Question

Question: Frank D. Drake, an investigator in the SETI (Search for Extra-Terrestrial Intelligence) program, once said that the large radio telescope in Arecibo, Puerto Rico (Fig.33-36), “can detect a signal which lays down on the entire surface of the earth a power of only one picowatt.” (a) What is the power that would be received by the Arecibo antenna for such a signal? The antenna diameter is $\mathbf{300}\mathbf{}\mathbf{m}$.(b) What would be the power of an isotropic source at the center of our galaxy that could provide such a signal? The galactic center is $\mathbf{2}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{ly}$away. A light-year is the distance light travels in one year.

Short answer

1. The power received by the Arecibo antenna is 1.4x10^-22w.
2. The power of the source at the center of our galaxy is $1.1\times {10}^{15}\text{W}$.

Step-by-step solution

Listing the given quantities

Antenna diameter, D_A=300m.

Distance of galactic center from source, R_S= 2.2x10⁴ly.

Understanding the concepts of intensity and power

We equatetheintensity of the receiver and the intensity of the earth to findthepower ofthereceiver. Inthesecond case, we equate the intensity of the source to the intensity of the earth to find the power ofthesource.

(a) Calculations of the power received by the Arecibo antenna

The intensity of the signal on the surface of the earth and the intensity of the signal received by the antenna have to be the same.

${\mathrm{I}}_{\mathrm{E}}={\mathrm{I}}_{\mathrm{A}}$ (1)

The intensity is nothing but power per unit area.

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$

Now, we can write equation 1 as,

$\frac{{\mathrm{P}}_{\mathrm{E}}}{{\mathrm{A}}_{\mathrm{E}}}=\frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{A}}_{\mathrm{A}}}$

Here,

$$\begin{gathered} {\mathrm{A}}_{\mathrm{E}}\left(\text{surface}\text{area}\text{of}\text{earth}\right)=4{\mathrm{\pi R}}_{\mathrm{E}}^2 \\ {\mathrm{A}}_{\mathrm{A}}\left(\text{area}\text{of}\text{antenna}\right)={\mathrm{\pi R}}_{\mathrm{A}}^2 \end{gathered}$$

${\mathrm{P}}_{\mathrm{A}}=\frac{{\mathrm{A}}_{\mathrm{A}}}{{\mathrm{A}}_{\mathrm{E}}}{\mathrm{P}}_{\mathrm{E}}$

Here ${\mathrm{P}}_{\mathrm{A}}$is the power radiated by the antenna.

Substitute the values in the above expression, and we get,

${\mathrm{P}}_{\mathrm{A}}=\frac{{\mathrm{R}}_{\mathrm{A}}^2}{4{\mathrm{R}}_{\mathrm{E}}^2}{\mathrm{P}}_{\mathrm{E}}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{A}}=\frac{1\times {10}^{-12}\mathrm{w}\left({\displaystyle \frac{{\left(300\mathrm{m}\right)}^2}{4}}\right)}{4\times {\left(6.37\times {10}^6\mathrm{m}\right)}^2} \\ =1.4\times {10}^{-22}\mathrm{w} \end{gathered}$$

Thus, the power received by the Arecibo antenna is $1.4\times {10}^{-22}\mathrm{w}$.

(b) Calculations of the power of the source at the center of our galaxy

The intensity of the source is equal to the intensity on earth:

I_E=I_S (2)

The intensity is nothing but power per unit area.

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$

Now, we can write equation 2 as,

$\frac{{\mathrm{P}}_{\mathrm{E}}}{{\mathrm{A}}_{\mathrm{E}}}=\frac{{\mathrm{P}}_{\mathrm{s}}}{{\mathrm{A}}_{\mathrm{s}}}$

Here

A_E (Surface area of earth)= $4\pi {R_E}^2$

A_S (area of source) =$4\pi {R_S}^2$

${\mathrm{P}}_{\mathrm{s}}=\frac{{\mathrm{A}}_{\mathrm{s}}}{{\mathrm{A}}_{\mathrm{E}}}{\mathrm{P}}_{\mathrm{E}}$

Here ${\mathrm{P}}_{\mathrm{s}}$is the power of the source.

Substitute the values in the above expression, and we get,

${\mathrm{P}}_{\mathrm{s}}=\frac{{\mathrm{R}}_{\mathrm{s}}^2}{{\mathrm{R}}_{\mathrm{E}}^2}{\mathrm{P}}_{\mathrm{E}}$ (3)

Here R_S is the radial distance of the galactic center.

1 year = 3.154x10⁷s.

So the distance traveled by the light in 1 year is given by,

$$\begin{gathered} 1\mathrm{ly}=3\times {10}^8\times 3.154\times {10}^7 \\ =9.462\times {10}^{15}\mathrm{m} \end{gathered}$$

role="math" localid="1662984011590" $$\begin{gathered} {\mathrm{R}}_{\mathrm{S}}=2.2\times {10}^4\mathrm{ly} \\ =2.2\times {10}^4\times 9.462\times {10}^{15}\mathrm{m} \\ =20.82\times {10}^{19}\mathrm{m} \end{gathered}$$

Substitute the values in the above expression (3), and we get,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{S}}=\frac{\left(1\times {10}^{-12}\mathrm{w}\right){\left(20.82\times {10}^{19}\mathrm{m}\right)}^2}{{\left(6.37\times {10}^6\mathrm{m}\right)}^2} \\ =1.1\times {10}^{15}\mathrm{w} \end{gathered}$$

Thus, the power of the source at the center of our galaxy is $1.1\times {10}^{15}\mathrm{w}$.