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Chapter 5

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.

Step-by-Step

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)

= – 0.1007 rad

Then using Kepler’s equation for the ellipse (Eq. 3.14), we find the mean anomaly,

M_{e_1}=E_1-e \sin E_1 = – 0.1007 – 0.4335 sin (- 0.1007) = -0.05715 rad

so that from Eq. (3.7), the time since perigee passage is

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

The minus sign means that, after the initial sighting, there are 256.1 s until perigee encounter.

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