.99\times {10}^{27}\text{J}$.

Conclusion:

The kinetic energy of the spacecraft is $5.99\times {10}^{27}\text{J}$.

To determine

The cost of energy.

Answer to Problem 43AP

The cost of energy is $\$2.16\times {10}^{20}$.

Explanation of Solution

Given Info:

The rate per kilowatt is $\$0.13/\text{kWh}$.

Formula to calculate cost of energy is,

$C=R\left(K.E\right)$

• $C$ is the cost of energy
• $R$ is the rate per kilowatt

Substitute $5.99\times {10}^{27}\text{J}$ for $K.E$ and $\$0.13/\text{kWh}$ for $R$ to find $C$.

$\begin{array}{c}C=\left(\$0.13/\text{kWh}\right)\left(5.99\times {10}^{27}\text{J}\right)\left(\frac{1\text{kWh}}{3.60\times {10}^6\text{J}}\right)\\ =\$2.16\times {10}^{20}\end{array}$

Thus, the cost of energy is $\$2.16\times {10}^{20}$.

Conclusion:

The cost of energy is $\$2.16\times {10}^{20}$