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## 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.

## 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,

$e_{1}=\frac{18 900 − 8100}{18 900 +8100}=0.4000$

From 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,

$T_{1}=\frac{2\pi }{\mu^{2} }(\frac{h_{1}}{\sqrt{1-e^{2}_{1} } } )^{3}=\frac{2\pi }{398600^{2} }(\frac{67232}{\sqrt{1-0.4^{2} } } )^{3}= 15 610 s$

In 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)

$t_{C}=\frac{T_{1}}{2\pi }(E_{C}-e_{1}\sin E_{C})=\frac{15610}{2\pi }(2.3646 − 0.4 ·\sin 2.3646)= 5178 s$

One hour later (Δt = 3600 s), the spacecraft will be in intercept position at C′,

$t_{C′}=t_{C}+\Delta t=5178+3600=8778 s$

The 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

$E_{C′}-e_{1}\sin E_{C′}=3.5331$

Applying 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)

## 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.