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Q. 5.2

The position of an earth satellite is first determined to be

$r _1=5000 \hat{ I }+10,000 \hat{ J }+2100 \hat{ K }( km )$

After 1 h the position vector is

$r _2=-14,600 \hat{ I }+2500 \hat{ J }+7000 \hat{ K }( km )$

Determine the orbital elements and find the perigee altitude and the time since perigee passage of the first sighting.

Verified Solution

We must first execute the steps of Algorithm 5.2 to find $v _1$ and $v _2$.
Step 1:

$r_1=\sqrt{5000^2+10,000^2+2100^2}=11,375 km$

$r_2=\sqrt{(-14,600)^2+2500^2+7000^2}=16,383 km$

Step 2: Assume a prograde trajectory.

$r _1 \times r _2=(64.75 \hat{ I }-65.66 \hat{ J }+158.5 \hat{ K })\left(10^6\right)\left( km ^2\right)$

$\cos ^{-1} \frac{r_1 \cdot r_2}{r_1 r_2}=100.29^{\circ} \text { or } 259.71^{\circ}$

Since the trajectory is prograde and the z component $r _1 \times r _2$ is positive, it follows from Eq. (5.26) that

Δθ = 100.29°

Step 3:

$A=\sin \Delta \theta \sqrt{\frac{r_1 r_2}{1-\cos \Delta \theta}}=\sin 100.29^{\circ} \sqrt{\frac{11,375 \times 16,383}{1-\cos 100.29^{\circ}}}=12,372 km$

Step 4:
Using this value of A and Δt = 3600 s, we can evaluate the functions F(z) and F'(z) given by Eqs. (5.40) and (5.43), respectively. Let us first plot F(z) to estimate where it crosses the z axis. As can be seen from Fig. 5.4, F(z) = 0 near z = 1.5. With z’ = 1.5 as our initial estimate, we execute Newton’s procedure (Eq. 5.45), $z_{i+1}=z_i-F\left(z_i\right) / F^{\prime}\left(z_i\right)$

$F(z)=\left[\frac{y(z)}{C(z)}\right]^{3 / 2} S(z)+A \sqrt{y(z)}-\sqrt{\mu} \Delta t$                             (5.40)

$F^{\prime}(z)=\left\{\begin{array}{r}{\left[\frac{y(z)}{C(z)}\right]^{3 / 2}\left\{\frac{1}{2 z}\left[C(z)-\frac{3}{2} \frac{S(z)}{C(z)}\right]+\frac{3 S(z)^2}{4C(z)}\right\}+\frac{A}{8}\left[3 \frac{S(z)}{C(z)} \sqrt{y(z)}+A \sqrt{\frac{C(z)}{y(z)}}\right](z \neq 0)} \\\frac{\sqrt{2}}{40} y(0)^{3 / 2}+\frac{A}{8}\left[\sqrt{y(0)}+A \sqrt{\frac{1}{2 y(0)}}\right] \quad(z=0)\end{array}\right.$           (5.43)

$z_1=1.5-\frac{-14,476.4}{362,642}=1.53991$

$z_2=1.53991-\frac{23.6274}{363,828}=1.53985$

$z_3=1.53985-\frac{6.29457 \times 10^{-5}}{363,826}=1.53985$

Thus, to five significant figures z = 1.5398. The fact that z is positive means the orbit is an ellipse.
Step 5:

$y=r_1+r_2+A \frac{z S(z)-1}{\sqrt{C(z)}}=11,375+16,383+12,372 \frac{1.5398 \overbrace{S(1.5398)}^{0.154296}}{\sqrt{\underbrace{C(1.5398)}_{0.439046}}}=13,523 km$

Step 6:
Eqs. (5.46a)–(5.46d) yields the Lagrange functions

$f=1-\frac{y}{r_1}=1-\frac{13,523}{11,375}=-0.18877$

$g=A \sqrt{\frac{\underline{y}}{\mu}}=12,372 \sqrt{\frac{13,523}{398,600}}=2278.9 s$

$\dot{g}=1-\frac{y}{r_2}=1-\frac{13,523}{16,383}=0.17457$

Step 7:

$v _1=\frac{1}{g}\left( r _2-f r _1\right)$

$=\frac{1}{2278.9}[(-14,600 \hat{ I }+2500 \hat{ J }+7000 \hat{ K })-(-0.18877)(5000 \hat{ I }+10,000 \hat{ J }+2100 \hat{ K })]$

$=-5.9925 \hat{ I }+1.9254 \hat{ J }+3.2456 \hat{ K } ( km / s )$

$v _2=\frac{1}{g}\left(\dot{ g } r _2- r _1\right)$

$=\frac{1}{2278.9}[(0.17457)(-14,600 \hat{ I }+2500 \hat{ J }+7000 \hat{ K })-(5000 \hat{ I }+10,000 \hat{ J }+2100 \hat{ K })]$

$=-3.3125 \hat{ I }-4.1966 \hat{ J }-0.38529 \hat{ K } ( km / s )$

Step 8:
Using $r _1$ and $v _1$, Algorithm 4.2 yields the orbital elements:

h = 80,470 km²/s
α = 20,000 km
e = 0.4335
Ω = 44.60°
i = 30.19°
ω = 30.71°
$θ _1$ = 350.8°

This elliptical orbit is plotted in Fig. 5.5. The perigee of the orbit is

$r_{ p }=\frac{h^2}{\mu} \frac{1}{1+e \cos (0)}=\frac{80,470^2}{398,600} \frac{1}{1+0.4335}=11,330 km$

Therefore, the perigee altitude is 11, 330 – 6378 = 4952 km .

To find the time of the first sighting, we first calculate the eccentric anomaly by means of Eq. (3.13b),

$E_1=2 \tan ^{-1}\left(\sqrt{\frac{1-e}{1+e}} \tan \frac{\theta}{2}\right)=2 \tan ^{-1}\left(\sqrt{\frac{1-0.4335}{1+0.4335}} \tan \frac{350.8^{\circ}}{2}\right)=2 \tan ^{-1}(-0.05041)$

$M_{e_1}=E_1-e \sin E_1$ = – 0.1007 – 0.4335 sin (- 0.1007) = -0.05715 rad
$t_1=\frac{h^3}{\mu^2} \frac{1}{\left(1-e^2\right)^{3 / 2}} M_{e_1}=\frac{80,470^3}{398,600^2} \frac{1}{\left(1-.4335^2\right)^{3 / 2}}(-0.05715)=-256.1 s$