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Chapter 3: Problem 12

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

Expert verified
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

01

(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.)
02

(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.)
03

(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: \(\frac{\Delta\phi_{rad}}{2\pi}\).)
04

(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: \(P_{req} = P_{total}\frac{A_{req}}{4 \pi R^2}\).)
05

(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, \(A_{earth}= 4 \pi d_{earth}^2\).)
06

(b) Calculate power density

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

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Most popular questions from this chapter

Spherical surfaces at \(r=2,4\), and \(6 \mathrm{~m}\) carry uniform surface charge densities of \(20 \mathrm{nC} / \mathrm{m}^{2},-4 \mathrm{n} \mathrm{C} / \mathrm{m}^{2}\), and \(\rho_{\mathrm{so}}\), respectively. \((a)\) Find \(\mathbf{D}\) at \(r=1\), 3 , and \(5 \mathrm{~m}\). (b) Determine \(\rho_{S 0}\) such that \(\mathbf{D}=0\) at \(r=7 \mathrm{~m}\).

The cylindrical surface \(\rho=8 \mathrm{~cm}\) contains the surface charge density, \(\rho_{S}=\) \(5 e^{-20|z|} \mathrm{nC} / \mathrm{m}^{2} .(a)\) What is the total amount of charge present? \((b)\) How much electric flux leaves the surface \(\rho=8 \mathrm{~cm}, 1 \mathrm{~cm}

Volume charge density is located in free space as \(\rho_{v}=2 e^{-1000 r} \mathrm{nC} / \mathrm{m}^{3}\) for \(0

Use Gauss's law in integral form to show that an inverse distance field in spherical coordinates, \(\mathbf{D}=A a_{r} / r\), where \(A\) is a constant, requires every spherical shell of \(1 \mathrm{~m}\) thickness to contain \(4 \pi A\) coulombs of charge. Does this indicate a continuous charge distribution? If so, find the charge density variation with \(r\).

An electric flux density is given by \(\mathbf{D}=D_{0} \mathbf{a}_{\rho}\), where \(D_{0}\) is a given constant. (a) What charge density generates this field? \((b)\) For the specified field, what total charge is contained within a cylinder of radius \(a\) and height \(b\), where the cylinder axis is the \(z\) axis?
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