Two-Body Orbital Mechanics
2.5 Elliptical Orbit
2.5.5 Geocentric Orbits
We will briefly discuss a few common Earth-centered or geocentric orbits in this subsec-tion. A low-Earth orbit (LEO) is a circular orbit with an altitude up to roughly 1,000 km.
The lower bound on LEO altitude is determined by interaction with the upper atmos-phere (and the subsequent aerodynamic drag) and the intended lifetime of the satellite.
For example, the Apollo lunar missions began by injecting the upper stage of the Saturn V rocket into a 190-km altitude circular parking orbit. After the ground control-lers verified that the spacecraft systems were operating as intended the upper stage was reignited to send the astronauts on a translunar orbit to the moon. Even at an altitude of 190 km, a satellite experiences enough aerodynamic drag such that it would lose energy over several days and eventually enter the dense atmosphere and be destroyed by aero-dynamic heating (we will discuss aeroaero-dynamic drag on LEO satellites in Chapter 5).
Therefore, a parking orbit is a temporary, intermediate orbit and not a final destination.
Interplanetary spacecraft (such as the Mars Science Laboratory) are inserted into parking orbits before an upper rocket stage is fired to send the spacecraft on a trajectory to its planetary target.
Orbiting science platforms, such as the International Space Station (ISS) and the Hubble Space Telescope (HST), occupy LEOs. For example, the ISS and HST are in nearly circular LEOs with altitudes of about 410 and 540 km, respectively. The US Space Shuttle achieved circular LEO with altitudes ranging from roughly 300 to 500 km and was used to construct and service the ISS. For a typical Shuttle orbit with an altitude of 320 km (i.e., r = 6,698 km), Eq. (2.82) shows that the circular velocity is vc= 7.714 km/s and Eq. (2.84) shows that its orbital period is about 91 min.
A medium-Earth orbit (MEO) has an altitude ranging from roughly 1,000 to 35,000 km. Navigation and communication satellites are often placed in MEO. One example is the Global Positioning System (GPS), which consists of a constellation of satellites in circular orbits at an altitude of 20,180 km (r = 26,558 km). Hence, a GPS satellite has a circular orbital velocity of 3.874 km/s and a period of 12 h.
A geostationary-equatorial orbit (GEO) is a circular orbit with angular velocity that matches the Earth’s rotation rate about its axis. Because the Earth completes one revo-lution in one sidereal day (23 h, 56 min, and 4 s) the circular radius required for GEO can be computed from Eq. (2.84)
rGEO= TGEO μ 2π
2/3
where the GEO period is TGEO= 86,164 s (or 23.934 h). GEO radius must be rGEO= 42,164 km (or an altitude of 35,786 km above the Earth). The orbital plane of a GEO satellite is coincident with the Earth’s equatorial plane (hence the “E” in the GEO acro-nym) so that it remains motionless (or stationary) to a ground-based observer and appears to“hover” overhead a particular geographic point on the equator. Communica-tion and weather satellites are placed in GEO because they are always visible from a ground-based station and can always monitor the same geographic region. Using Eq. (2.82), we find that the circular orbital velocity for GEO is 3.075 km/s.
The reader may wonder why the period of GEO is 23 h, 56 min, and 4 s and not simply 24 h. A sidereal day (23.934 h) is the time required for the Earth to complete one revolution relative to an inertial frame. Hence, the inertial spin rate of the Earth is one revolution per sidereal day orωE= 7.292(10−5) rad/s. Therefore, a GEO satellite’s rotation rate must beθ = vGEO/rGEO=ωEin order to match the Earth’s rotation. A solar dayis the time required for the sun to reappear directly over the same meridian (line of longitude). In other words, a solar day (24 h) is the period between 12 o’clock noon on one day and 12 o’clock noon on the next day. Because we measure the 24 h solar day relative to the sun (and the Earth is moving in its orbit about the sun), the Earth actually completes more than one revolution in 24 h when the sun reappears directly overhead at noon.
The Geostationary Operational Environmental Satellite (GOES) system is a collection of GEO satellites used by the National Weather Service for weather monitoring and fore-casting. The GOES system was established with the launch of GOES-1 in 1975. At the time of writing, GOES-13 (or GOES-East) is positioned at a longitude of 75οW (New York City is at 74οW) and GOES-15 (GOES-West) is located at 135ο W (Honolulu, Hawaii is at 157.5οW). Of course, the GOES satellites“hover” over the equator at their respective longitudes. At GEO altitude each GOES satellite can view about 42% of the Earth’s surface where the total viewing area is centered on its fixed longitude.
Russia uses the so-called Molniya orbit for its communication satellites. A Molniya orbit is highly elliptical with apogee and perigee altitudes of about 39,874 and 500 km, respectively. Eccentricity of a Molniya orbit is 0.741 and the period is 12 h.
Molniya orbits are oriented so that apogee is located at a very high geographic latitude (we will discuss the three-dimensional orientation of orbits in Chapter 3). Because satel-lites in Molniya orbits spend most of their time near apogee, they are well suited to view northern latitudes such as Russia.
