Chapter 6
Q. 6.9
Spacecraft B and C are both in the geocentric elliptical orbit (1) shown in Figure 6.20,from which it can be seen that the true anomalies are θ_{B} = 45^{◦} and θ_{C} = 150^{◦}. At the instant shown, spacecraft B executes a delta-v maneuver, embarking upon a trajector(2) which will intercept vehicle C in precisely one hour. Find the orbital parameters (e and h) of the intercept trajectory and the total delta-v required for the chasemaneuver.
Step-by-Step
Verified Solution
First, we must determine the parameters of orbit 1 in the usual way. The eccentricity is found using the orbit’s perigee and apogee, shown in Figure 6.20,
e_{1}=\frac{18 900 − 8100}{18 900 +8100}=0.4000From the orbit equation,
r_{p}=\frac{h^{2}_{1}}{\mu}\frac{1}{1+e_{1}\cos(0)}\Rightarrow8100=\frac{h^{2}_{1}}{398600}\frac{1}{1+0.4}\Rightarrow h_{1}= 67 232 km^{2} /s
Using Equation 2.72 yields the period,
T_{1}=\frac{2\pi }{\mu^{2} }(\frac{h_{1}}{\sqrt{1-e^{2}_{1} } } )^{3}=\frac{2\pi }{398600^{2} }(\frac{67232}{\sqrt{1-0.4^{2} } } )^{3}= 15 610 sIn perifocal coordinates (Equation 2.109) the position vector of B is
r=\frac{h^{2} }{\mu } \frac{1}{1+e\cos \theta }(\cos\theta \hat{p} +\sin\theta \hat{q} ) (2.109)
r_{B}=\frac{h^{2}_{1} }{\mu } \frac{1}{1+e_{1}\cos \theta _{B}}(\cos\theta _{B}\hat{p} +\sin\theta _{B}\hat{q} ) =\frac{67232^{2} }{398600 } \frac{1}{1+0.4\cos45^{◦}}(\cos45^{◦}\hat{p} +\sin45^{◦}\hat{q} )or
r_{B}=6250.6\hat{p}+ 6250.6\hat{q} (km) (a)
Likewise, according to Equation 2.115, the velocity at B on orbit 1 is
V=\frac{\mu }{h}[-\sin \theta \hat{p}+(e+\cos\theta)\hat{q}] (2.115)
V_{B_{1}}=\frac{\mu }{h}[-\sin \theta _{B}\hat{p}+(e+\cos\theta _{B})\hat{q}]=\frac{398600}{67232}[-\sin 45^{◦}\hat{p}+(0.4+\cos 45^{\circ})\hat{q}]so that
V_{B_{1}}=-4.1922\hat{p}+6.563 \hat{q} km/s (b)
Now we need to move spacecraft C along orbit 1 to the position C′ that it will occupy one hour later (Δt), when it will presumably be met by spacecraft B. To do that, we must first calculate the time since perigee passage at C. Since we know the true anomaly, the eccentric anomaly follows from Equation 3.10a,
\tan \frac{E}{2}=\sqrt{\frac{1-e}{1+e} }\tan \frac{\theta }{2} (3.10a)
\tan \frac{E_{C}}{2}=\sqrt{\frac{1-e_{1}}{1+e_{1}} }\tan \frac{\theta _{C}}{2}=\sqrt{\frac{1-0.4}{1+0.4} }\tan \frac{150^{◦}}{2}=2.4432\Rightarrow E_{C}=2.3646 rad
Substituting this value into Kepler’s equation (Equations 3.5 and 3.11) yields the time since perigee passage,
M_{e}=\frac{2\pi }{T}t (3.5)
M_{e}=E-e\sin E (3.11)
t_{C}=\frac{T_{1}}{2\pi }(E_{C}-e_{1}\sin E_{C})=\frac{15610}{2\pi }(2.3646 − 0.4 ·\sin 2.3646)= 5178 sOne hour later (Δt = 3600 s), the spacecraft will be in intercept position at C′,
t_{C′}=t_{C}+\Delta t=5178+3600=8778 sThe corresponding mean anomaly is
