941_PartUniversity Physics Solution

941_PartUniversity Physics Solution - 32-14 32.54. Chapter...

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32-14 Chapter 32 (c) P av = I / c = (9.75 × 10 &15 W/m 2 )/(3.00 × 10 8 m/s) = 3.25 × 10 &23 Pa (d) λ = c / f = (3.00 × 10 8 m/s)/(1575.42 × 10 6 Hz) = 0.190 m EVALUATE: The fields and pressures due to these waves are very small compared to typical laboratory quantities. 32.54. IDENTIFY: For a totally reflective surface the radiation pressure is 2 . I c Find the force due to this pressure and express the force in terms of the power output P of the sun. The gravitational force of the sun is sun g 2 . mM FG r = SET UP: The mass of the sum is 30 sun 1.99 10 kg. M 11 2 2 6.67 10 N m /kg . G EXECUTE: (a) The sail should be reflective, to produce the maximum radiation pressure. (b) rad 2 I FA c ⎛⎞ = ⎜⎟ ⎝⎠ , where A is the area of the sail. 2 4 P I r π = , where r is the distance of the sail from the sun. rad 22 2 42 A PP A F cr r c ππ ⎛ ⎞ == ⎜ ⎟ ⎝ ⎠ . rad g FF = so sun 2 PA mM G rc r = . 81 1 2 2 3 0 sun 26 2 2 (3.00 10 m/s)(6.67 10 N m /kg )(10,000 kg)(1.99 10 kg) 3.9 10 W cGmM A P ×× × × . 62 2 6.42 10 m 6.42 km A = . (c) Both the gravitational force and the radiation pressure are inversely proportional to the square of the distance from the sun, so this distance divides out when we set rad g . = EVALUATE: A very large sail is needed, just to overcome the gravitational pull of the sun. 32.55. IDENTIFY and SET UP: The gravitational force is given by Eq.(12.2). Express the mass of the particle in terms of its density and volume. The radiation pressure is given by Eq.(32.32); relate the power output L of the sun to the intensity at a distance r . The radiation force is the pressure times the cross sectional area of the particle. EXECUTE: (a) The gravitational force is g 2 . mM r = The mass of the dust particle is 3 4 3 . mV R ρρ Thus 3 g 2 4 . 3 GM R F r ρπ = (b) For a totally absorbing surface rad . I p c = If L is the power output of the sun, the intensity of the solar radiation a distance r from the sun is 2 . 4 L I r = Thus rad 2 . 4 L p cr = The force rad F that corresponds to rad p is in the direction of propagation of the radiation, so rad rad , Fp A = where 2 A R = is the component of area of the particle perpendicular to the radiation direction. Thus 2 2 rad () . 44 L LR FR cr cr (c) rad g = 32 4 34 R L R r = 43 and 1 6 L L RR cc G M 26 8 3 11 2 2 30 3(3.9 10 W) 16(2.998 10 m/s)(3000 kg/m )(6.673 10 N m / kg ) (1.99 10 kg) R × = × 7 1.9 10 m 0.19 m. R μ = EVALUATE: The gravitational force and the radiation force both have a 2 r dependence on the distance from the sun, so this distance divides out in the calculation of R . (d) rad rad 23 g 33 . 1 6 FL R r L F Fc rG m R c G M R is proportional to 2 R and g F is proportional to 3 , R so this ratio is proportional to 1/ R . If 0.20 m R < then rad g > and the radiation force will drive the particles out of the solar system. 32.56. IDENTIFY: The electron has acceleration 2 v a R = . SET UP: 19 1 eV 1.60 10 C . An electron has 19 1.60 10 C qe × .
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Electromagnetic Waves 32-15 EXECUTE: For the electron in the classical hydrogen atom, its acceleration is 22 1 9 1 22 2 2 31 11 1 2 2(13.6 eV)(1.60 10 J eV) 9.03 10 m/s .
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941_PartUniversity Physics Solution - 32-14 32.54. Chapter...

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