I have a question? e2e = 51.89 (Page 22).
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02-23-2023, 07:53 AM
(This post was last modified: 02-23-2023 07:56 AM by EdS2.)
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RE: I have a question? e2e = 51.89 (Page 22).
Just a few notes, no helpful answer from me sorry...
θ angle in orbital plane measured from perigee point; also, true anomaly subscript 2e equivalent selected position in first orbit (where equivalent means accounting for the rotation of the earth since launch) (we can even write the subscripts like so: θ₂ₑ) So the worked example on p20 of the document (pdf page 22) is Case A, eastward launch.- For case A, eastward launch, the given launch position is ϕ₁=28.50° N., λ₁=279.45° E.; the selected position is ϕ₂=34.00° N., λ₂=241.00° E.; the number of orbital passes n is 3; and the equivalent selected longitude λ₂ₑ is 309.689° E. The values of the first approximations are [t(θ₂ₑ) - t(θ₁) ] ≈ 7.693 Δλ₁₋₂ₑ = 32.162 θ₂ₑ = 51.89 t(θ₂ₑ) = 12.702 ψ₁ = 70.468 i = 34.081 From Appendix B, page 18, we have cos(θ₂ₑ - θ₁) = sinϕ₂ sinϕ₁ + cosϕ₂ cosϕ₁ cosΔλ₁₋₂ₑ |
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Messages In This Thread |
I have a question? e2e = 51.89 (Page 22). - tom234 - 02-23-2023, 12:52 AM
RE: I have a question? e2e = 51.89 (Page 22). - Steve Simpkin - 02-23-2023, 04:44 AM
RE: I have a question? e2e = 51.89 (Page 22). - tom234 - 02-23-2023, 07:21 PM
RE: I have a question? e2e = 51.89 (Page 22). - Massimo Gnerucci - 02-23-2023, 07:32 PM
RE: I have a question? e2e = 51.89 (Page 22). - tom234 - 02-23-2023, 10:35 PM
RE: I have a question? e2e = 51.89 (Page 22). - EdS2 - 02-23-2023 07:53 AM
RE: I have a question? e2e = 51.89 (Page 22). - tom234 - 02-23-2023, 12:11 PM
RE: I have a question? e2e = 51.89 (Page 22). - tom234 - 02-23-2023, 12:15 PM
RE: I have a question? e2e = 51.89 (Page 22). - EdS2 - 02-23-2023, 01:38 PM
RE: I have a question? e2e = 51.89 (Page 22). - tom234 - 02-23-2023, 02:56 PM
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