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Chapter 5

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.

Step-by-Step

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_3occurs 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