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## Q. 5.12

Starting with the state vector determined in Example 5.11, use Algorithm 5.6 to improve the vector to five significant figures.

## Verified Solution

Step 1:

$r_2=\left\| r _2\right\|=\sqrt{5659.1^2+6533.8^2+3270.1^2}=9241.8 km$

$v_2=\left\| v _2\right\|=\sqrt{(-3.8800)^2+5.1156+(-2.2397)^2}=6.7999 km / s$

Step 2:

$\alpha=\frac{2}{r_2}-\frac{v_2^2}{\mu}=\frac{2}{9241.8}-\frac{6.7999^2}{398,600}=1.0154\left(10^{-4}\right) km ^{-1}$

Step 3:

$\left.v_r\right)_2=\frac{ v _2 \cdot r _2}{r_2}=\frac{(-3.8800) \cdot 5659.1+5.1156 \cdot 6533.8+(-2.2397) \cdot 3270.1}{9241.8}=0.44829 km / s$

Step 4:

The universal Kepler equation at times $t_1$ and $t_3$, respectively, becomes

$\sqrt{398,600} \tau_1=\frac{9241.8 .0 \cdot 44829}{\sqrt{398,600}} \chi_1^2 C\left(1.0040 \times 10^{-4} \chi_1^2\right)$

$+\left(1-1.0040 \times 10^{-4} \cdot 9241.8\right) \chi_1^3 S\left(1.0040 \times 10^{-4} \chi_1^2\right)+9241.8 \chi_1$

$\sqrt{398,600} \tau_3=\frac{9241.8 \cdot 0.44829}{\sqrt{398,600}} \chi_3^2 C\left(1.0040 \times 10^{-4} \chi_3^2\right)$

$+\left(1-1.0040 \times 10^{-4} \cdot 9241.8\right) \chi_3^3 S\left(1.0040 \times 10^{-4} \chi_3^2\right)+9241.8 \chi_3$

or

$631.35 \tau_1=6.5622 \chi_1^2 C\left(1.0040 \times 10^{-4} \chi_1^2\right)+0.072085 \chi_1^3 S\left(1.0040 \times 10^{-4} \chi_1^2+9241.8 \chi_1\right)$

$631.35 \tau_3=6.5622 \chi_3^2 C\left(1.0040 \times 10^{-4} \chi_3^2\right)+0.072085 \chi_1^3 S\left(1.0040 \times 10^{-4} \chi_3^2+9241.8 \chi_3\right)$

Applying Algorithm 3.3 to each of these equations yields

$\chi_1=-8.0908 \sqrt{ km }$

$\chi_3=8.1375 \sqrt{ km }$

Step 5:

$f_1=1-\frac{\chi_1^2}{r_2} C\left(\alpha \chi_1^2\right)=1-\frac{(-8.0908)^2}{9241.8} \cdot \overbrace{C\left(1.0040 \times 10^{-4}[-8.0908]^2\right)}^{0.49973}=0.99646$

$g_1=\tau_1-\frac{1}{\sqrt{\mu}} \chi_1^3 S\left(\alpha \chi_1^2\right)$

$=-118.1-\frac{1}{\sqrt{398,600}}(-8.0908)^3 \cdot \overbrace{S\left(1.0040 \times 10^{-4}[-8.0908]^2\right)}^{0.16661}=-117.96 s$

and

$f_3=1-\frac{\chi_3^2}{r_2} C\left(\alpha \chi_3^2\right)=1-\frac{8.1375^2}{9241.8} \cdot \overbrace{C\left(1.0040 \times 10^{-4} \cdot 8.1375^2\right)}^{0.49972}=0.99642$

$g_3=\tau_3-\frac{1}{\sqrt{\mu}} \chi_3^3 S\left(\alpha \chi_3^2\right)$

$=-118.1-\frac{1}{\sqrt{398,600}} 8.1375^3 \cdot \overbrace{S\left(1.0040 \times 10^{-4} \cdot 8.1375^2\right)}^{0.16661}=119.33$

It turns out that the procedure converges more rapidly if the Lagrange coefficients are set equal to the average of those computed for the current step and those computed for the previous step. Thus, we set

