Question 5.10: At θ =300° local sidereal time a sea-level (H =0) tracking s...
At θ =300° local sidereal time a sea-level (H =0) tracking station at latitude φ=60° detects a space object and obtains the following data:
Slant range : ϱ = 2551km
Azimuth : A= 90°
Elevation : a = 30°
Range rate : \dot{\varrho} = 0
Azimuth rate: \dot{A}=1.973 \times 10^{-3} rad/s(0.1130°/s)
Elevation rate : \dot{a}=9.864 \times 10^{-4} rad/s(0.05651°/s)
What are the orbital elements of the object?
We must first employ Algorithm 5.4 to obtain the state vectors r and v in order to compute the orbital elements by means of Algorithm 4.1.
Step 1:
The equatorial radius of the earth is R_{e}=6378 km and the flattening factor is f =0.003353. It follows from Equation 5.56 that the position vector of the observer is
R =\left[\frac{R_{e}}{\sqrt{1-\left(2 f-f^{2}\right) \sin ^{2} \phi}}+H\right] \cos \phi(\cos \theta \hat{ I }+\sin \theta \hat{ J })+\left[\frac{R_{e}(1-f)^{2}}{\sqrt{1-\left(2 f-f^{2}\right) \sin ^{2} \phi}}+H\right] \sin \phi \hat{ K } (5.56)
R =1598 \hat{ I }-2769 \hat{ J }+5500 \hat{ K }( km )Step 2:
\delta=\sin ^{-1}(\cos \phi \cos A \cos a+\sin \phi \sin a)
=\sin ^{-1}\left(\cos 60^{\circ} \cos 90^{\circ} \cos 30^{\circ}+\sin 60^{\circ} \sin 30^{\circ}\right)
= 25.66°
Step 3:
Since the given azimuth lies between 0° and 180°, Equation 5.83b yields
h=360^{\circ}-\cos ^{-1}\left(\frac{\cos \phi \sin a-\sin \phi \cos A \cos a}{\cos \delta}\right)
=360^{\circ}-\cos ^{-1}\left(\frac{\cos 60^{\circ} \sin 30^{\circ}-\sin 60^{\circ} \cos 90^{\circ} \cos 30^{\circ}}{\cos 25.66^{\circ}}\right)
= 360° − 73.90° = 286.1°
Therefore, the right ascension is
α = θ − h = 300° − 286.1° = 13.90°
Step 4:
\hat{\varrho}=\cos 25.66\left(\cos 13.90^{\circ} \hat{ I }+\sin 13.90^{\circ} \hat{ J }\right)+\sin \delta \hat{ K }=0.8750 \hat{ I }+0.2165 \hat{ J }+0.4330 \hat{ K }Step 5:
r = R +\varrho \hat{\varrho}=(1598 \hat{ I }-2769 \hat{ J }+5500 \hat{ K })+2551(0.8750 \hat{ I }+0.2165 \hat{ J }+0.4330 \hat{ K })
r =3831 \hat{ I }-2216 \hat{ J }+6605 \hat{ K }( km )
Step 6:
Recalling from Equation 2.57 that the angular velocity \omega_{E} of the earth is 72.92 \times 10^{-6} rad / s,
\dot{ R }= \Omega \times R =\left(72.92 \times 10^{-6} \hat{ K }\right) \times(1598 \hat{ I }-2769 \hat{ J }+5500 \hat{ K })
=0.2019 \hat{ I }+0.1166 \hat{ J }( km / s )
Step 7:
\dot{\delta}=\frac{1}{\cos \delta}[-\dot{A} \cos \phi \sin A \cos a+\dot{a}(\sin \phi \cos a-\cos \phi \cos A \sin a)]\dot{\delta}=-1.2696 \times 10^{-4}( rad / s )
Step 8:
\dot{\alpha}-\omega_{E}=\frac{\dot{A} \cos A \cos a-\dot{a} \sin A \sin a+\dot{\delta} \sin A \cos a \tan \delta}{\cos \phi \sin a-\sin \phi \cos A \cos a}= −0.002184
Step 9:
\dot{\hat{\varrho}}=(-\dot{\alpha} \sin \alpha \cos \delta-\dot{\delta} \cos \alpha \sin \delta) \hat{ I }+(\dot{\alpha} \cos \alpha \cos \delta-\dot{\delta} \sin \alpha \sin \delta) \hat{ J }+\dot{\delta} \cos \delta \hat{ K }\dot{\hat{\varrho}}=(0.5104 \hat{ I }-1.834 \hat{ J }-0.1144 \hat{ K })\left(10^{-3}\right)( rad / s )
Step 10:
v =\dot{ R }+\dot{\varrho} \hat{\varrho}+\varrho \dot{\hat{\varrho}}Using the position and velocity vectors from steps 5 and 10,the reader can verify that Algorithm 4.1 yields the following orbital elements of the tracked object
a = 5170km
i = 113.4°
Ω = 109.8°
e = 0.6195
ω = 309.8°
θ = 165.3°
This is a highly elliptical orbit with a semimajor axis less than the earth’s radius, so the object will impact the earth (at a true anomaly of 216°).