Physics Solution Manual for 1100 and 2101

The beams striking the second and third sheets of

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Unformatted text preview: the power Pintercepted intercepted by Mercury divided by the power Psun radiated by the sun, or Fraction = Pintercepted/ Psun. According to Equation 16.8, the power intercepted by Mercury is the intensity S of the radiation at Mercury’s location times the area A that Mercury presents to the radiation. Since the sun radiates uniformly in all directions, we can use Equation 16.9 to evaluate the intensity at Mercury’s location. The planet Mercury looks like a circular disk to the radiation (just like the moon looks like a disk to anyone viewing it); 2 this area is A = π rMercury , where rMercury is the radius of Mercury. SOLUTION The fraction of the sun’s power that is intercepted by Mercury is Power intercepted Pintercepted by Mercury Fraction == Power radiated Psun by sun (1) According to Equation 16.8, the power Pintercepted by Mercury is equal to the intensity S of the radiation at Mercury times the area A of the Mercury disk, or Pintercepted = S A . Substituting this expression into Equation (1) gives = Fraction Pintercepted S A = Psun Psun (2) Since the sun radiates uniformly in all directions, the intensity of the radiation at Mercury’s location is S = Psun / 4π r 2 (Equation 16.9), where r is the mean distance from the sun to Mercury. As mentioned in the REASONING section, the area of the Mercury disk is 2 A = π rMercury . Substituting these expressions for S and A into Equation (2) yields Psun 2 ( π rMercury ) S A 4π r = Fraction = Psun Psun 2 = ( 2.44 ×106 m )2 = 2 4 ( 5.79 ×1010 m ) = 2 rMercury 4r 2 4.44 ×10−10 ______________________________________________________________________________ 1296 ELECTROMAGNETIC WAVES 34. REASONING The rms magnetic field Brms and the average intensity S1 of the gamma radiation emitted from the surface of the magnetar are related by S1 = c µ0 2 Brms (24.5c) where µ0 =×10 −7 T ⋅ m/A is the permeability of free space and c is the speed of light in a 4π vacuum. The average intensity S1 of the gamma radiation is defined as the average power that passes perpendicularly through an area A1. Therefore, if the radius of the magnetar is r1, then the average intensity must be = S1 P P 1 1 = A1 4π r12 (1) In Equation (1), A1 = 4π r12 is the surface area of the magnetar (assumed to be spherical). When the pulse of gamma radiation reaches earth, the average power P is spread out 1 uniformly over the surface of a spherical surface of radius R equal to the distance between earth and the magnetar. The area A of this sphere is A = 4π R 2 , and the average intensity P S2 of the pulse when it reaches the telescope is given by S2 = 1 , so we have that A = AS2 4π R 2 S2 = P 1 (2) The average power P2 intercepted by the telescope detector is proportional to the average intensity S2 and the surface area A2 of the detector: P2 = A2 S2 . Rearranging this relation, we obtain S2 = P2 A2 (3) We will determine the average power P2 from the energy delivered to the detector and the duration Δt of the pulse via Equation 6.10b: P2 = Energy ∆t SOLUTION Solving Equation 24.5c for Brms, we obtain (6.10b) Chapter 24 Problems 2 Brms = µ0 S1 or Brms = c µ0 S1 1297 (4) c Substituting Equation (1) into Equation (4) yields Brms = µ0 P 1 (5) 4π c r12 Substituting Equation (2) into Equation (5), we find that µ0 P 1 = 4π c r12 Brms = µ0 4π c r12 R ( 4π R2 S2 ) = r 1 µ0 S 2 (6) c Substituting Equation (3) into Equation (6) gives Brms = R µ0 S 2 R = r1 c r1 µ0 P2 (7) c A2 Replacing the average power P2 in Equation (7) with Equation 6.10b, we finally obtain R Brms = r1 µ0 ( Energy ) 4.5 ×1020 m = c A2 ∆t 9.0 × 103 m J) ( 4π ×10−7 T ⋅ m/A )(8.4 ×10−6 = (3.00 ×108 m/s )( 75 m2 ) ( 0.24 s ) 2.2 × 106 T ______________________________________________________________________________ 35. REASONING The Doppler effect for electromagnetic radiation is given by Equation 24.6; v = fs 1 ± rel fo c if vrel << c where f o is the observed frequency, fs is the frequency emitted by the source, and vrel is the speed of the source relative to the observer. As discussed in the text, the plus sign applies when the source and the observer are moving toward one another, while the minus sign applies when they are moving apart. According to Equation 16.1, the wavelength of these waves is λ = c/f. Therefore, the Doppler shift can be written in terms of wavelengths: 1 = λo vrel 1 1 ± c λs if vrel << c 1298 ELECTROMAGNETIC WAVES SOLUTION a. The wavelength λo of the light observed on earth is greater than the wavelength λs of the light when it is emitted from the distant galaxy (the source). Therefore, the frequency of the light observed on earth is less than the frequency of the light when it is emitted from the distant galaxy. Thus, the quantity in the brackets in Equation 24.6 must be less than one; it must be equal to 1 – (vrel / c) . Since the minus sign applies, we can conclude that the galaxy must be receding from the earth . b. We can find the speed of the galaxy r...
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