# lecture11 - Orbit propagation the Kepler problem With an...

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Orbit propagation: the Kepler problem With an understanding of Kepler’s equation, we can solve the position of a satellite for a given time. This is called orbit propagation . In the two-body model, it is sometimes called the Kepler problem . With perturbations and other forces it can get quite complicated, but for two-body motion, we have the necessary techniques in hand. L. Healy – ENAE404 – Spring 2007 – Lecture 11 (Mar. 1) 1

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Cartesian two-body propagation Ref.: Vallado Section 2.3. Assuming we wish to start and end with Carte- sian state r 0 , ˙ r 0 r , ˙ r , (use subscript 0 to indicate initial, or epoch values) there are two ways to approach the problem: Conversion to elements, application of Ke- pler’s equation, then conversion back to Cartesian state. Diagram of process. Direct propagation with Lagrange coeffi- cients f and g functions. L. Healy – ENAE404 – Spring 2007 – Lecture 11 (Mar. 1) 2
Worked example: convert elements Suppose we have a satellite which has Carte- sian IJK r = 26981 . 28850 . 13174 . km ˙ r = - 1 . 2784 0 . 83017 0 . 57114 km / s What is the position and velocity 24 hours later? L. Healy – ENAE404 – Spring 2007 – Lecture 11 (Mar. 1) 3

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Convert to orbital elements By now we are familiar with this process: com- pute the quantities h = r × ˙ r =5540 . 7 ˆ I - 32252 . ˆ J + 59281 . ˆ K km 2 / s e = ˙ r × h μ - ˆ r = - 0 . 47829 ˆ I - 0 . 49479 ˆ J - 0 . 22448 ˆ K ˆ n = K × h =0 . 98556 ˆ I + 0 . 16932 ˆ J - 0 . 0000 ˆ K and then compute the elements a =24164 . km e =0 . 72386 i =28 . 900 ω = - 140 . 08 Ω =9 . 7481 θ = - 179 . 03 L. Healy – ENAE404 – Spring 2007 – Lecture 11 (Mar. 1) 4
Mean anomaly Now let’s compute the mean anomaly and mean motion, which are necessary for doing the prop- agation. First, compute the eccentric anomaly E 0 = arctan 1 - e 2 sin θ, e + cos θ = - 3 . 0991 rad . Then, compute the mean anomaly (remember: radians only!) M 0 = E 0 - e sin E 0 = - 3 . 0683 rad . Finally, compute the mean motion from the semimajor axis, n = μ a 3 = 398600 (24164 . km) 3 = 1 . 68080 × 10 - 4 rad / s . L. Healy – ENAE404 – Spring 2007 – Lecture 11 (Mar. 1) 5

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Propagate forward The time span we wish to cover is 24 hours The change in the mean anomaly during this interval can be computed using this time and the mean motion, Δ M = n Δ t =1 . 68080 × 10 - 4 rad / s · 86400 s =14 . 522 rad . This is more than one revolution, so as usual we can normalize this, Δ M = 1 . 9557 rad.
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