Example 2.5 Consider a US communication satellite that is destined for a geostation-ary-equatorial orbit (GEO). Initially, the satellite is placed in a 200-km altitude circular low-Earth orbit (LEO) by a launch vehicle. In order to reach GEO, the satellite follows a geostationary transfer orbit(GTO) as shown in Figure 2.13. The GTO is an ellipse that is tangent to the inner and outer circular orbits and therefore has a perigee radius equal to the LEO radius and an apogee radius equal to the GEO radius. Determine:
a) Perigee velocity on the GTO and LEO circular velocity.
b) Apogee velocity on the GTO and GEO circular velocity.
c) Period of the GTO.
a) We begin by defining the perigee and apogee radii for the GTO:
GTO perigee rp= RE+ 200 km = 6,578 km GTO apogee ra= RE+ 35,786 km = 42,164 km
LEO altitude (200 km) is given in the problem while GEO altitude is known to be 35,786 km (see the previous section discussing GEO radius and the sidereal day). We can compute semimajor axis of the GTO from the perigee and apogee radii:
a=rp+ ra
2 = 24,371 km Total energy of the GTO is solely a function of a
ξ = − μ
2a=– 8 1778 km2/s2
Knowledge of orbital energy allows us to determine velocity at a known radial dis-tance using Eq. (2.29). Using perigee radius rp= 6,578 km, the perigee velocity on the GTO is
vp= 2 ξ + μ
rp = 10 2390 km/s The GTO perigee velocity vpis shown in Figure 2.13.
We compute circular velocity for LEO using Eq. (2.82) and radius rLEO= 6,578 km vLEO= μ
rLEO= 7 7843 km/s
A satellite in circular LEO and at GTO perigee has the same potential energy due to the common radius rLEO= rp= 6,578 km. However, the total energy of GTO is greater than LEO energy (to see this compare the GTO and LEO semimajor axes; see
Figure 2.13). Therefore, the kinetic energy at GTO perigee must be greater than the kinetic energy of circular LEO; the calculations of vpand vLEOdemonstrate this fact.
b) We can compute GTO apogee velocity using the energy equation with ra = 42,164 km:
va= 2 ξ + μ ra
= 1 5974 km/s
The GTO apogee velocity vais also shown in Figure 2.13. GTO apogee velocity is less than 16% of its perigee velocity.
We determine the circular velocity for GEO using radius rGEO= 42,164 km vGEO= μ
rGEO= 3 0747 km/s
This result makes sense because GEO energy is greater than GTO energy (again, see Figure 2.13 and compare the semimajor axes). Because GTO apogee and GEO have the same potential energy, the GEO kinetic energy must be greater than the kinetic energy at GTO apogee.
c) Orbital period is solely a function of semimajor axis. Using Eq. (2.80) the orbital period of the GTO is
TGTO= 2π
μa3/2= 37, 863 5 s = 10 52 h
In practice, an onboard rocket is fired in LEO to increase the satellite’s velocity from vLEO= 7.7843 km/s to the required GTO perigee velocity vp= 10.2390 km/s. The sat-ellite then coasts for one-half of a revolution (5.26 h) to reach GTO apogee where a
GTO perigee 200 km altitude GTO apogee
35,786 km altitude
vp
va
GTO
GEO
LEO
Figure 2.13 Geostationary transfer orbit (Example 2.5).
second rocket burn is needed to increase the velocity from va= 1.5974 km/s to GEO circular velocity vGEO= 3.0747 km/s. Orbital maneuvers and their associated propel-lant mass requirements are treated in Chapters 6 and 7.
Example 2.6 The Lunar Atmosphere and Dust Environment Explorer (LADEE) space-craft was launched into a highly elliptical orbit by a Minotaur V booster in September 2013. After completing the fifth-stage burn of the Minotaur V at an altitude of 200 km, the LADEE spacecraft entered an elliptical orbit with perigee and apogee alti-tudes of 200 km and 278,000 km, respectively (see Figure 2.14). The LADEE spacecraft completed one revolution in this elliptical orbit before firing an onboard rocket at perigee to increase its orbital energy. Determine the eccentricity of the elliptical orbit and the coasting time between the Minotaur booster burnout and the thrusting maneuver at perigee.
First, we determine the perigee and apogee radii from the altitude information:
Perigee rp= RE+ 200 km = 6,578 km Apogee ra= RE+ 278,000 km = 284,378 km Using Eq. (2.65), the eccentricity is determined to be
e=ra−rp
rp+ ra
= 0 9548
Hence, the“coasting orbit” for the LADEE spacecraft is highly elliptical.
We can compute the orbital period using Eq. (2.80) and knowledge of the semimajor axis. The semimajor axis is half of the sum of perigee and apogee radii:
a=rp+ ra
2 = 145,478 km The orbital period of the ellipse is
Tperiod= 2π
μa3/2= 552,214 s = 6 39 days Therefore, the total coasting time is one period, or
tcoast= Tperiod= 6 39 days
Perigee altitude 200 km Apogee altitude
278,000 km
Earth
Figure 2.14 Elliptical orbit for the LADEE spacecraft (Example 2.6).
The LADEE spacecraft was placed in this intermediate elliptical orbit because the Minotaur V booster could not provide enough energy to achieve a direct orbit to the moon. LADEE’s onboard propulsion system provided the final energy increase to send the spacecraft to the moon. We will analyze the LADEE orbital maneuvers in Chapter 7.