Question

The sun radiates a total power of about \(3.86 \times 10^{26}\) watts \((\mathrm{W})\). If we imagine the sun's surface to be marked off in latitude and longitude and assume uniform radiation, \((a)\) what power is radiated by the region lying between latitude \(50^{\circ} \mathrm{N}\) and \(60^{\circ} \mathrm{N}\) and longitude \(12^{\circ} \mathrm{W}\) and \(27^{\circ} \mathrm{W} ?(b)\) What is the power density on a spherical surface \(93,000,000\) miles from the sun in \(\mathrm{W} / \mathrm{m}^{2} ?\)

Short answer

And what is the power density on a spherical surface at a distance of 93,000,000 miles from the sun? Answer: To find the power radiated by the specified region and the power density on a spherical surface at a distance of 93,000,000 miles from the sun, we calculated the area within the given coordinates on the sun's surface and the surface area of the sphere at the given distance. We then used these areas to calculate the power and power density, resulting in: \(P_{req}\) = [power radiated by the region, please solve and insert the numerical value] \(\rho\) = [power density, please solve and insert the numerical value] \(W/m^2\)

Step-by-step solution

(a) Calculate area of the sun's surface

(To solve for the power radiated by the specified region, we will first calculate the area on the sun's surface that lies within the given coordinates of latitude and longitude. We can consider the sun as a sphere and compute area using the spherical cap formula: \(A_s = 2 \pi Rh\), where \(R\) is the sun's radius and \(h\) is the height of the cap.)

(a) Calculate height of the cap

(We will calculate the height of the cap using the difference in the latitudes: \(\Delta\theta=60°-50°=10°\). We can convert the angle to radians: \(\Delta\theta_{rad}=\frac{\pi}{180}\times10° =\frac{\pi}{18}\)rad. Now we'll determine the height of the cap through the following formula: \(h = R(1-\cos\Delta\theta_{rad})\). Using the sun's radius, \(R = 6.96\times10^{8} m\), we can compute h.)

(a) Calculate area between longitudes

(Now that we have the area of the spherical cap, let's find the area between the longitudes \(12^{\circ} \mathrm{W}\) and \(27^{\circ} \mathrm{W}\). The difference in longitudes, \(\Delta\phi=27°-12°=15°\). Converting to radians, \(\Delta\phi_{rad}=\frac{\pi}{180}\times15° =\frac{\pi}{12}\) rad. We'll then find the fractional area between these longitudes: \(\frac{\Delta\phi_{rad}}{2\pi}\).)

(a) Calculate power radiated by the region

(Once we have the fractional area between longitudes, we can compute the area of the sun's surface within the given coordinates: \(A_{req} = A_s \times \frac{\Delta\phi_{rad}}{2\pi}\). We can then find the power radiated by the region by multiplying the total power radiated by the sun \((P_{total}=3.86 × 10^{26} \mathrm{W})\) by the ratio of the required area to the total surface area of the sun: \(P_{req} = P_{total}\frac{A_{req}}{4 \pi R^2}\).)

(b) Calculate spherical surface area at given distance

(To find the power density on a spherical surface, 93,000,000 miles from the sun, we first need to find the surface area of this sphere. We can convert the given distance to meters: \(d_{earth} = 93000000\,\text{miles} \times \frac{1609.34\,\text{m}}{1\,\text{mile}}\). Now, the surface area can be calculated using the formula, \(A_{earth}= 4 \pi d_{earth}^2\).)

(b) Calculate power density

(Once we have the surface area, we can compute the power density as \(\rho = \frac{P_{total}}{A_{earth}}\) in \(\mathrm{W}/\mathrm{m}^2\).)

Key concepts

Spherical Cap

When discussing electromagnetic radiation, especially from celestial bodies like the sun, the concept of a 'spherical cap' becomes crucial. A spherical cap refers to a geometric shape that resembles a "cap" covering the top of a sphere. Imagine slicing through a sphere parallel to its base; the section above the cut is the spherical cap.
This is particularly useful in the context of calculating areas on the sun's surface between specific latitudes.To determine the area of a spherical cap, we can use the formula:

• \(A_s = 2 \pi R h \)

Here, \(R\) represents the radius of the sphere (or sun, in this context), and \(h\) stands for the height of the cap. The height \(h\) can be calculated using the angle difference in radians between the two latitudes you’re interested in. By using basic trigonometric identities, the formula for \(h\) becomes:

• \(h = R(1-\cos \Delta \theta_{rad})\)

Here, \(\Delta \theta_{rad}\) is the difference of the latitudes converted from degrees to radians.

Power Density

Power density is a measure of how much power (in watts) is distributed over a unit area (in square meters). When it comes to electromagnetic radiation from the sun, power density becomes especially relevant in understanding how much solar energy reaches a specific area, such as the Earth.In essence, power density explains how the solar energy that the sun radiates z from a spherical surface at a certain distance,

• Calculated as \(\rho = \frac{P_{total}}{A_{ ext{earth}}}\)

where \(\rho\) stands for power density, \(P_{total}\) is the total power output from the sun, and \(A_{ ext{earth}}\) is the surface area of a sphere with a radius equivalent to the distance from the sun to Earth.As a practical application, knowing how much energy reaches each square meter of Earth is pivotal for disciplines like solar energy research and climate studies.

Latitude and Longitude

Latitude and longitude are coordinate systems that allow us to pinpoint locations on a spherical body, like Earth or the sun. Latitude measures how far north or south a point is from the equator, with the equator itself being 0° latitude.
It ranges from 0° to 90° either north or south. Longitude, on the other hand, measures how far east or west a point is from the prime meridian, which is defined as 0° longitude. Longitude extends from 0° to 180° both east and west. In the presented problem, the surface of the sun is divided according to these coordinates. To define a specific region on the sun's surface using latitude and longitude lines involves calculating differences in these values:

• Latitude difference: 60°N - 50°N = 10°
• Longitude difference: 27°W - 12°W = 15°

These calculations help in determining the exact part of the sun's surface contributing to certain radiation values.

Spherical Surface Area

The spherical surface area is an essential concept that applies to spheres of any size, from microscopic cells to astronomical giants like stars. To calculate the surface area of a sphere, the formula used is:

• \(A = 4 \pi R^2 \)

This equation is essential when determining how radiation or any other interaction takes place on spherical surfaces.For example, when figuring out how much solar energy strikes a planet, converting distance from the source (like the sun) into an appropriate radius is necessary. Then, this radius can be substituting into the surface area formula to determine the total area over which energy is spread.In our context, if you want to find the spherical surface area at a certain distance such as the Earth from the sun (which is about 93 million miles), the formula can be adjusted:

• \(A_{ ext{earth}}= 4\pi d_{ ext{earth}}^2\)

Converting this distance to meters and plugging it into the formula will give you the area over which the sun's power is spread, aiding in calculating power density.