Question 4.3: Given the state vector, r = -6045I - 3490J + 2500K (km) v = ......

Step 1:

r=\sqrt{ r \cdot r }=\sqrt{(-6045)^2+(-3490)^2+2500^2}=7414  km                 (a)

Step 2:

v=\sqrt{ v \cdot v }=\sqrt{(-3.457)^2+6.618^2+2.533^2}=7.884  km / s                  (b)

Step 3:

(c)

Since v_r > 0, the satellite is flying away from perigee.
Step 4:

h = r \times V =\left|\begin{array}{ccc}\hat{ I } & \hat{ J } & \hat{ K } \\-6045 & -3490 & 2500 \\-3.457 & 6.618 & 2.533\end{array}\right|=-25,380 \hat{ I }+6670 \hat{ J }-52,070 \hat{ K }\left( km ^2 / s \right)                         (d)

Step 5:

h=\sqrt{ h \cdot h }=\sqrt{(-25,380)^2+6670^2+(-52,070)^2}=\boxed{58,310  km ^2 / s}                     (e)

Step 6:

i=\cos ^{-1} \cfrac{h_Z}{h}=\cos ^{-1}\left(\cfrac{-52,070}{58,310}\right)=\boxed{153.2^{\circ}}                         (f)

Since i is greater than 90^{\circ}, this is a retrograde orbit.
Step 7:

N =\hat{ K } \times h =\left|\begin{array}{ccc}\hat{ I } & \hat{ J } & \hat{ K } \\0 & 0 & 1 \\-25,380 & 6670 & -52,070\end{array}\right|=-6670 \hat{ I }-25,380 \hat{ J }\left( km ^2 / s \right)                                              (g)

Step 8:

N=\sqrt{ N \cdot N }=\sqrt{(-6670)^2+(-25,380)^2}=26,250  km ^2 / s                             (h)

Step 9:

From Eqn (g) we know that N_Y < 0; therefore, Ω must lie in the third quadrant,

\boxed{Q=255.3^{\circ}}                         (i)

Step 10:

Step 11:

e=\sqrt{ e \cdot e}=\sqrt{(-0.09160)^2+(-0.1422)^2+(0.02644)^2}= \boxed{0.1712}                                 (k)

Clearly, the orbit is an ellipse.
Step 12:

ω lies in the first quadrant if e_Z > 0, which is true in this case, as we see from Eqn (j). Therefore,

\boxed{\omega=20.07^{\circ}}                         (l)

Step 13:

From Eqn (c) we know that v_r > 0, which means 0 \leq \theta<180^{\circ}. Therefore,

Having found the orbital elements, we can go on to compute other parameters. The perigee and apogee radii are

From these it follows that the semimajor axis of the ellipse is

This leads to the period,

The orbit is illustrated in Figure 4.8.