Question 5.12: Starting with the state vector determined in Example 5.11, u...
Starting with the state vector determined in Example 5.11, use Algorithm 5.6 to improve the vector to five significant figures.
Step 1:
r_{2}=\left\| r_{2}\right\|=\sqrt{5659.1^{2}+6533.8^{2}+3270.1^{2}}=9241.8 km
Step 2:
\alpha=\frac{2}{r_{2}}-\frac{v_{2}^{2}}{\mu}=\frac{2}{9241.8}-\frac{6.7666^{2}}{398600}=1.0154 \times 10^{-4} km ^{-1}
Step 3:
v_{r 2}=\frac{ v _{2} \cdot r _{2}}{r_{2}}=\frac{(-3.9080) \cdot 5659.1+5.0573 \cdot 6533.8+(-2.2222) \cdot 3270.1}{9241.8} = 0.39611km/s
Step 4:
The universal Kepler’s equation at times t_{1} \text { and } t_{3}, respectively, becomes
or
631.35 \tau_{1}=5.7983 \chi_{1}^{2} C\left(1.0154 \times 10^{-4} \chi_{1}^{2}\right)+0.061594 \chi_{1}^{3} S\left(1.0154 \times 10^{-4} \chi_{1}^{2}\right)+9241.8 \chi_{1}
631.35 \tau_{3}=5.7983 \chi_{3}^{2} C\left(1.0154 \times 10^{-4} \chi_{3}^{2}\right)+0.061594 \chi_{1}^{3} S\left(1.0154 \times 10^{-4} \chi_{3}^{2}\right)+9241.8 \chi_{3}
Applying Algorithm 3.3 to each of these equations yields
\chi_{3}=8.1404 \sqrt{ km }
Step 5:
f_{1}=1-\frac{\chi_{1}^{2}}{r_{2}} C\left(\alpha \chi_{1}^{2}\right)=1-\frac{(-8.0882)^{2}}{9241.8} \cdot \overbrace{C\left[1.0154 \times 10^{-4}(-8.0882)^{2}\right.}]^{0.49972}=0.99646g_{1}=\tau_{1}-\frac{1}{\sqrt{\mu}} \chi_{1}^{3} S\left(\alpha \chi_{1}^{2}\right)=-118.1-\frac{1}{\sqrt{398600}}(-8.0882)^{3}\times \overbrace{S\left[1.0154 \times 10^{-4}(-8.0882)^{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{8.1404^{2}}{9241.8} \cdot \overbrace{C\left[1.0154 \times 10^{-4} \cdot 8.1404^{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{398600}} 8.1404^{3} \times \overbrace{S\left[1.0154 \times 10^{-4}(-8.0882)^{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}=0.99647g_{1}=\frac{-117.97+(-117.96)}{2}=-117.96 s
f_{3}=\frac{0.99642+0.99641}{2}=0.99641
g_{3}=\frac{119.3+119.3}{2}=119.3 s
Step 6:
c_{1}=\frac{119.3}{(0.99647)(119.3)-(0.99641)(-117.96 s )}=0.50467c_{3}=-\frac{-117.96}{(0.99647)(119.3)-(0.99641)(-117.96)}=0.49890
Step 7:
\varrho_{2}=\frac{1}{-0.0015198}(-0.50467 \cdot 1646.5+1651.5-0.49890 \cdot 1656.6)=3877.2 km
Step 8:
r _{1}=(3489.8 \hat{ I }+3430.2 \hat{ J }+4078.5 \hat{ K })+3650.7(0.71643 \hat{ I }+0.68074 \hat{ J }-0.15270 \hat{ K })=6105.3 \hat{ I }+5915.4 \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 })=5662.1 \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 119.3-0.99641(-117.96)} \times[-0.99641(6105.3 \hat{ I }+5915.4 \hat{ J }+3521.1 \hat{ K })+0.99647(5181.4 \hat{ I }+7132.4 \hat{ J }+2987.6 \hat{ K })] =-3.8918 \hat{ I }+5.1307 \hat{ J }-2.2472 \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 step1.The results of the first and subsequent iterations are shown in Table5.2. Convergence to five significant figures in the slant ranges \varrho_{1}, \varrho_{2} \text { and } \varrho_{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 )
| Table 5.2 Key results at each step of the iterative procedure | |||||||||
| Step | \chi_{1} | \chi_{3} | f_{1} | g_{1} | f_{3} | g_{3} | \varrho_{1} | \varrho_{2} | \varrho_{3} |
| 1 | −8.0882 | 8.1404 | 0.99647 | −117.97 | 0.99641 | 119.33 | 3650.7 | 3877.2 | 4186.2 |
| 2 | −8.0818 | 8.1282 | 0.99647 | −117.96 | 0.99642 | 119.33 | 3643.8 | 3869.9 | 4178.3 |
| 3 | −8.0871 | 8.1337 | 0.99647 | −117.96 | 0.99642 | 119.33 | 3644.0 | 3870.1 | 4178.6 |
| 4 | −8.0869 | 8.1336 | 0.99647 | −117.96 | 0.99642 | 119.33 | 3644.0 | 3870.1 | 4178.6 |
Using Algorithm 4.1 we find that the orbital elements are
a = 10000km (h = 62818km²/s)
e = 0.1000
i = 30°
Ω = 270°
ω = 90°
θ = 45.01°