ROCKET PROPELLANTS

Liquids Solids Hybrids Tables of Properties

Propellant is the chemical mixture burned to produce thrust in rockets and consists of a fuel and an oxidizer. A fuel is a substance which burns when combined with oxygen producing gas for propulsion. An oxidizer is an agent that releases oxygen for combination with a fuel. The ratio of oxidizer to fuel is called the mixture ratio. Propellants are classified according to their state - liquid, solid, or hybrid.

The gauge for rating the efficiency of rocket propellants is specific impulse, stated in seconds. Specific impulse indicates how many pounds (or kilograms) of thrust are obtained by the consumption of one pound (or kilogram) of propellant in one second. Specific impulse is characteristic of the type of propellant, however, its exact value will vary to some extent with the operating conditions and design of the rocket engine.

Liquid PropellantsIn a liquid propellant rocket, the fuel and oxidizer are stored in separate tanks, and are fed through a system of pipes, valves, and turbopumps to a combustion chamber where they are combined and burned to produce thrust. Liquid propellant engines are more complex than their solid propellant counterparts, however, they offer several advantages. By controlling the flow of propellant to the combustion chamber, the engine can be throttled, stopped, or restarted.

A good liquid propellant is one with a high specific impulse or, stated another way, one with a high speed of exhaust gas ejection. This implies a high combustion temperature and exhaust gases with small molecular weights. However, there is another important factor which must be taken into consideration: the density of the propellant. Using low density propellants means that larger storage tanks will be required, thus increasing the mass of the launch vehicle. Storage temperature is also important. A propellant with a low storage temperature, i.e. a cryogenic, will require thermal insulation, thus further increasing the mass of the launcher. The toxicity of the propellant is likewise important. Safety hazards exist when handling, transporting, and storing highly toxic compounds. Also, some propellants are very corrosive, however, materials that are resistant to certain propellants have been identified for use in rocket construction.

Liquid propellants used in rocketry can be classified into three types: petroleum, cryogens, and hypergols.

Petroleum fuels are those refined from crude oil and are a mixture of complex hydrocarbons, i.e. organic compounds containing only carbon and hydrogen. The petroleum used as rocket fuel is a type of highly refined kerosene, called RP-1 in the United States. Petroleum fuels are usually used in combination with liquid oxygen as the oxidizer. Kerosene delivers a specific impulse considerably less than cryogenic fuels, but it is generally better than hypergolic propellants.

Specifications for RP-1 where first issued in the United States in 1957 when the need for a clean burning petroleum rocket fuel was recognized. Prior experimentation with jet fuels produced tarry residue in the engine cooling passages and excessive soot, coke and

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other deposits in the gas generator. Even with the new specifications, kerosene-burning engines still produce enough residues that their operational lifetimes are limited.

Liquid oxygen and RP-1 are used as the propellant in the first-stage boosters of the Atlas and Delta II launch vehicles. It also powered the first-stages of the Saturn 1B and Saturn V rockets.

Cryogenic propellants are liquefied gases stored at very low temperatures, most frequently liquid hydrogen (LH2) as the fuel and liquid oxygen (LO2 or LOX) as the oxidizer. Hydrogen remains liquid at temperatures of -253 oC (-423 oF) and oxygen remains in a liquid state at temperatures of -183 oC (-297 oF) .

Because of the low temperatures of cryogenic propellants, they are difficult to store over long periods of time. For this reason, they are less desirable for use in military rockets that must be kept launch ready for months at a time. Furthermore, liquid hydrogen has a very low density (0.071 g/ml) and, therefore, requires a storage volume many times greater than other fuels. Despite these drawbacks, the high efficiency of liquid oxygen/liquid hydrogen makes these problems worth coping with when reaction time and storability are not too critical. Liquid hydrogen delivers a specific impulse about 30%-40% higher than most other rocket fuels.

Liquid oxygen and liquid hydrogen are used as the propellant in the high efficiency main engines of the Space Shuttle. LOX/LH2 also powered the upper stages of the Saturn V and Saturn 1B rockets, as well as the Centaur upper stage, the United States' first LOX/LH2 rocket (1962).

Another cryogenic fuel with desirable properties for space propulsion systems is liquid methane (-162 oC). When burned with liquid oxygen, methane is higher performing than state-of-the-art storable propellants but without the volume increase common with LOX/LH2 systems, which results in an overall lower vehicle mass as compared to common hypergolic propellants. LOX/methane is also clean burning and non-toxic. Future missions to Mars will likely use methane fuel because it can be manufactured partly from Martian in-situ resources. LOX/methane has no flight history and very limited ground-test history.

Liquid fluorine (-188 oC) burning engines have also been developed and fired successfully. Fluorine is not only extremely toxic; it is a super-oxidizer that reacts, usually violently, with almost everything except nitrogen, the lighter noble gases, and substances that have already been fluorinated. Despite these drawbacks, fluorine produces very impressive engine performance. It can also be mixed with liquid oxygen to improve the performance of LOX-burning engines; the resulting mixture is called FLOX. Because of fluorine's high toxicity, it has been largely abandoned by most space-faring nations.