M_{e})_{C′}=2\pi \frac{t_{C′}}{T_{1}}=2\pi \frac{8778}{15610}=3.5331 rad
With this value of the mean anomaly, Kepler’s equation becomes
E_{C′}-e_{1}\sin E_{C′}=3.5331Applying Algorithm 4.1 to the solution of this equation we get
E_{C′}=3.4223 rad
Substituting this result into Equation 3.10a yields the true anomaly at C′,
\tan \frac{\theta _{C′}}{2}=\sqrt{\frac{1+0.4}{1-0.4} }\tan \frac{3.4223 }{2}=-10.811\Rightarrow \theta _{C′}=190.57^{◦}We are now able to calculate the perifocal position and velocity vectors at C′ on orbit 1:
r_{C′}=\frac{ 67 232^{2}}{398600}\frac{1}{1+0.4\cos 190.57^{◦}}(\cos 190.57^{◦}\hat{p}+\sin 4190.57^{◦}\hat{q})= −18 372\hat{p} − 3428.1\hat{q} (km)
V_{C′_{1}}=\frac{398600}{67232}[-\sin 190.57^{◦}\hat{p}+(0.4+\cos 190.57^{◦} )\hat{q}]=1.0875\hat{p}-3.4566\hat{q} (km/s) (c)
The intercept trajectory connecting points B and C′ are found by solving Lambert’s problem. Substituting r_{B} and r_{C′} along with Δt = 3600 s into Algorithm 5.2 yields
V_{B_{2}}= −8.1349\hat{p} + 4.0506\hat{q} (km/s) (d)
V_{C′_{2}}= −3.4745\hat{p} − 4.7943\hat{q} (km/s) (e)
MATLAB Verified Solution
Script Files
These velocities are most easily obtained by running the following MATLAB script, which executes Algorithm 5.2 by means of the function M-file lambert.m (Appendix D.11).
clear
global mu
deg = pi/180;
mu = 398600;
e = 0.4;
h = 67232;
theta1 = 45*deg;
theta2 = 190.57*deg;
delta_t = 3600;
rB = hˆ2/mu/(1 + e*cos(theta1))...
*[cos(theta1),sin(theta1),0];
rC_prime = hˆ2/mu/(1 + e*cos(theta2))...
*[cos(theta2),sin(theta2),0];
string = 'pro';
[vB2 vC_prime_2] = lambert(rB, rC_prime,...
delta_t, string)
From (b) and (d) we find
\Delta V_{B}=V_{{B}_{2}}-V_{{B}_{1}}=-3.426\hat{p}− 2.5132\hat{q} (Km/s)
whereas (c) and (e) yield
\Delta V_{C'}=V_{C'_{1}}-V_{C'_{2}}= 4.5620\hat{p}+ 1.3376\hat{q} (Km/s)
The anticipated, extremely large, delta-v requirement for this chase maneuver is the sum of the magnitudes of these two vectors,
\Delta V=\left\|\Delta V_{B}\right\|+\left\|\Delta V_{C'}\right\|= 4.6755 + 4.7540 = 9.430 km/s
We know that orbit 2 is an ellipse. To pin it down a bit more, we can use r_{B} and v_{{B}_{2}}to obtain the orbital elements from Algorithm 4.1, which yields
h_{2} = 76 167 km^{2}/s
e_{2} = 0.8500
a_{2} = 52 449 km
θ_{{B}_{2} }= 319.52^{◦}
These may be found quickly by running the following MATLAB script, in which the M-function coe_from_sv.m is Algorithm 4.1 (see Appendix D.8):
clear
global mu
mu = 398600;
rB = [6250.6 6250.6 0];
vB2 = [-8.1349 4.0506 0];
orbital_elements = coe_from_sv(rB, vB2)
The details of the intercept trajectory and the delta-v maneuvers are shown in Figure 6.21. A far less dramatic though more leisurely(and realistic)way for B to catch up with C would be to use a phasing maneuver.