Chapter 2, Problem 15P

<a>

To determine

Relativistic factor for speed $95\text{ km / h}$.

Answer to Problem 15P

The relativistic factor for speed $95\text{ km / h}$. is $1+3.86\times {10}^{-15}$.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ =\frac{1}{\sqrt{1-{\beta }^2}}\end{array}$

Because $\beta =\frac{v}{c}$, where, v is the speed of the object and c is the speed of the light.

Convert, speed kilometer per hour to meter per second. Thus

$\begin{array}{c}v=95\text{ km / h}\\ =95\text{ km / h}\times \left(\frac{1000\text{m}}{1.0\text{km}}\right)\left(\frac{1.0\text{h}}{3600\text{s}}\right)\\ =\frac{950}{36}\text{ m / s}\\ =26.38\text{ m / s}\end{array}$

Here, $v<<c$, thus $\beta =\frac{v}{c}$ will be very small, therefore, for smaller value of $\beta$ use binomial expansion for equation $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$. Therefore,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-{\beta }^2}}\\ \gamma =\frac{1}{{\left(1-{\beta }^2\right)}^{\frac{1}{2}}}\\ \gamma ={\left(1-{\beta }^2\right)}^{-\frac{1}{2}}\\ \gamma =1-\left(-\frac{1}{2}\right)\times {\beta }^2\end{array}$

Further simplify the above equation. Thus,

$\begin{array}{l}\gamma =1-\left(-\frac{1}{2}\right)\times {\beta }^2\\ \gamma =1+\frac{{\beta }^2}{2}\end{array}$

Substitute the value of the $\beta =\frac{v}{c}$ in the above equation. Thus,

$\begin{array}{l}\gamma =1+\frac{{\left(\frac{v}{c}\right)}^2}{2}\\ \gamma =1+\frac{1}{2}{\left(\frac{v}{c}\right)}^2\end{array}$

Substitute $v=26.38\text{ m / s}$ and $c=3\times {10}^8\text{ m / s}$ in above equation to calculate relativistic factor. Thus,

$\begin{array}{c}\gamma =1+\frac{1}{2}{\left(\frac{26.38\text{ m / s}}{3\times {10}^8\text{ m / s}}\right)}^2\\ =1+\frac{1}{2}{\left(8.79\times {10}^{-8}\right)}^2\\ =1+\frac{1}{2}\times 77.26\times {10}^{-16}\\ =1+3.86\times {10}^{-15}\end{array}$

Conclusion:

Therefore, relativistic factor for speed $95\text{ km / h}$. is $1+3.86\times {10}^{-15}$.

<b>

To determine

Relativistic factor for $v=240\text{ m / s}$.

Answer to Problem 15P

The relativistic factor for $v=240\text{ m / s}$ is $1+3.2\times {10}^{-13}$.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ =\frac{1}{\sqrt{1-{\beta }^2}}\end{array}$

Because $\beta =\frac{v}{c}$, where, v is the speed of the object and c is the speed of the light.

Here, $v=240\text{ m / s}$ which leads $v<<c$, thus $\beta =\frac{v}{c}$ will be very small, therefore, for smaller value of $\beta$ use binomial expansion for equation $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$. Therefore,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-{\beta }^2}}\\ \gamma =\frac{1}{{\left(1-{\beta }^2\right)}^{\frac{1}{2}}}\\ \gamma ={\left(1-{\beta }^2\right)}^{-\frac{1}{2}}\\ \gamma =1-\left(-\frac{1}{2}\right)\times {\beta }^2\end{array}$

Further simplify the above equation. Thus,

$\begin{array}{l}\gamma =1-\left(-\frac{1}{2}\right)\times {\beta }^2\\ \gamma =1+\frac{{\beta }^2}{2}\end{array}$

