## Tof

5.256 hr

3.457 hr

previous example, the total velocity change required to go from low-Earth orbit to geosynchronous is 4.71 km/s using a spiral transfer. We obtain this by subtracting the results of step 3 from the results of step 2 in the above example.

### 6.3.2 Orbit Plane Changes

To change the orientation of the satellite's oibital plane, typically the inclination, we must change the direction of the velocity vector. This maneuver requires a component of A V to be perpendicular to the oibital plane and, therefore, perpendicular to the initial velocity vector. If the size of the oibit remains constant, we call the maneuver a simple plane change (Fig. 6-9A). We can find the required change in velocity by using the law of cosines. For the case in which Vj is equal to V,, this expression reduces to

where V. is the velocity before and after the burn, and 0 is angle change required.

Fig. 6-9. Vector Representation of Simple and Combined Changes in Orbital Plane. For the simple plane change, the magnitude of initial and final velocities are equal.

### A. Simple Plane Change B. Combined Plane Change

Fig. 6-9. Vector Representation of Simple and Combined Changes in Orbital Plane. For the simple plane change, the magnitude of initial and final velocities are equal.

For example, the change in velocity required to transfer from a low-altitude (h = 185 km) inclined (/=28 deg) orbit to an equatorial oibit (i = 0) at the same altitude is:

From Eq. (6-38) we see that if the angular change equals 60 deg, the required change in velocity equals the current velocity. Plane changes are very expensive in terms of the required velocity change and resulting fuel consumption. To minimize this, we should change the plane at a point where the velocity of the satellite is a minimum: at apogee for an elliptical oibit In some cases, it may even be cheaper to boost the satellite into a higher orbit, change the oibital plane at apogee, and return the satellite to its original oibit

Typically, oibital transfers require changes in both the size and the plane of the orbit, such as transferring from an inclined parking oibit at low altitude to a zero-inclination orbit at geosynchronous altitude. We do this transfer in two steps: a Hohmann transfer to change the size of the oibit and a simple plane change to make the orbit equatorial. A more efficient method (less total change in velocity) would be to combine the plane change with the tangential burn at apogee of the transfer orbit (Fig. 6-9B). As we must change both the magnitude and direction of the velocity vector, we can find the required change in velocity using the law of cosines:

where Vi is the initial velocity, Vj\s the final velocity, and 0 is the angle change required.

For example, we find the total change in velocity to transfer from a Shuttle parking oibit to a geosynchronous equatorial orbit as follows:

Completing a Hohmann transfer followed by a simple plane change would require a velocity change of 5.44 km/s, so the Hohmann transfer with a combined plane change at apogee of the transfer orbit represents a savings of 1.15 km/s. As we see from Gq. (6-39), a small plane change (0 £ 0) can be combined with an energy change for almost no cost in AV or propellant. Consequently, in practice, we do geosynchronous transfer with a small plane change at perigee and most of the plane change at apogee.

Another option is to complete the maneuver using three bums. The first bum is a coplanar maneuver placing the satellite into a transfer orbit with an apogee much higher than the final orbit. When the satellite reaches apogee of the transfer orbit, it does a combined plane change maneuver. This places the satellite in a second transfer orbit which is coplanar with the final orbit and has a perigee altitude equal to the altitude of the final orbit. Finally, when the satellite reaches perigee of the second transfer orbit, another coplanar maneuver places the satellite into the final oibit. This three-burn maneuver may save fuel, but the fuel savings comes at the expense of the total time required to complete the maneuver.

### 633 Orbit Rendezvous

Orbital transfer becomes more complicated when the objective is to rendezvous with or intercept another object in space: both the interceptor and target must arrive at the rendezvous point at the same time. This precise timing demands a phasing oibit to accomplish the maneuver. A phasing orbit is any orbit which results in the interceptor achieving the desired geometry relative to the target to initiate a Hohmann transfer. If the initial and final orbits are circular, coplanar, and of different sizes, then the phasing orbit is simply the initial interceptor orbit (Fig. 6-10). The interceptor remains in the initial orbit until the relative motion between the interceptor and target results in the desired geometry. At that point, we inject the interceptor into a Hohmann transfer orbit The equation to solve for die wait time in the initial orbit is:

where ^is the phase angle (angular separation of target and interceptor) needed for rendezvous, is the initial phase angle, k is the number of rendezvous opportunities, (for the first opportunity, k = 0), co,„, is the angular velocity of the interceptor, and <olgt is the angular velocity of the target We calculate the lead angle, aL, by multiplying oa,gl by the time of flight for the Hohmann transfer and is 180 deg minus aL.

Target

Target

Fig. 6-10. Geometry Depleting Rendezvous Between Two Circular, Coplanar Orbits. The phase angle is the angular separation between the target and interceptor at the start of the rendezvous and the lead angle Is the distance the target travels from the start until rendezvous occurs.

Fig. 6-10. Geometry Depleting Rendezvous Between Two Circular, Coplanar Orbits. The phase angle is the angular separation between the target and interceptor at the start of the rendezvous and the lead angle Is the distance the target travels from the start until rendezvous occurs.

