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 J 2 on a Satellite Orbit of the Earth #10.6 Variational Equations using the Disturbing Function Recall the variational equatioms ∂s ∂s d α ds 0 = + = Fs + ad dt ∂t ∂ α dt =⇒ ∂s d α 0 = ad ∂ α dt which we can write as T ∂r dα =0 ∂α dt or T T ∂r ∂r ∂v dα = ad ∂α ∂α dt ∂α If we use the gradient of the disturbing function R for the disturbing acceleration ∂r dα =0 ∂α dt ∂v dα = ad ∂α dt ∂v ∂α adT = then we have so that ∂r ∂α T ∂r ad = ∂α T ∂R ∂r ∂R ∂r T ∂R ∂r = ∂r ∂ α T ∂R = ∂α T T T ∂v ∂R ∂v ∂r dα = − ∂α ∂α ∂α dt ∂α Lagrange Matrix L ∂r ∂α T Therefore, the variational equation using the matrix L is T ∂R dα = L dt ∂α The Lagrange Matrix is skew-symmetric, i.e., L = −L T . 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 1 dΩ ∂R = dt nab sin i ∂i 1 cos i ∂R di ∂R =− + dt nab sin i ∂Ω nab sin i ∂ω cos i ∂R b ∂R dω =− + 3 dt nab sin i ∂i na e ∂e 16.346 Astrodynamics 2 ∂R da = dt na ∂λ b ∂R b2 ∂R de =− 3 + 4 dt na e ∂ω na e ∂λ 2 ∂R b2 ∂R dλ =− − 4 dt na ∂a na e ∂e Lecture 30 Effect of the J2 Term on Satellite Orbits Gravitational potential function of the earth ∞ Gm Gm req k V (r, φ) = − Jk Pk (cos φ) r r r k=2 =R where the angle φ is the colatitude with cos φ = ir · iz and ir = [cos Ω cos(ω + f ) − sin Ω sin(ω + f ) cos i] ix + [sin Ω cos(ω + f ) + cos Ω sin(ω + f ) cos i] iy (8.92) Problem 3–21 + sin(ω + f ) sin i iz Hence cos φ = sin(ω + f ) sin i so that 2 GmJ2 req (1 + e cos f )3 [3 sin2 (ω + f ) sin2 i − 1] + O[(req /r)3 ] 2p3 2π 1 R= R dM where dM = n dt and r2 df = h dt 2π 0 2π 2 µJ2 req n 2 1 2 = Rr df = 3 (2 − 3 sin i) 3 2 2π 0 h 4a (1 − e ) 2 R=− Averaged Variational Equations 1 dΩ ∂R = dt nab sin i ∂i di =0 dt cos i ∂R b ∂R dω =− + 3 dt nab sin i ∂i na e ∂e For the Earth da =0 dt de =0 dt 2 ∂R b2 ∂R dλ =− − 4 dt na ∂a na e ∂e Problem 10–12 r 3.5 3 req 2 dΩ eq = − J2 n cos i = −9.96 (1 − e2 )−2 cos i degrees/day dt 2 p a r 3.5 3 req 2 dω eq = J2 n(5 cos2 i − 1) = 5.0 (1 − e2 )−2 (5 cos2 i − 1) degrees/day dt 4 p a Coefficients of the earth’s gravitational potential ( ×106 ) J2 = 1, 082.28 ± 0.03 J3 = −2.3 ± 0.2 J4 = −2.12 ± 0.05 16.346 Astrodynamics J5 = −0.2 ± 0.1 J6 = 1.0 ± 0.8 Lecture 30