Question 5.1: The geocentric position vectors of a space object at three s...

Step 1:

r_{1}=\sqrt{(-294.32)^{2}+4265.1^{2}+5986.7^{2}}=7356.5 km
r_{2}=\sqrt{(-1365.5)^{2}+3637.6^{2}+6346.8^{2}}=7441.7 km
r_{3}=\sqrt{(-2940.3)^{2}+2473.7^{2}+6555.8^{2}}=7598.9 km

Step 2:

C _{12}=\left|\begin{array}{ccc}\hat{ I } & \hat{ J } & \hat{ K } \\-294.32 & 4265.1 & 5986.7 \\-1365.5 & 3637.6 & 6346.8\end{array}\right|
=(5.292 \hat{ I }-6.3066 \hat{ J }+4.7531 \hat{ K }) \times 10^{6}\left( km ^{2}\right)
C _{23}=\left|\begin{array}{ccc}\hat{ I } & \hat{ J } & \hat{ K } \\-1365.5 & 3637.6 & 6346.8 \\-2940.3 & 2473.7 & 6555.8\end{array}\right|

C _{31}=\left|\begin{array}{ccc}\hat{ I } & \hat{ J } & \hat{ K } \\-2940.3 & 2473.7 & 6555.8 \\-294.32 & 4265.1 & 5986.7\end{array}\right|
=(-1.3151 \hat{ I }+1.5673 \hat{ J }-1.1812 \hat{ K }) \times 10^{6}\left( km ^{2}\right)

Step 3:

Therefore

\hat{ u }_{r_{1}} \cdot \hat{ C }_{23}=\frac{-294.32 \hat{ I }+4265.1 \hat{ J }+5986.7 \hat{ K }}{7356.5} \cdot(0.55667 \hat{ I }-0.66341 \hat{ J }+0.5000 \hat{ K })
=6.9200 \times 10^{-20}

This certainly is close enough to zero for our purposes. The three vectors r _{1}, r _{2} and r _{3} are coplanar.

Step 4:

or

so that

N=\sqrt{\left[2.2807^{2}+(-2.7181)^{2}+2.0486^{2}\right] \times 10^{18}}
=4.0971 \times 10^{9}\left( km ^{3}\right)
D = C _{12}+ C _{23}+ C _{31}

or

so that

D =\sqrt{\left[2.8797^{2}+(-3.4319)^{2}+2.5866^{2}\right] \times 10^{12}}
=5.1731 \times 10^{5}\left( km ^{2}\right)

Lastly,

or

Step 5:

or

Step 6:
Using r _{2} \text { and } v _{2}, Algorithm 4.1 yields the orbital elements:

a = 8000km
e = 0.1
i = 60°
Ω = 40°
ω = 30°
θ = 50°  (for position vector r _{2})
The orbit is sketched in Figure 5.2.