$f_1=\frac{0.99648+0.99646}{2}=0.99647$

$g_1=\frac{-117.97+(-117.96)}{2}=-117.96 s$

$f_3=\frac{0.99642+0.99641}{2}=0.99641$

$g_3=\frac{119.33+119.33}{2}=119.34 s$

Step 6:

$c_1=\frac{119.33}{(0.99647)(119.33)-(0.99641)(-117.96)}=0.50467$

$c_3=\frac{-117.96}{(0.99647)(119.33)-(0.99641)(-117.96)}=0.49890$

Step 7:

$\rho_1=\frac{1}{-0.0015198}\left(-782.15+\frac{1}{0.50467} 787.72-\frac{0.49890}{0.50467} 787.31\right)=3650.6 km$

$\rho_2=\frac{1}{-0.0015198}(-0.50467 \cdot 1646.5+1651.5-0.49890 \cdot 1656.6)=3877.2 km$

$\rho_3=\frac{1}{-0.0015198}\left(-\frac{0.50467}{0.49890} 887.10+\frac{1}{0.49890} 889.60-892.13\right)=4186.2 km$

Step 8:

$r _1=(3489.8 \hat{ I }+3430.2 \hat{ J }+4078.5 \hat{ K })+3650.6(0.71643 \hat{ I }+0.68074 \hat{ J } -0.15270 \hat{ K })$

$=6105.2 \hat{ I }+5915.3 \hat{ J }+3521.1 \hat{ K } ( km )$

$r _2=(3460.1 \hat{ I }+3460.1 \hat{ J }+4078.5 \hat{ K })+3877.2(0.56897 \hat{ I }+0.79531 \hat{ J }-0.20917 \hat{ K })$

$=5666.6 \hat{ I }+6543.7 \hat{ J }+3267.5 \hat{ K } ( km )$

$r _3=(3429.9 \hat{ I }+3490.1 \hat{ J }+4078.5 \hat{ K })+4186.2(0.41841 \hat{ I }+0.87007 \hat{ J } -0.26059 \hat{ K })$

$=5181.4 \hat{ I }+7132.4 \hat{ J } +2987.6 \hat{ K } ( km )$

Step 9:

$v _2=\frac{1}{0.99647 \cdot 119.33-0.99641(-117.96)}$

$\times[-0.99641(6105.2 \hat{ I }+5915.3 \hat{ J }+3521.1 \hat{ K })+0.99647(5181.4 \hat{ I }+7132.4 \hat{ J }+2987.6 \hat{ K })]$

$=-3.8856 \hat{ I }+5.1214 \hat{ J }-2.2434 \hat{ K } ( km / s )$

This completes the first iteration.
The updated position $r_2$ and velocity $v_2$ are used to repeat the procedure beginning at Step 1. The results of the first and subsequent iterations are shown in Table 5.2. Convergence to five significant figures in the slant ranges $\rho_1$, $\rho_2$, and $\rho_3$occurs in four steps, at which point the state vector is

$r _2=5662.1 \hat{ I }+6538.0 \hat{ J }+3269.0 \hat{ K } ( km )$

$v _2=-3.8856 \hat{ I }+5.1214 \hat{ J }-2.2433 \hat{ K } ( km / s )$

Using $r_2$ and $v_2$ in Algorithm 4.2, we find that the orbital elements are

a = 10,000 km (h = 62,818km²/s)
e = 0.1000
i = 30°
Ω = 270°
ω = 90°
θ = 45.01°

Table 5.2 Key results at each step of the iterative procedure

 Step $\chi_1$ $\chi_2$ $f_1$ $g_1$ $f_3$ $g_3$ $\rho_1$ $\rho_2$ $\rho_3$ 1 -8.0908 8.1375 0.99647 -117.97 0.99641 119.33 3650.6 3877.2 4186.2 2 -8.0818 8.1282 0.99647 -117.96 0.996 42 119.33 3643.8 3869.9 4178.3 3 -8.0871 8.1337 0.99647 -117.96 0.996 42 119.33 3644.0 3870.1 4178.6 4 -8.0869 8.1336 0.99647 -117.96 0.996 42 119.33 3644.0 3870.1 4178.6