An earth satellite has the following orbital parameters: rp = 6700 km Perigee ra = 10 000 km Apogee θ0 = 230° True anomaly Ω0 = 270° Right ascension of the ascending node i0 = 60° Inclination ω0 = 45° Argument of perigee Calculate the right ascension (longitude east of x′) and declination

Question

An earth satellite has the following orbital parameters:

[latex]\begin{array}{ll}r_{ p }=6700 km & ext { Perigee } \r_{ a }=10 000 km & ext { Apogee } \ heta_0=230^{\circ} & ext { True anomaly } \\Omega_0=270^{\circ} & ext { Right ascension of the ascending node } \i_0=60^{\circ} & ext { Inclination } \\omega_0=45^{\circ} & ext { Argument of perigee }\end{array}[/latex]

Calculate the right ascension (longitude east of [latex]x^{\prime}[/latex]) and declination (latitude) relative to the rotating earth 45 min later.

Accepted Answer

First, we compute the semimajor axis a, eccentricity e, the angular momentum h, the semimajor axis a, and the period T. For the semimajor axis, we recall that

[latex]a=\cfrac{r_{ p }+r_{ a }}{2}=\cfrac{6700+10,000}{2}=8350 km[/latex]

From Eqn (2.84) we get

[latex]e=\cfrac{r_{ a }-r_{ p }}{r_{ a }+r_{ p }}=\cfrac{10,000-6700}{10,000+6700}=0.19760[/latex]

Equation (2.50) yields

[latex]r_p=\cfrac{h^2}{\mu} \cfrac{1}{1+e}[/latex] (2.50)

[latex]h=\sqrt{\mu r_{ p }(1+e)}=\sqrt{398,600 imes 6700 imes(1+0.19760)}=56,554 km ^2 / s[/latex]

Finally, we obtain the period from Eqn (2.83),

[latex]T=\cfrac{2 \pi}{\sqrt{\mu}} a^{3 / 2}=\cfrac{2 \pi}{\sqrt{398,600}} 8350^{3 / 2}=7593.5 s[/latex]

Now we can proceed with Algorithm 4.6.
Step 1:

[latex]\begin{aligned}& \dot{\Omega}=-\left[\frac{3}{2} \frac{\sqrt{\mu} J_2 R_{ ext {earth }}^2}{\left(1-e^2 ight) a^{7 / 2}} ight] \cos i=-\left[\frac{3}{2} \frac{\sqrt{398,600} imes 0.0010836 imes 6378^2}{\left(1-0.19760^2 ight) 8350^{7 / 2}} ight] \cos 60^{\circ}=-2.3394 imes 10^{-7} {}^{\circ} / s \\& \dot{\omega}=\dot{\Omega} \frac{5 / 2 \sin ^2 i-2}{\cos i}=-2.3394 imes 10^{-5}\left(\frac{5 / 2 \sin ^2 60^{\circ}-2}{\cos 60^{\circ}} ight)=5.8484 imes 10^{-6} {}^{\circ} / s\end{aligned}[/latex]

Step 2:
a.

[latex]E=2 an ^{-1}\left( an \cfrac{ heta}{2} \sqrt{\cfrac{1-e}{1+e}} ight)=2 an ^{-1}\left( an \cfrac{230^{\circ}}{2} \sqrt{\cfrac{1-0.19760}{1+0.19760}} ight)=-2.1059 rad[/latex]

b.

[latex]M=E-e \sin E=-2.1059-0.19760 \sin (-2.1059)=-1.9360 rad[/latex]

c.

[latex]t_0=\cfrac{M}{2 \pi} T=\cfrac{-1.9360}{2 \pi} \cdot 7593.5=-2339.7 s \quad(2339.7 s ext { until perigee })[/latex]

Step 3: [latex]t=t_0+45 ext{min} =-2339.7+45 imes 60=360.33 s \quad(360.33 s ext { after perigee })[/latex]

a.

[latex]\begin{aligned}M & =2 \pi \cfrac{t}{T}=2 \pi \cfrac{360.33}{7593.5}=0.29815 ext{rad} \\E-0.19760 \sin E & =0.29815 \overbrace{\Rightarrow}^{ ext {Algorithm 3.1 }} E=0.36952 rad \\ heta & =2 an ^{-1}\left( an \cfrac{E}{2} \sqrt{\cfrac{1+e}{1-e}} ight)=2 an ^{-1}\left( an \cfrac{0.36952}{2} \sqrt{\cfrac{1+0.19760}{1-0.19760}} ight)=25.723^{\circ}\end{aligned}[/latex]

b.

