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AE 245 homework #6 solutions Tim Smith 15 March 2000 1 Problem 1 The Hubble Space Telescope was placed in a circular orbit about the Earth at an altitude of 330 mi. Identify the following orbital characteristics: 1. What is the circular velocity of the Hubble Telescope? 2. What is the orbital period of the Hubble Telescope? 3. What are the specific total energy and specific angular momentum of the Hubble Telescope? 1.1 Circular velocity In a circular orbit, the gravitational acceleration is a centripetal acceleration; thus, µ r2 = v2 r (1) where µ = GM is the Earth’s gravitational parameter. Rearranging Eqn. 1 gives the circular speed v= rµ (2) r Since the Earth’s mean equatorial radius r = 3963:2 miles, the orbital radius r = (3963:2 + 330) mi = 2:2668 107 ft, so s v= 1:4076 1016 ft 2:2668 107 s = 24; 919 ft s 1.2 Orbital period Since a circular path has a length 2πr, a steady speed of v gives an orbital period T = 2πr v 1 (3) Substituting Eqn. 2 into 3 and rearranging gives T = 2πr3=2 pµ = 2π(2:2668 107)3=2 p s = 5715:6 s = 1 hr 35 min 15:6 s 1:4076 1016 (4) 1.3 Specific total energy and angular momentum The specific total energy for any orbit is E v2 2 , µr (5) Substituting Eqn. 2 for a circular orbit, the Hubble Space Telescope specific total energy µ E =, 2r = 1016 , 21(24076 2668 107) : : ft2 s2 2 = 3:1048 108 fts2 (6) Likewise, the specific angular momentum for any orbit is h ~r ~v ~ (7) which, for our circular orbit, has the constant scalar value h = rv = 2:2668 107(1:4076 1016) ft2 =s = 3:7555 1023 ft2 =s (8) 2 Problem 2 Consider a satellite in circular orbit about the Earth. 1. Plot the gravitational acceleration of the satellite due to Earths gravity as a function of the altitude of the satellite as measured from the surface of the Earth. Scale your plot so that the altitude goes from 100 km to 10,000 km. 2. On the same graph, plot the gravitational acceleration of the satellite due to Earths gravity and the (maximum) gravitational acceleration of the satellite due to the Suns gravity as a function of the altitude of the satellite as measured from the surface of the Earth. Scale your plot so that the altitude goes from 100 km to 10 million km. 3. Use your graphs in part (b) to decide (approximately) how large the radius of acircular orbit can be so that the two body model of the Earth and satellite is accurate to within 2%. 2.1 Earth’s gravitational acceleration The gravitational acceleration due to the Earth on the satellite is g = 2 µ r2 (9) where r is the sum of the altitude z and the mean equatorial radius of the Earth r . Figure 1 is produced by the following IDL code: 10 acceleration (m/s^2) 8 6 4 2 0 0 2000 4000 6000 altitude (km) 8000 10000 Figure 1: Gravitational acceleration of satellite due to the Earth as a function of altitude. pro hw6_2a ; Define fundamental constants & conversions. ; Source: Bate, Mueller & White, _Fundamentals of Astrodynamics_, 1971. mu_e r_e = 3.986012e14 = 6.378145e6 ; Earth’s GM, mˆ3/sˆ2 ; mean Earth radius, m ; Create altitude array. h_min h_max bins range h h = = = = = = 1.e5 1.e7 401 alog(h_max/h_min) range*findgen(bins)/(bins-1) h_min*exp(h) ; minimum altitude, m ; maximum altitude, m ; number of array bins ; altitude, m ; Calculate and plot acceleration due to Earth’s gravity. g_e xtitle ytitle file scale xsz = mu_e/(r_e + h)ˆ2 = = = = = ’altitude (km)’ ’acceleration (m/sˆ2)’ ’fig6_2a.eps’ 1.0 4.0*scale ; x-axis title ; y-axis title ; file name 3 ysz = 3.0*scale set_plot, ’ps’ device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, h/1.e3, g_e, xtitle=xtitle, $ ytitle=ytitle, font=0 ; device, /close ; set plot device to PostScript ; open file & save EPS ; close the file end 2.2 Solar gravitational acceleration The maximum solar gravitational acceleration for a given altitude from the Earth’s surface occurs when the satellite is closest to the Sun. This happens when the satellite is on a line between the Earth and the Sun, so that the radius of the satellite from the Sun’s center r = a , r where a is the distance between the Earth’s and Sun’s centers (1 AU, or 1:4960 1011 m). Thus, the solar gravitational acceleration g = µ (a , r)2 Figure 2 is produced by the following IDL code: pro hw6_2b ; Define fundamental constants & conversions. ; Source: Bate, Mueller & White, _Fundamentals of Astrodynamics_, 1971. mu_e r_e mu_s au = = = = 3.986012e14 6.378145e6 1.3271544e20 1.4959965e11 ; ; ; ; Earth’s GM, mˆ3/sˆ2 mean Earth radius, m solar GM, mˆ3/sˆ2 Earth-Sun distance, m ; Create altitude array. h_min h_max bins range h h = = = = = = 1.e5 1.e10 401 alog(h_max/h_min) range*findgen(bins)/(bins-1) h_min*exp(h) ; minimum altitude, m ; maximum altitude, m ; number of array bins ; altitude, m ; Calculate and plot acceleration due to Earth’s & solar gravity. g_e = mu_e/(r_e + h)ˆ2 ; Earth’s accel, m/sˆ2 4 (10) 102 acceleration (m/s^2) 100 10-2 10-4 10-6 102 103 104 105 altitude (km) 106 107 Figure 2: Terrestrial (solid line) and solar (dotted line) gravitational acceleration as a function of altitude. g_s xtitle ytitle file scale xsz ysz = mu_s/(au - r_e - h)ˆ2 = = = = = = ; solar accel, m/sˆ2 ’altitude (km)’ ’acceleration (m/sˆ2)’ ’fig6_2b.eps’ 1.0 4.0*scale 3.0*scale ; x-axis title ; y-axis title ; file name set_plot, ’ps’ device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, h/1.e3, g_e, xtitle=xtitle, $ ytitle=ytitle, font=0, /xlog, /ylog oplot,h/1.e3, g_s, linestyle=1 device, /close ; set plot device to PostScript ; open file & save EPS ; close the file end 2.3 Validity of two-body model The combined terrestrial and solar gravity model, g , g , is more accurate than the two-body model, g . The percent error is 5 ε= g , (g , g ) g , g = g g , g (11) Letting ε = 0:02 in Eqn. 11 gives the requirement that g = 51:0g (12) at our desired location. Substituting Eqn. 9 and 10 into 12 yields µ r2 = 51:0µ , r)2 (13) (a which can be rearranged to give a radius from Earth r= a p 1 + 51 0µ : (14) =µ Since r = r + z, the altitude for 2% error is 0 1 11 1 4960 10 q 51 0µ , r = @ q 51 0 1 371510 , 6 3781 106A m = 29 326 km a z= 1+ : : µ 1+ : ( : 3:9860105 11) : ; (15) 3 Problem 3 Assume that each of the planets Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune is in a circular orbit about the Sun. 1. Make a table showing the mean orbital radius (in km), the mean circular velocity (in km/sec), and the mean orbital period (in Earth years) for each of these planets. Check that the assumptions of a circular orbit for these planets is justified. 2. It is not accurate to assume that the planets Mercury and Pluto are in circular orbit about the Sun. Show this from data for their mean orbital radius (in km), mean circular velocity (in km/sec), and mean orbital period (in Earth years) for Mercury and Pluto. 3.1 Circular planetary orbits Table 1 compares tabulated data for the planets1 to values calculated by the circular orbit equations for circular velocity vc = rµ r and period 1 R. R. Bate, D. D. Mueller, and J. E. White, Fundamentals of Astrodynamics, Dover Publications, New York, 1971, p. 361. 6 (16) s T = 2π r3 µ (17) where the Sun’s gravitational parameter µ GM = 1:3272 1011 km3 /s2 . Planet Venus Earth Mars Jupiter Saturn Uranus Neptune Table 1: Mean circular orbits for the given planets. From Bate, Mueller & White Calculated Mean orbital Mean circular Mean orbital Mean circular Mean orbital radius (km) speed (km/s) period (yr) speed (km/s) period (yr) 1:081 108 35.04 0.615 35.04 0.6147 1:495 108 29.79 1.000 29.80 0.9997 2:278 108 24.14 1.881 24.14 1.880 7:780 108 13.06 11.86 13.06 11.87 1:426 109 9.65 29.46 9.647 29.45 2:868 109 6.80 84.01 6.803 84.00 4:494 109 5.49 164.8 5.434 164.8 The calculated values are all within 0.05% of the tabulated values, indicating that a circular orbit assumption works pretty well for all the planets except Mercury and Pluto. 3.2 Validity of circular orbits Let’s continue Table 1 for Mercury and Pluto: Planet Mercury Pluto Table 2: Mean circular orbits for Mercury and Pluto. From Bate, Mueller & White Calculated Mean orbital Mean circular Mean orbital Mean circular Mean orbital radius (km) speed (km/s) period (yr) speed (km/s) period (yr) 5:790 107 47.87 0.241 47.88 0.2410 5:896 109 4.740 247.7 4.745 247.6 At first glance, this seems to show that the circular orbit assumption is also valid for Mercury and Pluto! One reason for this may be that the standard mean radii are calculated from the orbital period. Mercury and Pluto don’t have very circular orbits because their orbital eccentricities are too far from zero. Table 3 lists the eccentricities for all the planets. 7 Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Table 3: Mean orbital parameters for the planets. Mean orbital Mean circular Mean orbital Eccentricity radius (km) speed (km/s) period (yr) 7 5:790 10 47.87 0.241 0.2056 1:081 108 35.04 0.615 0.0068 1:495 108 29.79 1.000 0.0167 24.14 1.881 0.0934 2:278 108 7:780 108 13.06 11.86 0.0482 1:426 109 9.65 29.46 0.0539 2:868 109 6.80 84.01 0.0514 5.49 164.8 0.0050 4:494 109 5:896 109 4.740 247.7 0.2583 8