z = 0. [8] z = 0. R.. 1 [7]. ( λ o oxyz 1 ox ( oy ( oz SN ω ω = iω cos λ + kω sin λ ω = rad/s (24. 1 m ν 0 o f = mkv k mẍ = 2mωẏ s

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34 4 2012 8 1 2 ( 710048.... -- O313.1 A 1000-0879(201204-049-08 DOI 10.6052/1000-0879-12-016 Abstract RESEARCH ON A REMOTE PROJECTILE CONSIDERING THE AIR RESISTANCE AND THE EARTH ROTATION 1 WANG Anxiang 2 ZHANG Xiaojun ZHAI Xuejun CHANG Hongfang (School of Science, Xi an Polytechnic University, Xi an 710048, China Remote projectile landing time and location on the earth s surface were calculated using two approaches, becase of the calculation defects in the rectangular Cartesian coordinate system connected to the earth s surface. First, we established a rectangular Cartesian coordinate system on the earth s surface, and then calculated the analytical solution for arbitrary projection angles considering the air resistance and the earth rotation. The conversion relation between the rectangular Cartesian coordinate system and the geocentric coordinate system was obtained. The landing time and location of the remote projectile were calculated under the condition that the distance between the projectile and the earth s core is equal to the earth radius R. Second, the position, velocity and acceleration of the projectile was described in the spherical coordinates system, the landing information of the projectile was obtained by numerical calculation. The calculation results from the two methods are very close, indicating that both the two methods are feasible for obtaining information of the projectile. Key words projectile, Coriolis force, Runge-Kutta method, air resistance [1-7] [1-3]. [4-5]. [6-7].. 2012 01 10 1 2012 05 02. 1 (11JK0557, (2010JG48 (2012JM1011. 2. E-mail: waxiang0921@126.com

50 2012 34 z = 0. [8] z = 0. R.. 1 [7]. ( λ o oxyz 1 ox ( oy ( oz SN ω ω = iω cos λ + kω sin λ ω = 7.3 10 5 rad/s (24. 1 m ν 0 o f = mkv k mẍ = 2mωẏ sin λ mkẋ (1a mÿ = 2mω(ẋ sin λ + ż cos λ mkẏ (1b m z = 2mωẏ cos λ mkż mg (1c t = 0 x = y = z = 0 ẋ = ν 0 cos α ẏ = ν 0 cos β ż = ν 0 cos γ. α, β γ cos 2 α + cos 2 β + cos 2 γ = 1 α, β γ. (1 ẋ = 2ωy sin λ kx + ν 0 cos α 3 2 ẏ = 2ω(x sin λ + z cos λ ky + ν 0 cos β ż = 2ωy cos λ kz gt + ν 0 cos γ (2a (2b (2c (2a (2c (1b (2b (x sin λ + z cos λ ÿ + 2kẏ + (k 2 + 4ω 2 y = 2ωgt cos λ 2ων 0 cos α sin λ 2ων 0 cos γ cos λ + kν 0 cos β(3 (3 y = e kt (c 1 cos 2ωt + c 2 sin 2ωt+ 2gω cos λ ( k 2 + 4ω 2 t + 4gkω cos λ q k 2 + 4ω 2 q = kν 0 cos β 2ων 0 cos α sin λ 2ων 0 cos γ cos λ (4 ( 4gkω cos λ / c 1 = k 2 + 4ω 2 q (k 2 + 4ω 2 (5 c 2 = 1 [ ( 4gkω cos λ / ν 0 cos β + k 2ω k 2 + 4ω 2 q (k 2 + 4ω 2 2gω cos λ ] k 2 + 4ω 2 (6 (4 (2a (2c [ x = c 1 sin λ sin 2ωt c 2 sin λ cos 2ωt+ 4ω 2 g sin λ cos λ ( te kt k 2 + 4ω 2 k 2ω sin λ k(k 2 + 4ω 2 ν 0 k cos α(ekt 1 + c 2 sin λ+ ekt k 2 + ( 4gkω cos λ q k 2 + 4ω 2 (e kt 1+ 1 k 2 4ω2 g sin λ cos λ ] k 2 + 4ω 2 e kt (7 [ z = c 1 cos λ sin 2ωt c 2 cos λ cos 2ωt+ 4ω 2 g cos 2 λ ( te kt k 2 + 4ω 2 k 2ω cos λ k(k 2 + 4ω 2 ekt k 2 + ( 4gkω cos λ q k 2 + 4ω 2 (e kt 1

