91影视

Antenna diameter, D_A=300m.

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

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.

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,

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

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

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,

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

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

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)= $4蟺{R_E}^2$

A_S (area of source) =$4蟺{R_S}^2$

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

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,

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

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

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

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