Substitute the value of the $\beta =\frac{v}{c}$ in the above equation. Thus,

$\begin{array}{l}\gamma =1+\frac{{\left(\frac{v}{c}\right)}^2}{2}\\ \gamma =1+\frac{1}{2}{\left(\frac{v}{c}\right)}^2\end{array}$

Substitute $v=240\text{ m / s}$ and $c=3\times {10}^8\text{ m / s}$ in above equation to calculate relativistic factor. Thus,

$\begin{array}{c}\gamma =1+\frac{1}{2}{\left(\frac{240\text{ m / s}}{3\times {10}^8\text{ m / s}}\right)}^2\\ =1+\frac{1}{2}{\left(\text{8}\text{.0}\times {10}^{-7}\right)}^2\\ =1+\frac{1}{2}\times 64\times {10}^{-14}\\ =1+3.2\times {10}^{-13}\end{array}$

Conclusion:

Therefore, relativistic factor for $v=2.3\text{ Mach}$ is $1+3.2\times {10}^{-13}$.

<c>

To determine

The ratio $\beta =\frac{v}{c}$, for an airplane travelling with 2.3 Mach.

Answer to Problem 15P

The relativistic factor for an airplane travelling with 2.3 Mach is $\text{3}\text{.2}\times {10}^{-12}$.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ =\frac{1}{\sqrt{1-{\beta }^2}}\end{array}$

Because $\beta =\frac{v}{c}$, where, v is the speed of the object and c is the speed of the light.

Velocity of the sound in the is $v_s=330\text{ m / s}$.

Write the formula for the conversion of the speed of the supersonic plane from Mach to meter per second. Thus,

$v=\text{M}{v}_s$, where, $v_s$ is the velocity of the sound in the air and M is the Mach number.

Substitute $v_s\text{=330 m / s}$ and $\text{M}=2.3$ in in $v=\text{M}{v}_s$ to  get v. Thus,

$\begin{array}{c}v=2.3\times 330\text{ m / s}\\ \text{=759 m / s}\end{array}$

Here, $v=759\text{ m / s}$ which leads $v<<c$, thus $\beta =\frac{v}{c}$ will be very small, therefore, for smaller value of $\beta$ use binomial expansion for equation $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$. Therefore,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-{\beta }^2}}\\ \gamma =\frac{1}{{\left(1-{\beta }^2\right)}^{\frac{1}{2}}}\\ \gamma ={\left(1-{\beta }^2\right)}^{-\frac{1}{2}}\\ \gamma =1-\left(-\frac{1}{2}\right)\times {\beta }^2\end{array}$

Further simplify the above equation. Thus,

$\begin{array}{l}\gamma =1-\left(-\frac{1}{2}\right)\times {\beta }^2\\ \gamma =1+\frac{{\beta }^2}{2}\end{array}$

Substitute the value of the $\beta =\frac{v}{c}$ in the above equation. Thus,

$\begin{array}{l}\gamma =1+\frac{{\left(\frac{v}{c}\right)}^2}{2}\\ \gamma =1+\frac{1}{2}{\left(\frac{v}{c}\right)}^2\end{array}$

Substitute $v=759\text{ m / s}$ and $c=3\times {10}^8\text{ m / s}$ in above equation to calculate relativistic factor. Thus,

$\begin{array}{c}\gamma =1+\frac{1}{2}{\left(\frac{759\text{ m / s}}{3\times {10}^8\text{ m / s}}\right)}^2\\ =1+\frac{1}{2}{\left(253\times {10}^{-8}\right)}^2\\ =1+\frac{1}{2}{\left(2.53\times {10}^{-6}\right)}^2\\ =1+\frac{1}{2}\times 6.4\times {10}^{-12}\end{array}$,

Simplify the above equation. Thus,

$\begin{array}{c}\gamma =1+\frac{1}{2}\times 6.4\times {10}^{-12}\\ =1+3.2\times {10}^{-12}\end{array}$

Conclusion:

Therefore, the relativistic factor for speed $2.3\text{ Mach}$ is $1+3.2\times {10}^{-12}$.

