Chapter 27, Problem 26Q

To determine

(a)

The speed of earth in its orbit.

Answer to Problem 26Q

The speed of earth in its axis is $29.8\text{km}/\text{s}$.

Explanation of Solution

Given:

The radius of Earth orbit is, $r=149\cdot 6\times {10}^6\text{km}$.

Formula Used:

The expression of speed of earth in its axis is given by,

$v=\frac{\left(2\pi r\right)}{\text{length}\text{of}\text{year}}$

Calculation:

The speed of earth in its axis is calculated as,

$\begin{array}{c}v=\frac{\left(2\pi r\right)}{\text{length}\text{of}\text{year}}\\ =\frac{2\pi \times \left(149\cdot 6\times {10}^6\text{km}\right)}{3.155815\times {10}^7\text{s}}\\ =29.8\text{km}/\text{s}\end{array}$

Conclusion:

The speed of earth in its axis is $29.8\text{km}/\text{s}$.

To determine

(b)

The wavelength of transmission which Earth receives.

Answer to Problem 26Q

The wavelength of transmission which Earth receives is $0\cdot 1\text{m}$.

Explanation of Solution

Given:

The frequency is $f=3000\text{MHz}$.

Formula Used:

The expression of wavelength is given by,

$\lambda =\frac{c}{f}$

Calculation:

The wavelength is calculated as,

$\begin{array}{c}\lambda =\frac{c}{f}\\ =\frac{3\times {10}^8\text{m}/\text{s}}{\left(3000\text{MHz}\times \frac{1\text{Hz}}{{10}^{-6}\text{MHz}}\right)}\\ =\frac{3\times {10}^8}{30\times {10}^8}\text{m}\\ =0\cdot 1\text{m}\end{array}$

Conclusion:

The wavelength of transmission which Earth receives is $0\cdot 1\text{m}$.

To determine

(c)

The shift in wavelength using Doppler Effect.

Answer to Problem 26Q

The shift in wavelength is $9.9\times {\text{10}}^{-6}\text{m}$ and the percent shift in wavelength is $0.0099\%$.

Explanation of Solution

Formula Used:

The expression of shift in wavelength is given by,

$\Delta \lambda =\frac{v}{c}\times {\lambda }_0$

The expression of percent shift in wavelength is given by,

$\%\text{shift}\text{in}\text{wavelength}=\frac{\Delta \lambda }{{\lambda }_0}\times 100\%$

Calculation:

The shift in wavelength is calculated as,

$\begin{array}{c}\Delta \lambda =\frac{v}{c}\times {\lambda }_0\\ =\frac{29.8\text{km}/\text{s}}{3\times {10}^8\text{m}/\text{s}}\times 0\cdot 1\text{m}\\ =\frac{29.8\text{km}/\text{s}\left(\frac{{10}^3\text{m}/\text{s}}{1\text{km}/\text{s}}\right)}{3\times {10}^8\text{m}/\text{s}}\times 0\cdot 1\text{m}\\ =9.9\times {\text{10}}^{-6}\text{m}\end{array}$

The percent shift in wavelength is calculated as,

$\begin{array}{c}\%\text{shift}\text{in}\text{wavelength}=\frac{\Delta \lambda }{{\lambda }_0}\times 100\%\\ =\frac{9.9\times {\text{10}}^{-6}\text{m}}{0.1\text{m}}\times 100\%\\ =0.0099\%\end{array}$

Conclusion:

The shift in wavelength is $9.9\times {\text{10}}^{-6}\text{m}$ and the percent shift in wavelength is $0.0099\%$.

To determine

(d)

The importance that SETI radio receivers be able to measure frequency and wavelength to very high precision.

Explanation of Solution

Introduction:

SETI pioneer Bernard Oliver was the first one who draws attention to a range of relatively noise-free frequency in the neighbor-hood of the microwave emission lines of hydrogen and hydroxide.

It is essential to track all the frequency and wavelength sent by civilization in another planetary system. So, there is a requirement of high precision technology to detect that frequency to communicate with them.

Conclusion:

Therefore, there is a requirement of a high precision radio receiver which able to measure frequency and wavelength.