4 51 ( te kt g k ekt k 2 + ν 0 cos γ (e kt 1+ k c 2 cos λ + 4ω2 g cos 2 λ k 2 + 4ω 2 1 k 2 g k 2 ] e kt (8 (4 (7 (8. O XY Z X = R cos λ cos ϕ + x Y = R cos λ sin ϕ + y Z = R sin λ + z (11 2 O XY Z 2. oxyz O XY Z. (4 A O XY Z [ X = R cos λ cos ϕ + x cos θ+ ] z cos 2 θ cos ϕ + y cos 2 + ϕ [ Y = R cos λ sin ϕ + y cos ϕ + x cos θ+ ] z cos 2 θ cos 2 ϕ Z = R sin λ + z cos θ + x cos 2 + θ (12 2 A oxyz (x y z. (1 1 oxyz oy θ ox y z A x = x cos θ + z cos 2 θ y = y (9 z = z cos θ + x cos 2 + θ λ = 45 ϕ = 45 h = 0 m, ν 0 = 5.0 km/s α = 45, β = 90, γ = 45 k = 1 mn s/(kg m 3 6. z = 0 t = 650.76 s. R t = 737.52 s.. 6 z = 0 t = 650.76 s. (2 2 ox y z oz ϕ ox y z x = x cos ϕ + y cos 2 + ϕ y = y cos ϕ + x cos 2 ϕ z = z = z cos θ + x cos 2 + θ (10 (3 3 ox y z 3

52 2012 34 4 6 z = 0 5 R 1 3 λ = 45 ϕ = 45 h = 0 m.... 1 1 k = 1 mn s/(kg m R / z = 0 (km s 1 /s /( /( /m /s /( /( 5.0 650.76 31.307 613 44.325 145 220 543.74 737.52 28.192 156 44.229 425 4.0 530.22 35.312 789 44.585 795 105 401.34 578.16 33.770 227 44.543 359 3.0 405.42 38.953 431 44.790 432 39 119.62 427.52 38.352 887 44.776 594 2.0 275.88 42.011 780 44.925 770 9 128.28 283.12 41.863 922 44.923 240 1.0 140.98 44.170 122 44.989 001 683.87 142.0 44.158 433 44.988 891 2 k = 10 mn s/(kg m R / z = 0 (km s 1 /s /( /( /m /s /( /( 5.0 455.90 41.935 460 44.949 380 9 593.56 466.20 41.850 112 44.950 079 4.0 379.88 42.565 270 44.960 419 5 994.35 386.58 42.509 846 44.960 728 3.0 300.82 43.214 619 44.972 465 3 184.58 304.68 43.183 227 44.972 534 2.0 216.18 43.885 689 44.984 985 1 223.70 217.92 43.872 366 44.984 966 1.0 120.62 44.556 295 44.995 877 193.70 121.02 44.553 842 44.995 868

4 53 3 k = 100 mn s/(kg m R / z = 0 (km s 1 /s /( /( /m /s /( /( 5.0 370.76 44.683 013 44.999 526 98.89 371.76 44.682 137 44.999 528 4.0 298.60 44.746 279 44.999 619 64.21 299.26 44.745 699 44.999 620 3.0 226.46 44.809 595 44.999 710 35.41 226.82 44.809 280 44.999 711 2.0 154.30 44.872 999 44.999 802 16.41 154.46 44.872 858 44.999 803 1.0 82.12 44.936 489 44.999 895 5.03 82.18 44.936 435 44.999 895 v = 5.0 km/s 0.095 72 ( 10.646 6 km. 2 3... k 0. 7 8 λ = 45 ϕ = 45 h = 0 m ν 0 = 5.0 km/s. 8 (k = 100 mn s/(kg m. 3. 9. r = r(te r (13 v = ṙe r + e θ r θ + e ϕ r ϕ sin θ = ν r e r + ν θ e θ + ν ϕ e ϕ (14 a = ( r r θ 2 r ϕ 2 sin 2 θe r + (r θ + 2ṙ θ r ϕ 2 sin θ cos θe θ + (r ϕ sin θ + 2ṙ ϕ sin θ + 2r θ ϕ cos θe ϕ (15 ma = F mge r 2mω ν (16 7 (k = 10 mn s/(kg m

