Question 5.2: The position of an earth satellite is first determined to be...

We first must execute the steps of Algorithm 5.2 in order to find v _{1} \text { and } v _{2}.
Step 1:

r_{1}=\sqrt{5000^{2}+10000^{2}+2100^{2}}=11375 km
r_{2}=\sqrt{(-14600)^{2}+2500^{2}+7000^{2}}=16383 km

Step 2: assume a prograde trajectory:

r _{1} \times r _{2}=(64.75 \hat{ I }-65.66 \hat{ J }+158.5 \hat{ K }) \times 10^{6}
\cos ^{-1} \frac{ r _{1} \cdot r _{2}}{r_{1} r_{2}}=100.29^{\circ}

Since the trajectory is prograde and the z component of r _{1} \times r _{2} is positive, it follows from Equation 5.26 that

Δθ = 100.29°

Step 3:

Step 4:
Using this value of A and Δt =3600s, we can evaluate the functions F(z) and F'(z) given by Equations 5.40 and 5.43, respectively. Let us first plot F(z) to get at least a rough idea of where it crosses the z axis. As can be seen from Figure 5.4, F(z)=0 near z =1.5. With z_{0}=1.5 as our initial estimate, we execute Newton’s procedure, Equation 5.45,

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

z_{i+1}=z_{i}-\frac{F\left(z_{i}\right)}{F^{\prime}\left(z_{i}\right)}                    (5.45)
z_{1}=1.5-\frac{-14476.4}{362642}=1.53991
z_{2}=1.53991-\frac{23.6274}{363828}=1.53985
z_{3}=1.53985-\frac{6.29457 \times 10^{-5}}{363826}=1.53985

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

Step 6:
Equations 5.46 yield the Lagrange functions

f=1-\frac{y}{r_{1}}=1-\frac{13523}{11375}=-0.18877
g=A \sqrt{\frac{y}{\mu}}=12372 \sqrt{\frac{13523}{398600}}=2278.9 s
\dot{g}=1-\frac{y}{r_{2}}=1-\frac{13523}{16383}=0.17457

Step 7:

v _{1}=\frac{1}{g}\left( r _{2}-f r _{1}\right)=\frac{1}{2278.9}[(-14600 \hat{ I }+2500 \hat{ J }+7000 \hat{ K })-(-0.18877)(5000 \hat{ I }+10000 \hat{ J }+2100 \hat{ K })]
v _{1}=-5.9925 \hat{ I }+1.9254 \hat{ J }+3.2456 \hat{ K }( km )

v _{2}=\frac{1}{g}\left(\dot{g} r _{2}- r _{1}\right)=\frac{1}{2278.9}[(0.17457)(-14600 \hat{ I }+2500 \hat{ J }+7000 \hat{ K })-(5000 \hat{ I }+10000 \hat{ J }+2100 \hat{ K })]
v _{2}=-3.3125 \hat{ I }-4.1966 \hat{ J }-0.38529 \hat{ K }( km )

Step 8:
Using r _{1} \text { and } v _{1}, Algorithm 4.1 yields the orbital elements:
h = 80 470km²/s
a = 20 000km
e = 0.4335
Ω = 44.60°
i = 30.19°
ω = 30.71°
\theta_{1} = 350.8°
This elliptical orbit is plotted in Figure 5.5. The perigee of the orbit is

Therefore the perigee altitude is 11 330 − 6378 = 4952km.
To find the time of the first sighting, we first calculate the eccentric anomaly by means of Equation 3.10b,

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 (Equation 3.11), the mean anomaly is found to be

so that from Equation 3.4,the time since perigee passage is

M_{e}=\frac{\mu^{2}}{h^{3}}\left(1-e^{2}\right)^{\frac{3}{2}} t                   (3.4)

The minus sign means there are 256.1 seconds until perigee encounter after the initial sighting.