MIT OpenCourseWare http://ocw.mit.edu 16.346 Astrodynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Lecture 30 Effect of J2 on a Satellite Orbit of the Earth #10.6 Variational Equations using the Disturbing Function Recall the variational equatioms ds ∂s ∂s dα 0 ∂s dα 0 = + = Fs += = dt ∂t ∂ α dt ad ⇒ ∂ α dt ad which we can write as T ∂r dα ∂v ∂r dα = 0= 0 ∂α dt ∂α ∂α dt or T T∂v dα ∂r ∂v dα ∂r ∂α dt = ad ∂α ∂α dt = ∂α ad If we use the gradient of the disturbing function R for the disturbing acceleration ∂R aT = d ∂r then we have T T T T T∂r ∂r ∂R ∂R ∂r ∂R ∂ α ad = ∂ α ∂r = ∂r ∂ α = ∂ α so that T T T∂r ∂v ∂v ∂r dα ∂R ∂α ∂α − ∂α ∂α dt = ∂ α Lagrange Matrix L Therefore, the variational equation using the matrix L is Tdα ∂RL = dt ∂ α The Lagrange Matrix is skew-symmetric, i.e., L = −LT . Because of the skew-symmetry, there are only 15 elements to calculate and only 6 of these are different from zero. Lagrange’s Planetary Equations dΩ1 ∂R da 2 ∂R == dt nab sin i ∂i dt na ∂λ di 1 ∂R cos i∂R de b∂Rb2 ∂R dt = − nab sin i ∂Ω+ nab sin i ∂ω dt = − na3e ∂ω + na4e ∂λ dω cos i∂R b∂R dλ 2 ∂R b2 ∂R dt = − nab sin i ∂i + na3e ∂e dt = − na ∂a − na4e ∂e 16.346 Astrodynamics Lecture 30 Effect of the J2 Term on Satellite Orbits Gravitational potential function of the earth r kGm Gm ∞eqV (r, φ)= r − rJk rPk(cos φ) (8.92) k=2 = R where the angle φ is the colatitude with cos φ = i r · i z and i = [cos Ω cos(ω + f) − sin Ω sin(ω + f) cos i] i r x + [sin Ω cos(ω + f) + cos Ω sin(ω + f) cos i] i y Problem 3–21 + sin(ω + f) sin i i z Hence cos φ = sin(ω + f) sin i so that GmJ2r2 R = − 2p3 eq (1 + e cos f )3[3 sin2(ω + f) sin2 i − 1] + O[(r eq /r)3] 2π1 R = RdM where dM = ndt and r 2df = hdt 2π 0 1 2π n µJ2r2 =2π 0 hRr2df =4a3(1 − eq e2)32 (2 − 3 sin2 i) Averaged Variational Equations dΩ1 ∂R da = =0 dt nab sin i ∂i dt di de =0 =0 dt dt dω cos i∂R b∂R dλ 2 ∂R b2 ∂R dt = − nab sin i ∂i + na3e ∂e dt = − na ∂a − na4e ∂e For the Earth Problem 10–12 rrdΩ= − 3 J2 eq 2 n cos i = −9.96 eq 3.5 (1 − e 2)−2 cos i degrees/day dt 2 p a dω 3 r 2 r 3.5 dt =4J2 p eq n(5 cos2 i − 1)=5.0 a eq (1 − e 2)−2(5 cos2 i − 1) degrees/day Coefficients of the earth’s gravitational potential (×106 ) J2 =1, 082.28 ± 0.03 J5 = −0.2 ± 0.1 J3 = −2.3 ± 0.2 J6 =1.0 ± 0.8 J4 = −2.12 ± 0.05 16.346 Astrodynamics Lecture
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