Question

The radio telescope in Arecibo, Puerto Rico, has a diameter of $305 \mathrm{m}$. It can detect radio waves from space with intensities as small as \(10^{-26} \mathrm{W} / \mathrm{m}^{2} .\) (a) What is the average power incident on the telescope due to a wave at normal incidence with intensity \(1.0 \times 10^{-26} \mathrm{W} / \mathrm{m}^{2} ?\) (b) What is the average power incident on Earth's surface? (c) What are the rms electric and magnetic fields?

Short answer

Question: Calculate the average power incident on the telescope, the average power incident on Earth's surface, and the root-mean-square (rms) electric and magnetic fields for an intensity of \(1.0 \times 10^{-26} \mathrm{W} / \mathrm{m}^2\) and a radio telescope diameter of 305 m. Answer: Follow the steps outlined in the solution. After calculating the areas for the telescope and Earth, determine the average power incident on both using the given intensity. Then, use the formulas for the rms electric and magnetic fields to compute their values.

Step-by-step solution

Calculate the radio telescope area

We are given the diameter of the radio telescope, which is 305 m. For a circular shape, we can calculate the area, A, with the formula \(A = \pi r^2\), using the radius, r, where \(r = \frac{D}{2}\): $$ A = \pi \left(\frac{305}{2}\right)^2 $$

Calculate the average power on the telescope

Now, we have the intensity at \(1.0 \times 10^{-26} \mathrm{W} / \mathrm{m}^2\). Using the formula \(P = I \cdot A\), we can calculate the average power on the telescope: $$ P_{telescope}= (1.0 \times 10^{-26} \mathrm{W} / \mathrm{m}^2) \cdot A $$

Estimate Earth's surface area

To calculate the average power on Earth's surface, we need to estimate Earth's surface area. We will assume Earth to be a perfect sphere with a radius of 6371 km. Surface area, A, is given by \(A = 4 \pi R^2\): $$ A_{earth} = 4 \pi (6.371 \times 10^6)^2 \mathrm{m}^2 $$

Calculate the average power on Earth's surface

Now, we can calculate the average power incident on Earth's surface using the formula \(P = I \cdot A\): $$ P_{earth} = (1.0 \times 10^{-26} \mathrm{W} / \mathrm{m}^2) \cdot A_{earth} $$

Determine rms electric field value

To calculate the rms electric field value, use the formula \(I = \frac{1}{2} \varepsilon_0 c E^2_{rms}\). We will solve for \(E_{rms}\): $$ E_{rms} = \sqrt{\frac{2 \cdot I}{\varepsilon_0 c}} $$ where \(\varepsilon_0\) is the vacuum permittivity, approximately \(8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{N} \cdot \mathrm{m}^2\) and c is the speed of light, approximately \(3.00 \times 10^8 \mathrm{m} / \mathrm{s}\).

Determine rms magnetic field value

Similarly, for the rms magnetic field value, use the formula \(I = \frac{1}{\mu_0 c}B^2_{rms}\), and solve for \(B_{rms}\): $$ B_{rms} = \sqrt{\frac{I \cdot \mu_0 c}{2}} $$ where \(\mu_0\) is the vacuum permeability, approximately \(4\pi \times 10^{-7} \mathrm{H} / \mathrm{m}\). Now you can calculate the rms electric and magnetic fields using the intensity given in the problem statement.