Question 6.4: Spacecraft at A and B are in the same orbit (1). At the inst......
Spacecraft at A and B are in the same orbit (1). At the instant shown in Figure 6.11 the chaser vehicle at A executes a phasing maneuver so as to catch the target spacecraft back at A after just one revolution of the chaser’s phasing orbit (2). What is the required total delta-v?
We must find the angular momenta of orbits 1 and 2 so that we can use Eqn (2.31) to find the velocities on orbits 1 and 2 at point A. (We can alternatively use energy, Eqn (2.81), to find the speeds at A.) These velocities furnish the delta-v required to leave orbit 1 for orbit 2 at the beginning of the phasing maneuver and to return to orbit 1 at the end.
h=r ν_{\perp} (2.31)
\cfrac{ν^2}{2}-\cfrac{\mu}{r}=-\cfrac{\mu}{2 a} (2.81)
Angular momentum of orbit 1
From Figure 6.11 we observe that perigee and apogee radii of orbit 1 are, respectively,
It follows from Eqn (6.2) that the orbit’s angular momentum is
h=\sqrt{2 \mu} \sqrt{\cfrac{r_{ a } r_{ p }}{r_{ a }+r_{ p }}} (6.2)
h_1=\sqrt{2 \mu} \sqrt{\cfrac{r_A r_C}{r_A+r_C}}=\sqrt{2 \cdot 398,600} \sqrt{\cfrac{6800 \cdot 13,600}{6800+13,600}}=60,116 km / s ^2Angular momentum of orbit 2
The phasing orbit must have a period T_2 equal to the time it takes the target vehicle at B to coast around to point A on orbit 1. That flight time equals the period of orbit 1 minus the flight time t_{AB} from A to B. That is,
T_2=T_1-t_{A B} (a)
The period of orbit 1 is found by computing its semimajor axis,
a_1=\cfrac{1}{2}\left(r_A+r_C\right)=10,200 kmand substituting that result into Eqn (2.83),
T=\cfrac{2 \pi}{\sqrt{\mu}} a^{\frac{3}{2}} (2.83)
T_1=\cfrac{2 \pi}{\sqrt{\mu}} a_1^{3 / 2}=\cfrac{2 \pi}{\sqrt{398,600}} 10,200^{3 / 2}=10,252 s (b)
The flight time from the perigee A of orbit 1 to point B is obtained from Kepler’s equation (Eqns (3.8) and (3.14)),
M_e=\cfrac{2 \pi}{T} t (3.8)
\boxed{M_e=E-e \sin E} (3.14)
t_{A B}=\cfrac{T_1}{2 \pi}\left(E_B-e_1 \sin E_B\right) (c)
Since the eccentricity of orbit 1 is
e_1=\cfrac{r_C-r_A}{r_C+r_A}=0.33333 (d)
and the true anomaly of B is 90^{\circ}, it follows from Eqn (3.13b) that the eccentric anomaly of B is
E=2 \tan ^{-1}\left(\sqrt{\cfrac{1-e}{1+e}} \tan \cfrac{\theta}{2}\right) (3.13b)
E_B=2 \tan ^{-1}\left(\sqrt{\cfrac{1-e_1}{1+e_1}} \tan \frac{\theta_B}{2}\right)=2 \tan ^{-1}\left(\sqrt{\cfrac{1-0.33333}{1+0.33333}} \tan \cfrac{90^{\circ}}{2}\right)=1.2310 rad (e)
Substituting Eqns (b), (d), and (e) into Eqn (c) yields
t_{A B}=\cfrac{10,252}{2 \pi} (1.231-0.33333 \cdot \sin 1.231)=1495.7 sIt follows from Eqn (a) that
T_2=10,252-1495.7=8756.3 sThis, together with the period formula, Eqn (2.83), yields the semimajor axis of orbit 2,
a_2=\left(\cfrac{\sqrt{\mu} T_2}{2 \pi}\right)^{2 / 3}=\left(\cfrac{\sqrt{398,600} \cdot 8756.2}{2 \pi}\right)^{2 / 3}=9182.1 kmSince 2 a_2=r_A+r_D, we find that the apogee of orbit 2 is
r_D=2 a_2-r_A=2 \cdot 9182.1-6800=11,564 kmFinally, Eqn (6.2) yields the angular momentum of orbit 2,
h_2=\sqrt{2 \mu} \sqrt{\cfrac{r_A r_D}{r_A+r_D}}=\sqrt{2 \cdot 398,600} \sqrt{\cfrac{6800 \cdot 11,564}{6800+11,564}}=58,426 km / s ^2Velocities at A
Since A is the perigee of orbit 1, there is no radial velocity component there. The speed, directed entirely in the transverse direction, is found from the angular momentum formula,
Likewise, the speed at the perigee of orbit 2 is
\left.v_A\right)_2=\cfrac{h_2}{r_A}=\cfrac{58,426}{6800}=8.5921 km / sAt the beginning of the phasing maneuver, the velocity change required to drop into the phasing orbit 2 is
\left.\left.\Delta v_A=v_A\right)_2-v_A\right)_1=8.5921-8.8406=-0.24851 km / sAt the end of the phasing maneuver, the velocity change required to return to orbit 1 is
\left.\left.\Delta v_A=v_A\right)_1-v_A\right)_2=8.8406-8.5921=0.24851 km / sThe total delta-v required for the chaser to catch up with the target is
\Delta v_{\text {total }}=|-0.24851|+|0.24851|=\boxed{0.4970 km / s}