<d>

To determine

The relativistic factor for $v=27000\text{ km / h}$.

Answer to Problem 15P

The relativistic factor for $v=27000\text{ km / h}$ is $1+3.13\times {10}^{-10}$.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ =\frac{1}{\sqrt{1-{\beta }^2}}\end{array}$

Because $\beta =\frac{v}{c}$, where, v is the speed of the object and c is the speed of the light.

Convert, speed kilometer per hour to meter per second. Thus

$\begin{array}{c}v=27000\text{ km / h}\\ \text{=}27000\text{ km / h}\times \left(\frac{1000\text{m}}{1.0\text{km}}\right)\left(\frac{1.0\text{h}}{3600\text{s}}\right)\\ =\frac{270\times {10}^3}{36}\text{ m / s}\\ =7.5\times {10}^3\text{ m / s}\end{array}$

Here, $v=7.5\times {10}^3\text{ m / s}$ which leads $v<<c$, thus $\beta =\frac{v}{c}$ will be very small, therefore, for smaller value of $\beta$ use binomial expansion for equation $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$. Therefore,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-{\beta }^2}}\\ \gamma =\frac{1}{{\left(1-{\beta }^2\right)}^{\frac{1}{2}}}\\ \gamma ={\left(1-{\beta }^2\right)}^{-\frac{1}{2}}\\ \gamma =1-\left(-\frac{1}{2}\right)\times {\beta }^2\end{array}$

Further simplify the above equation. Thus,

$\begin{array}{l}\gamma =1-\left(-\frac{1}{2}\right)\times {\beta }^2\\ \gamma =1+\frac{{\beta }^2}{2}\end{array}$

Substitute the value of the $\beta =\frac{v}{c}$ in the above equation. Thus,

$\begin{array}{l}\gamma =1+\frac{{\left(\frac{v}{c}\right)}^2}{2}\\ \gamma =1+\frac{1}{2}{\left(\frac{v}{c}\right)}^2\end{array}$

Substitute $v=7.5\times {10}^3\text{ m / s}$ and $c=3\times {10}^8\text{ m / s}$ in above equation to calculate relativistic factor. Thus,

$\begin{array}{c}\gamma =1+\frac{1}{2}{\left(\frac{7.5\times {10}^3\text{ m / s}}{3\times {10}^8\text{ m / s}}\right)}^2\\ =1+\frac{1}{2}{\left(2.5\times {10}^{-5}\right)}^2\\ =1+\frac{1}{2}\times 6.25\times {10}^{-10}\\ =1+3.13\times {10}^{-10}\end{array}$,

Conclusion:

Therefore, the relativistic factor for $v=27000\text{ km / h}$ is $1+3.13\times {10}^{-10}$.

<e>

To determine

The relativistic factor, when an electron covering a distance 25 cm in 2 ns.

Answer to Problem 15P

The relativistic factor, when an electron covering a distance 25 cm in 2 ns is 1.10.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ =\frac{1}{\sqrt{1-{\beta }^2}}\end{array}$

Because $\beta =\frac{v}{c}$, where, v is the speed of the object and c is the speed of the light.

Write the equation to calculate the speed of the electron. Thus

$v=\frac{d}{t}$, where, d is the distance covered in time t.

Distance covered by the electron is 25 cm. Convert this distance in meter, thus,

$\begin{array}{c}25\text{ cm}=25.0\text{ cm}\times \frac{{\text{10}}^{-2}\text{ m}}{1.0\text{ cm}}\\ =25.0\times {10}^{-2}\text{ m}\end{array}$

Time taken by the electron to cover 25 cm is 2 ns. Convert this time in second, thus,

$\begin{array}{c}2\text{ ns}=2\text{ ns}\times \frac{{10}^{-9}\text{ s}}{1\text{ ns}}\\ =2\times {10}^{-9}\text{ s}\end{array}$

