S11 PHY114 Poblem Set 8 S. G. Rajeev Apil 4, 2011 Due Monday 4 Ap 2011 1. Find the wavelength of the adio waves emitted by the Univesity of Rocheste adio station (88.5 Mhz)? The numbe of bits pe second tansmitted on a adio signal is about ten times less than the fequency of the wave. A wieless local aea netwok (WLAN) equies about 100 Mbits/sec. What fequency and wavelength of adio waves ae needed? Visible light has wavelength in the ange of 380-750nm (violet to ed). Find the coesponding fequencies. The Voyage 1 spacecaft is now at a distance of 117 AU (astonomical unit-the distance between the Eath and the Sun). If you send a signal to Voyage 1, how long would you have to wait fo a esponse? 2. A geosynchonous satellite obits at the same as ate as the Eath s otation, so that to an obseve on the Eath it will appea stationay in sky. At what latitude should this satellite be located in the sky? At what angle to the hoizontal should a satellite antenna at Rocheste have to be pointed, assuming that the longitude of the satellite is the same as ous? At what latitude would communication with a geosynchonous satellite become impossible?what is the time delay in a telephone signal which is bounced off a satellite in geoscynchonous obit? A quate-second delay would be noticeable. 3. Estimate the aveage optical powe output of the Sun, knowing that about 1000 Watts pe squae mete eaches the Eath. Aadiotelescopecandetectsignalsaslowas10 23 Wm 2. Assuming that the maximum powe of an alien adio tansmitte is about the same as that of a sta, what is the maximum distance fom which we can expect to ecieve a signal? 1
The United States consumes about 30 10 15 Wh (Watt-hous) of enegy in a yea. How lage a field of sola panels would it take to powe this usage? Assume a 10% efficiency fo sola panels. 4. Calculate the displacement cuent between the squae plates of side 9.0cm, ofacapacitoiftheelecticfieldischangingataateof1.8 10 6 V (ms) 1. Solutions 1. The wavelength of a adio wave with fequency 88.5MHz is, λ = c f = 3 108 ms 1 88.5 10 6 3.4m s 1 To tansmit 100Mbits/sec we need a fequency of oughly 1GHz which coesponds to a wavelength of λ = 3 108 10 0.3m 9 The fequency of light of wavelength 380nm (violet) is and of 750 nm (ed) is f = 3 108 380 10 9 7.8 1014 Hz f = 3 108 750 10 9 4.0 1014 Hz An AU is about 150 million km o about 150 10 9 m. The ound tip distance to the Voyage is 234AU 35 10 12 m The time it takes fo light to tavel this distance is 35 1012 3 10 12 10 4 s 33 hous. So it takes moe than a day to get 8 aesponsefomvoyage1. 2. The adius of obit of a geosynchonous satellite is detemined by Newtonian mechanics. Equating centipetal acceleation to the gavitational acceleation, GM E 2 = ω 2, ω = 2π T 3 = GM E 4π 2 T 2 The gavitational acceleation at the suface of the Eath is g E = GM E R 2 E 3 = g 2 E 4π 2 T 2 2
This basically Keple s thid law. Since we get = fo a peiod of obit of one day, = 2 ge R 1 3 E 4π 2 T 2 3 g E 9.8ms 2, 6.4 10 6 m T = 24 60 60 = 86.4 10 3 s 9.8 6.4 10 6 2 4π 2 1 3 86.4 10 3 2 3 = 9.8 (6.4) 2 1 3 86.4 4π 2 10 6 m 42 10 3 km This obit must be ove the equato in ode that the angula velocity of the satellite be oiented along the angula velocity of the Eath. Thus fom Rocheste antennas have to point south to see a satellite ove the Equato at ou longitude. The angle χ to the hoizontal is given by some tigonomety. Imagine a tiangle with vetices at the cente of the Eath (O), a point (P) on the Eath and the satellite (Q). The angle POQ is the latitude θ. The angle OPQ= φ is 90 plus the angle that the line of sight PQ makes with the hoizontal at P. φ = 90 + χ. The side OP is the adius of the Eath.The side OQ is the adius of the satellite s obit. By the law of sines of tigonomety But OQ sin OPQ = OP sin OQP. OQP = 180 (OPQ + POQ) = 180 (θ + φ). Thus sin OQP =sin[θ + φ]. 3
sin φ = sin[θ + φ] Thus φ is detemined as the solution to the equation That is sin φ = sin φ = sin[θ + φ] {sin θ cos φ + cos θ sin φ} =sinθ cot φ + cos θ cot φ = cos θ sin θ. cot[90 + χ] = cos θ sin θ Since tan χ = cos θ sin θ 0.15, θ 43 tan χ = 0.73 0.15 0.68 =0.85 χ = 40 At a latitude of accos[ ] 81 the satellite antenna would have to be hoizontal to see the satellite; at any highe latitude it becomes impossible, as the line of sight to the satellite is blocked by the Eath. In pactice, because of hills and buildings, communication becomes difficult above about sixty degees latitude. The distance to the satellite is given by the cosine fomula d = RE 2 + 2 2 cos φ = 1 2R 1 E cos φ + R2 2 E 2 4
Since and the delay is 0.15 we can appoximate 2 c d 2 42 106 3 10 8 =0.28s which is noticeable. 3. The adius of the Eath s obit is R 150 10 9 m. Hence the powe output P of the Sun is at least 4πR 2 times the powe pe unit aea of sunlight at the Eath. P =4π 150 10 9 2 4 10 26 W The Sun is a typical sta. If the aliens ae able to somehow tansmit as much powe as a sta, at a distance the powe ecieved pe squae mete will be 2 R2 Setting this equal to the powe that can be detected by a adio telescope p = 2 R2 = R = 150 10 9 p 10 23 2 1021 m 2 10 5 ly The size of ou galaxy is compaable, at a hunded thousand light yeas. So we should be able to detect such tansmission fom within the galaxy. Conveting fom Watt-hou pe yea to Watts, the powe consumption of the US is 3 10 16 1 24 365 =3.4 1012 W At 10% efficiency we would need an optical powe of 34TW to geneate enough powe. Assuming a squae of side L fo the sola panel, this gives L 2 =3.4 10 12 L = 160km Thus a squae sola panel field of side a hunded miles can powe the whole United States duing the day. The cost is pohibitive cuently, though. 5
4. The displacement cuent is, fo a capacito of side L I D 0 dφ E dt = 0 L 2 de dt I D =8.9 10 12 (0.09) 2 1.8 10 6 = 13 10 8 A This is too small to be obeved diectly. 6