The total time to rendezvous equals the wait time from Eq. (6-40) plus the time of flight of the Hohmann transfer orbit

The denominator in Eq. (6-40) represents the relative motion between the interceptor and target As the size of the interceptor orbit approaches the size of the taiget orbit, the relative motion approaches zero, and the wait time approaches infinity. If the two orbits are exactly the same, then the interceptor must enter a new phasing orbit to rendezvous with the taiget (Fig. 6-11). For this situation, the rendezvous occurs at the point where the interceptor enters the phasing oibit. The period of the phasing oibit equals the time it takes the taiget to get to the rendezvous point. Once we know the period, we can calculate the semimajor axis. The two orbits are tangential at their point of intersection, so the velocity change is the difference in magnitudes of the two velocities at the point of intersection of the two orbits. Because we know the size of the oibits, and therefore, the energies, we can use the energy Eq. (6-4) to solve for the current and needed velocities.

Frequently operators must adjust the relative phasing for satellites in circular orbits. They accomplish this by making the satellite drift relative to its initial position. The drift rate in deg/orbit is given by

A. Interceptor slightly behind target: Interceptor

B. Interceptor slightly ahead of target: Interceptor

Fig. 6-11. Rendezvous from Same Orbit Showing the Target both Leading and Trailing

A. Interceptor slightly behind target: Interceptor

B. Interceptor slightly ahead of target: Interceptor

Fig. 6-11. Rendezvous from Same Orbit Showing the Target both Leading and Trailing where Fis the nominal orbital velocity and A V is the velocity change required to start or stop the drift

The techniques described above move the target vehicle close to the interceptor. Once the two vehicles are close to each other they begin proximity operations by solving a set of relative motion equations to achieve the final rendezvous. Vallado [1997] contains an excellent discussion of the solution to the nearby relative motion problem, as addressed by the Clohessy-Wiltshire or Hill's equations of relative motion.

Similar to the rendezvous problem is the launch-window problem, or determining the appropriate time to launch from the surface of the Earth into the desired orbital plane. Because the orbital plane is fixed in inertial space, the launch window is the time when the launch site on the surface of the Earth rotates through the orbital plane. As Fig. 6-12 shows, the launch time depends on the launch site's latitude and longitude and the satellite orbit's inclination and right ascension of the ascending node.

For a launch window to exist, the launch site must pass through the orbital plane. This requirement places restrictions on the orbital inclinations, /, possible from a given launch latitude, L:

• No launch windows exist if L > i for direct orbit or L > 180 deg - / for retrograde orbits.

• One launch window exists if L = i or L = 180 deg - i.

• Two launch windows exist if L < i or L < 180 deg - i.

The launch azimuth, f), is the angle measured clockwise from north to the velocity vector. If a launch window exists, then the launch azimuth required to achieve an inclination, i, from a given launch latitude, L, is given by:

the Interceptor.

6.4 Launch Windows

Fig. 6-12. Launch Window Geometry for Launches near the Ascending Node (1) and Descending Node (2). The angles shown are the orbital inclination (I), launch site latitude (/.), and launch azimuth (ft).

where VL is the inertial velocity of the launch site given by Eq. (6-46) below, Vgg = 464.5 mis is the velocity of Earth's rotation at the equator, and V0 ~ 7.8 km/s is the velocity of the satellite immediately after launch, ft is the inertial launch azimuth and y is a small correction to account for the velocity contribution caused by Earth's rotation. For launches to low-Earth orbit, y ranges from 0 for a due east launch to 3.0 deg for launch into a polar orbit The approximation for y in Eq. (6-42c) is accurate to within 0.1 deg for low-Earth orbits. For launches near the ascending node, P is in the first or fourth quadrant and the plus sign applies in Eq. (6-42a). For launches near the descending node, /? is in the second or third quadrant and the minus sign applies in Eq. (642a).

Let S, shown in Fig. 6-12, be the angle in the equatorial plane from the nearest node to the longitude of the launch site. We can determine <5 from:

where <5 is positive for direct orbits and negative for retrograde orbits. Finally, the local sidereal time, 1ST, of launch is the angle from the vernal equinox to the longitude of the launch site at the time of launch:

= £2+ 180 deg - <5 (launch at the descending node) (6-44) where iiisthe right ascension of the ascending node of the resulting orbit

Having calculated the launch azimuth required to achieve the desired orbit, we can now calculate the velocity needed to accelerate the payload from rest at the launch site to the required burnout velocity. To do so, we use topocentric-horizon coordinates with velocity components Vs, VE, Vz:

where Fto is the velocity at burnout (usually equal to the circular orbital velocity at the prescribed altitude), <j) is the flight path angle at burnout, pb is the launch azimuth at burnout, and VL is the velocity of the launch site at a given latitude, L, as given by:

Equations (6-45c) do not include losses in the velocity of the launch vehicle because of atmospheric drag and gravity—approximately 1,500 m/s for a typical launch vehicle. Also, in Eq. (6-45c) we assume that the azimuth at launch and the azimuth at burnout are the same. Changes in the latitude and longitude of the launch vehicle during powered flight will introduce small errors into the calculation of the burnout conditions. We can calculate the velocity required at burnout from the energy equation if we know the semimajor axis and radius of burnout of the orbit [Eq. (6-4)].

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