Substitute $d=25.0\times {10}^{-2}\text{ m}$ and $t=2\times {10}^{-9}\text{ s}$ in $v=\frac{d}{t}$ to calculate the v. Thus,

$\begin{array}{c}v=\frac{25.0\times {10}^{-2}\text{ m}}{2\times {10}^{-9}\text{ s}}\\ =12.5\times {10}^7\text{ m / s}\\ =1.25\times {10}^8\text{ m / s}\end{array}$

Here, $v=1.25\times {10}^8\text{ m / s}$ is comparable to speed of light c. Thus $\beta =\frac{v}{c}$ will not be very small, therefore, calculate the $\beta =\frac{v}{c}$, by substituting the value of the v and c.

Substitute $v=1.25\times {10}^8\text{ m / s}$ in $\beta =\frac{v}{c}$, where, $c=3\times {10}^8\text{ m / s}$ is the speed of the light. Thus,

$\begin{array}{c}\beta =\frac{1.25\times {10}^8\text{ m / s}}{3\times {10}^8\text{ m / s}}\\ =0.416\\ =0.42\end{array}$

Substitute $\beta =0.42$ in equation $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$ to calculate the relativistic factor. Thus,

$\begin{array}{c}\gamma =\frac{1}{\sqrt{1-{\left(0.42\right)}^2}}\\ =1.10\end{array}$

Conclusion:

Therefore, the relativistic factor for $v=1.25\times {10}^8\text{ m / s}$ is 1.10.

<f>

To determine

The relativistic factor, when an electron covering a distance ${10}^{-14}\text{ m}$ in $3.5\times {10}^{-23}\text{ s}$.

Answer to Problem 15P

The relativistic factor, when an electron covering a distance ${10}^{-14}\text{ m}$ in $3.5\times {10}^{-23}\text{ s}$ is 3.2.

Explanation of Solution

Write the equation for the relativistic factor. Thus,

$\begin{array}{l}\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ =\frac{1}{\sqrt{1-{\beta }^2}}\end{array}$

Because $\beta =\frac{v}{c}$, where, v is the speed of the object and c is the speed of the light.

Write the equation to calculate the Speed of the electron. Thus

$v=\frac{d}{t}$, where, d is the distance covered in time t.

Distance covered by the electron is ${10}^{-14}\text{ m}$ and time taken to cover this distance is $3.5\times {10}^{-23}\text{ s}$.

Substitute $d={10}^{-14}\text{ m}$ and $t=3.5\times {10}^{-23}\text{ s}$ in $v=\frac{d}{t}$ to calculate the v. Thus,

$\begin{array}{c}v=\frac{{10}^{-14}\text{ m}}{3.5\times {10}^{-23}\text{ s}}\\ =0.285\times {10}^{-14+23}\text{ m / s}\\ =0.285\times {10}^9\text{ m / s}\\ =2.86\times {10}^8\text{ m / s}\end{array}$

Here, $v=2.86\times {10}^8\text{ m / s}$ is comparable to speed of light c. Thus $\beta =\frac{v}{c}$ will not be very small, therefore, calculate the $\beta =\frac{v}{c}$, by substituting the value of the v and c.

Substitute $v=2.86\times {10}^8\text{ m / s}$ in $\beta =\frac{v}{c}$, where, $c=3\times {10}^8\text{ m / s}$ is the speed of the light. Thus,

$\begin{array}{c}\beta =\frac{2.86\times {10}^8\text{ m / s}}{3\times {10}^8\text{ m / s}}\\ =0.95\end{array}$

Substitute $\beta =0.95$ in equation $\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$ to calculate the relativistic factor. Thus,

$\begin{array}{c}\gamma =\frac{1}{\sqrt{1-{\left(0.95\right)}^2}}\\ =3.2\end{array}$

Conclusion:

Therefore, the relativistic factor for $v=2.86\times {10}^8\text{ m / s}$ is 3.2.