[latex]\begin{aligned}& \Omega=\Omega_0+\dot{\Omega} \Delta t=270^{\circ}+\left(-2.3394 imes 10^{-5 \%} / s ight)(2700 s )=269.94 \& \omega=\omega_0+\dot{\omega} \Delta t=45^{\circ}+\left(5.8484 imes 10^{-6} / s ight)(2700 s )=45.016^{\circ}\end{aligned}[/latex]

c.

[latex]\{ r \}_X \stackrel{ ext { Algorithm } 4.5}{\widehat{=}}\left\{\begin{array}{c}3212.6 \-2250.5 \5568.6\end{array} ight\}( km )[/latex]

d.

[latex]\begin{aligned}& heta=\omega_E \Delta t=\frac{360^{\circ}\left(1+\frac{1}{365.26} ight)}{24 imes 3600 s } imes 2700 s =11.281^{\circ}\\&\begin{aligned}& {\left[ R _3( heta) ight]=\left[\begin{array}{ccc}\cos 11.281^{\circ} & \sin 11.281^{\circ} & 0 \-\sin 11.281^{\circ} & \cos 11.281^{\circ} & 0 \0 & 0 & 1\end{array} ight]=\left[\begin{array}{ccc}0.98068 & 0.19562 & 0 \-0.19562 & 0.98068 & 0 \0 & 0 & 1\end{array} ight]} \\& \{ r \}_{X^{\prime}}=\left[ R _3( heta) ight]\{ r \}_X=\left[\begin{array}{ccc}0.98068 & 0.19562 & 0 \-0.19562 & 0.98068 & 0 \0 & 0 & 1\end{array} ight]\left\{\begin{array}{c}3212.6 \-2250.5 \5568.6\end{array} ight\}=\left\{\begin{array}{c}2710.3 \-2835.4 \5568.6\end{array} ight\}( km )\end{aligned}\end{aligned}[/latex]

e.

[latex]r =2710.3 I ^{\prime}-2835.4 \hat{\jmath}^{\prime}+5568.6 \hat{ K }^{\prime}\overbrace{\Rightarrow }^{ ext{Algorithm 4:1}} \boxed{\alpha=313.7^{\circ} \quad \delta=54.84^{\circ}}[/latex]

Question 4.12: An earth satellite has the following orbital parameters: rp ......

Related Answered Questions

Verified Answer:

Verified Answer:

Given the state vector, r = -6045I – 3490J + 2500K (km) v = -3.457I + 6.618J + 2.533K (km/s) find the orbital elements h, i, Ω, e, ω, and θ using Algorithm 4.2. ...

Verified Answer:

Determine the perigee and apogee for an earth satellite whose orbit satisfies all the following conditions: it is sunsynchronous, its argument of perigee is constant, and its period is 3 h. ...

Verified Answer:

A satellite is to be launched into a sun-synchronous circular orbit with period of 100 min. Determine the required altitude and inclination of its orbit. ...

Verified Answer:

For a given earth orbit, the elements are h = 80,000 km²/s, e = 1.4, i = 30°, Ω = 40°, ω = 60°, and θ = 30°. Using Algorithm 4.5, find the state vectors r and v in the geocentric equatorial frame. ...

Verified Answer:

At time t0, the state vector of an earth satellite is r0 = 1600I + 5310J + 3800 K (km) (a) v0 = – 7.350I + 0.4600J + 2.470 K (km/s) (b) Determine the position and velocity 3200 s later and plot the orbit in three dimensions ...

Verified Answer:

A spacecraft is in a 280 km by 400 km orbit with an inclination of 51.43°. Find the rates of node regression and perigee advance. ...

Verified Answer:

If the direct cosine matrix for the transformation from xyz to x ′ y′ z′ is the same as it was in Example 4.5, [Q] = [ 0.64050 0.75319 -0.15038 0.76736 – 0.63531 0.086824 -0.030154 -0.17101 -0.98481] find the angles α , β and γ of the yaw, pitch, and roll sequence ...

Verified Answer:

If the position vector of the International Space Station in the geocentric equatorial frame is r = – 5368I – 1784J + 3691 K(km) what are its RA and Dec? ...

Verified Answer: