lecture12 - Recalculate example with f and g functions Once...

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Recalculate example with f and g functions Once again, suppose we have a satellite which has Cartesian IJK r 0 = 26981 . 28850 . 13174 . km ˙ r 0 = - 1 . 2784 0 . 83017 0 . 57114 km / s What is its state 24 hours later? L. Healy – ENAE404 – Spring 2007 – Lecture 12 (Mar. 6) 1
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Solve with f and g We will use the eccentric anomaly formulation of the f and g functions. First, we know the semimajor axis and eccentric anomalies at the start and end, from the previous calculation; if they weren’t known, they would have to be calculated: a =24164km E 0 = - 3 . 099rad E = - 1 . 8150rad However, we cannot use the value of E as is. Recall that on the way to calculating E , we normalized the change in mean anomaly from Δ M = 14 . 522 to Δ M = 1 . 9557, reasoning that multiples of 2 π can be removed. Unfor- tunately, here we cannot do that because E appears secularly, that is, not as the argument to a trig function, but in a polynomial. L. Healy – ENAE404 – Spring 2007 – Lecture 12 (Mar. 6) 2
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Recompute E ; solve for f Recompute E : re-solve Kepler’s equation Newton-Raphson with Δ M = 14 . 522, or restore 4 π to the previous E calculation. Either way, we get E = 10 . 7513rad. With this E , the change is Δ E = E - E 0 = 13 . 8504rad. We can use these in the f and g functions. So we compute f f =1 - a r 0 (1 - cosΔ E ) =1 - 24160 41640 (1 - cos(13 . 85)) =0 . 5838 . Note for Δ E we keep 4 digits after the decimal point because the multiples of 2 π we’re keeping don’t aid precision. L. Healy – ENAE404 – Spring 2007 – Lecture 12 (Mar. 6) 3
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Compute g Next compute g , which additionally needs Δ t = 86400sec, g t - v u u t a 3 μ E - sinΔ E ) =86400 - s 24160 3 398600 (13 . 85 - sin13 . 85) =9703 . s Here’s where the non-normalization of Δ E is very important: if you remove 4 π , you get g = 84467 . 1s, which is wildly incorrect! At this point, we must calculate the new posi- tion vector, because its magnitude r is required to compute ˙ f and ˙ g . L. Healy – ENAE404 – Spring 2007 – Lecture 12 (Mar. 6) 4
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Compute r r = f r 0 + g ˙ r 0 =0 . 5838 26981 28850 13174 9703 - 1 . 2784 0 . 83017 0 . 57114 = 3348 . 24900
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lecture12 - Recalculate example with f and g functions Once...

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