54 2012 34 (20 r r θ 2 r ϕ 2 sin 2 θ = bṙ g + 2ωr ϕ sin 2 θ r θ + 2ṙ θ r ϕ 2 sin θ cos θ = 2ωr ϕ cos θ sin θ br θ r ϕ sin θ + 2ṙ ϕ sin θ + 2r θ ϕ cos θ = br ϕ sin θ 2(ωr θ cos θ + ωṙ sin θ] (21 9 2mω ν = 2m(e r ω cos θ e θ ω sin θ (ṙe r + r θe θ + r ϕ sin θe ϕ = 2m[(ωr θ cos θ + ωṙ sin θe ϕ ωr ϕ cos θ sin θe θ ωr ϕ sin 2 θe r ] (17 F = mbν = mb(ṙe r + r θe θ + r ϕ sin θe ϕ = mbṙe r mbr θe θ mbr ϕ sin θe ϕ (18 F mge r 2mω ν = -- -- [9]. 4 λ = 45 ϕ = 45 h = 0 m α = 45 β = 90 γ = 45 --. 4 1 3 R k -- z = 0. -- R. 10 11 ν 0 = 5.0 km/s k = 1 mn s/(kg m k = 100 mn s/(kg m. ( mbṙ mg + 2mωr ϕ sin 2 θe r + 4 -- (2mωr ϕ cos θ sin θ mbr θe θ + [ mbr ϕ sin θ 2m(ωr θ cos θ + ωṙ sin θ]e ϕ (19 (14, (15 (19 (16 m( r r θ 2 r ϕ 2 sin 2 θ = mbṙ mg + 2mωr ϕ sin 2 θ m(r θ + 2ṙ θ r ϕ 2 sin θ cos θ = 2mωr ϕ cos θ sin θ mbr θ m(r ϕ sin θ + 2ṙ ϕ sin θ + 2r θ ϕ cos θ = mbr ϕ sin θ 2m(ωr θ cos θ + ωṙ sin θ] (20 / / (mn s/(kg m (km s 1 /s /( /( 5.0 722.01 30.456 309 44.323 825 4.0 572.57 34.730 184 44.579 642 1 3.0 426.04 38.668 124 44.786 660 2.0 282.90 41.929 148 44.924 786 1.0 141.99 44.162 800 44.988 948 5.0 465.72 41.978 444 44.953 108 4.0 386.34 42.585 520 44.962 391 10 3.0 304.58 43.220 383 44.973 260 2.0 217.89 43.885 025 44.985 166 1.0 121.03 44.555 426 44.995 885 5.0 371.76 44.683 841 44.999 535 4.0 299.25 44.746 792 44.999 624 100 3.0 226.82 44.809 883 44.999 713 2.0 154.47 44.873 115 44.999 804 1.0 82.17 44.936 504 44.999 896

4 55 10 (k = 1 mn s/(kg m 11 λ λ -- λ 1 4 k ν 0 --. k ν 0 λ λ -- 5 k k = 100 mn s/(kg m k = 10 mn s/(kg m.. k = 1 mn s/(kg m ν = 3.0 km/s ν < 3.0 km/s. ν 0 k. (k = 100 mn s/(kg m 5 / / /( /( (mn s/(kg m (km s 1 ( -- ( /( /% 5.0 30.456 309 28.192 156 45 15.568 4.0 34.730 184 33.770 227 45 9.347 1 3.0 38.668 124 38.352 887 45 4.979 2.0 41.929 148 41.863 922 45 2.124 1.0 44.162 800 44.158 433 45 0.522 5.0 41.978 444 41.850 112 45 4.247 4.0 42.585 520 42.509 846 45 3.134 10 3.0 43.220 383 43.183 227 45 2.088 2.0 43.885 025 43.872 366 45 1.135 1.0 44.555 426 44.553 842 45 0.356 5.0 44.683 841 44.682 137 45 0.539 4.0 44.746 792 44.745 699 45 0.432 100 3.0 44.809 883 44.809 280 45 0.317 2.0 44.873 115 44.872 858 45 0.203 1.0 44.936 504 44.936 435 45 0.109

56 2012 34 4. R.. 1 2009 28(12 5-7 (Yin Xinguo, Gong Pifeng, Zhu Mengzheng, et al Influence of air resistance on a falling body defecting eastward College Physics, 2009 28(12 5-7(in Chinese 2 2001 23(2 68-69 (Lou Zhimei. A solution of expression for drift of a free-falling body in the northern hemisphere after taking the influence of air drag force into account. Mechanics and Engineering, 2001, 23(2: 68-69 (in Chinese 3 2005 24(8 61-63 (Li Tuo A new drift solution of the free-falling body and the up-cast College Physics, 2005 24(8 61-63 (in Chinese 4 2003 28(4 576-579 (Lou Zhimei The approximative solution of differential kinematics equation containing a cubic veloccity nonlinear term for a projectile in the northern hemisphere Journal of Southwest China Normal University (Natural Science, 2003 28(4 576-579 (in Chinese 5 2002 16(3 29-34 (Zhu Rong The systematic study of the influence of rofation of the earth and the air resistance to the projectile Journal of Shanxi Teacher s University Natural Science Edition, 2002 16(3 29-34 (in Chinese 6 2010 23(4 509-513 (Wang Anxiang, Li Jijun, Zhai Xuejun, et al The effect of Coriolis force on vertical throw-up motion Basic Sciences Journal of Textile Universities, 2010 23(4 509-513 (in Chinese 7 Matlab 2010 28(6 106-110 (Wang Anxiang, Li Jijun, Zhai Xuejun, et al Study on the effect of coriolis force on projectile motion based on matlab Journal of Shanxi University of Science & Technology, 2010 28(6 106-110 (in Chinese 8 2007 31(5 377-386(Fan Changzeng, Wang Wenkui. Review on the electromagnetic railgun. Journal of Yanshan University, 2007 31(5 377-386 (in Chinese 9 -- 2008 28(2: 62-65 (Li Xiayun, Chen Chuanmiao. Solving nonliner equations by using runge-kutta method. Mathematicaltheory and Applications, 2008 28(2: 62-65 (in Chinese ( : (..... ( 98, 2012 8.