Chapter 27, Problem 26Q

(a)

To determine

The speed of Earth in its orbit by using the data in appendix 1, if the Earth’s orbit is assumed to be circular.

Answer to Problem 26Q

Solution:

$29.8\text{ km/s}$

Explanation of Solution

Given data:

From appendix 1, the radius of the Earth’s orbit is $1.496\times {10}^8\text{ km}$ and the time period of its revolution is $1\text{ year}$ or $365.256\text{ days}$.

Formula used:

Write the expression for speed.

$v=\frac{S}{t}$

Here, $v$ is the speed of Earth in its orbit, $S$ is the distance covered by the Earth during revolution, and $t$ is the time period of revolution.

Explanation:

Since it is given that Earth’s orbit is to be assumed to be circular, the distance covered by the Earth in a revolution is equal to the perimeter of the circle.

Recall the expression for speed.

$v=\frac{S}{t}$

Substitute $2\pi R$ for $S$ in the above expression,

$v=\frac{2\pi R}{t}$

Here, is the radius of Earth’s orbit.

Substitute $1.496\times {10}^8\text{ km}$ for $R$ and $365.256\text{ days}$ for $t$ in the above expression,

$\begin{array}{c}v=\frac{2\pi \left(1.496\times {10}^8\text{ km}\right)}{365.256\text{ days}\left(\frac{24\times 60\times 60\text{ s}}{1\text{ day}}\right)}\\ =29.78\text{ km/s}\\ \approx \text{29}\text{.8 km/s}\end{array}$

Conclusion:

Hence, the speed of Earth in its orbit calculated by using the data from appendix 1 is close to $29.8\text{ km/s}$.

(b)

To determine

The wavelength (in meters) of the signal received by the Earth if the aliens are transmitting it at a frequency of $3000\text{ MHz}$, and the Earth is neither moving toward nor away from the alien’s planet.

Answer to Problem 26Q

Solution:

$0.1\text{ m}$

Explanation of Solution

Given data:

The frequency of the signal is $3000\text{ MHz}$.

Formula used:

Write the expression for the wavelength of a wave.

${\lambda }_o=\frac{c}{\upsilon }$

Here, ${\lambda }_o$ is the wavelength of the signal wave, $c$ is the speed of light, and $\upsilon$ is the frequency of the signal.

Explanation:

Consider the speed of light to be $3\times {10}^8\text{ m/s}$.

As the Earth is neither moving toward nor away from the alien’s planet, so there is no relative motion between the planets. Therefore, there would be no Doppler effect.

Recall the expression for the wavelength of a wave.

${\lambda }_o=\frac{c}{\upsilon }$

Substitute $3\times {10}^8\text{ m/s}$ for $c$ and $3000\text{ MHz}$ for $\upsilon$ in the above expression,

$\begin{array}{c}{\lambda }_o=\frac{3\times {10}^8\text{ m/s}}{3000\text{ MHz}\left(\frac{{10}^6\text{ Hz}}{1\text{ MHz}}\right)}\\ =0.1\text{ m}\end{array}$

Conclusion:

Thus, the wavelength of the signal received by Earth from the aliens under the given condition is calculated to be $0.1\text{ m}$.

(c)

To determine

The maximum wavelength shift in meters as well as in percentage of unshifted wavelength, calculated in part (b), if the maximum Doppler shift occurs when moving directly toward or away from the alien planet.

Answer to Problem 26Q

Solution:

$9.9\times {10}^{-6}\text{ m}$ and $0.0099\%$

Explanation of Solution

Given data:

The frequency of the signal is $3000\text{ MHz}$. The Earth is moving directly toward or away from the planet.

Formula used:

Write the expression for the Doppler shift if the receiver is directly approaching the transmitter.

$\frac{\Delta \lambda }{{\lambda }_o}=\frac{v}{c}$

Here, $\Delta \lambda$ is the Doppler shift, ${\lambda }_o$ is the wavelength of the signal wave, $c$ is the speed of light, and $v$ is the speed of Earth in its orbit.

Explanation:

Since the Earth (receiver) is moving directly toward the alien planet (transmitter), there is a relative motion between them. Therefore, a Doppler shift would be observed.

Recall the value of the wavelength of the signal received by the Earth from the aliens as calculated in part (b).

${\lambda }_o=0.1\text{ m}$

Recall the value of Earth’s speed in its orbit under the given condition form part (a).

$v=\text{29}\text{.8 km/s}$

Consider the speed of light to be $3\times {10}^8\text{ m/s}$.

Recall the expression for the Doppler shift if the receiver is directly approaching toward the transmitter.

$\frac{\Delta \lambda }{{\lambda }_o}=\frac{v}{c}$

Rearrange the expression in terms of $\Delta \lambda$.

$\Delta \lambda =\frac{v}{c}{\lambda }_o$

Substitute $0.1\text{ m}$ for ${\lambda }_o$, $3\times {10}^8\text{ m/s}$ for $c$, and $\text{29}\text{.8 km/s}$ for $v$ in the above expression,

$\begin{array}{c}\Delta \lambda =\frac{\text{29}\text{.8 km/s}\left(\frac{1000\text{ m/s}}{1\text{ km/s}}\right)}{\left(3\times {10}^8\text{ m/s}\right)}\left(0.1\text{ m}\right)\\ =9.9\times {10}^{-6}\text{ m}\end{array}$

The maximum wavelength shift in percentage of unshifted wavelength is given by the expression:

$\%\lambda =\frac{\Delta \lambda }{{\lambda }_o}\times 100$

Here, ${\lambda }_0$ is the maximum wavelength shift in percentage of unshifted wavelength.

Substitute $9.9\times {10}^{-6}\text{ m}$ for $\Delta \lambda$ and $0.1\text{ m}$ for ${\lambda }_o$ in the above expression,

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

Conclusion:

The maximum wavelength shifts in meters as well as in percentage of unshifted wavelength is calculated as $9.9\times {10}^{-6}\text{ m}$ and $0.0099\%$, respectively.

(d)

To determine

The reason behind the utmost importance given to the precision of the SETI radio receiver to measure frequency and wavelength.

Answer to Problem 26Q

Solution:

The Doppler shift calculated is very small $\left(0.0099\%\right)$, so the SETI radio receiver must be very precise to detect this negligible shift in the wavelength.

Explanation of Solution

Introduction:

Doppler Effect is defined as the phenomenon in which there is a sudden noticeable change in the pitch or wavelength when there is a relative motion between the transmitter and the receiver.

The expression for the Doppler shift if the receiver is directly approaching the transmitter is:

$\frac{\Delta \lambda }{{\lambda }_o}=\frac{v}{c}$

Here, $\Delta \lambda$ is the Doppler shift, ${\lambda }_o$ is the wavelength of the signal wave, $c$ is the speed of light, and $v$ is the speed of the Earth in its orbit.

Explanation:

Recall the value of the maximum wavelength shift in meters, calculated in the previous part, that is, part (c).

$\Delta \lambda =9.9\times {10}^{-6}\text{ m}$

Also recall the value of the maximum wavelength shift in percentage of unshifted wavelength.

$\%\lambda =0.0099\%$

It can be clearly observed from these two values that the wavelength of the receiving signal is very less. Therefore, to identify the Doppler shift, the SETI radio receiver must be highly precise. If the receiver is not precise and accurate, it would not detect such a small magnitude change and the signal from the alien planet would not be verified.

Conclusion:

The SETI radio receiver must be very precise to detect the very small shift in the wavelength due to Doppler effect.