Some fluorine containing compounds, such as chlorine pentafluoride, have also been considered for use as an 'oxidizer' in deep-space applications.

Hypergolic propellants are fuels and oxidizers that ignite spontaneously on contact with each other and require no ignition source. The easy start and restart capability of hypergols make them ideal for spacecraft maneuvering systems. Also, since hypergols remain liquid at normal temperatures, they do not pose the storage problems of cryogenic propellants. Hypergols are highly toxic and must be handled with extreme care.

Hypergolic fuels commonly include hydrazine, monomethyl hydrazine (MMH) and unsymmetrical dimethyl hydrazine (UDMH). Hydrazine gives the best performance as a rocket fuel, but it has a high freezing point and is too unstable for use as a coolant. MMH is more stable and gives the best performance when freezing point is an issue, such as spacecraft propulsion applications. UDMH has the lowest freezing point and has enough

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thermal stability to be used in large regeneratively cooled engines. Consequently, UDMH is often used in launch vehicle applications even though it is the least efficient of the hydrazine derivatives. Also commonly used are blended fuels, such as Aerozine 50 (or "50-50"), which is a mixture of 50% UDMH and 50% hydrazine. Aerozine 50 is almost as stable as UDMH and provides better performance.

The oxidizer is usually nitrogen tetroxide (NTO) or nitric acid. In the United States, the nitric acid formulation most commonly used is type III-A, called inhibited red-fuming nitric acid (IRFNA), which consists of HNO3 + 14% N2O4 + 1.5-2.5% H2O + 0.6% HF (added as a corrosion inhibitor). Nitrogen tetroxide is less corrosive than nitric acid and provides better performance, but it has a higher freezing point. Consequently, nitrogen tetroxide is usually the oxidizer of choice when freezing point is not an issue; however, the freezing point can be lowered with the introduction nitric oxide. The resulting oxidizer is called mixed oxides of nitrogen (MON). The number included in the description, e.g. MON-3 or MON-25, indicates the percentage of nitric oxide by weight. While pure nitrogen tetroxide has a freezing point of about -9 oC, the freezing point of MON-3 is -15 oC and that of MON-25 is -55 oC.

USA military specifications for IRFNA were first published in 1954, followed in 1955 with UDMH specifications.

The Titan family of launch vehicles and the second stage of the Delta II rocket use NTO/Aerozine 50 propellant. NTO/MMH is used in the orbital maneuvering system (OMS) and reaction control system (RCS) of the Space Shuttle orbiter. IRFNA/UDMH is often used in tactical missiles such as the US Army's Lance (1972-91).

Hydrazine is also frequently used as a monopropellant in catalytic decomposition engines. In these engines, a liquid fuel decomposes into hot gas in the presence of a catalyst. The decomposition of hydrazine produces temperatures up to about 1,100 oC (2,000 oF) and a specific impulse of about 230 or 240 seconds. Hydrazine decomposes to either hydrogen and nitrogen, or ammonia and nitrogen.

Other propellants have also been used, a few of which deserve mentioning:

Alcohols were commonly used as fuels during the early years of rocketry. The German V-2 missile, as well as the USA Redstone, burned LOX and ethyl alcohol (ethanol), diluted with water to reduce combustion chamber temperature. However, as more efficient fuels where developed, alcohols fell into general disuse.

Hydrogen peroxide once attracted considerable attention as an oxidizer and was used in Britain's Black Arrow rocket. In high concentrations, hydrogen peroxide is called high-test peroxide (HTP). The performance and density of HTP is close to that of nitric acid, and it is far less toxic and corrosive, however it has a poor freezing point and is unstable. Although HTP never made it as an oxidizer in large bi-propellant applications, it has found widespread use as a monopropellant. In the presence of a catalyst, HTP decomposes into oxygen and superheated steam and produces a specific impulse of about 150 s.

Nitrous oxide has been used as both an oxidizer and as a monopropellant. It is the oxidizer of choice for many hybrid rocket designs and has been used frequently in amateur high-powered rocketry. In the presence of a catalyst, nitrous oxide will decompose exothermically into nitrogen and oxygen and produce a specific impulse of about 170 s.

Solid PropellantsSolid propellant motors are the simplest of all rocket designs. They consist of a casing, usually steel, filled with a mixture of solid compounds (fuel and oxidizer) which burn at a

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rapid rate, expelling hot gases from a nozzle to produce thrust. When ignited, a solid propellant burns from the center out towards the sides of the casing. The shape of the center channel determines the rate and pattern of the burn, thus providing a means to control thrust. Unlike liquid propellant engines, solid propellant motors can not be shut down. Once ignited, they will burn until all the propellant is exhausted.

There are two families of solids propellants: homogeneous and composite. Both types are dense, stable at ordinary temperatures, and easily storable.

Homogeneous propellants are either simple base or double base. A simple base propellant consists of a single compound, usually nitrocellulose, which has both an oxidation capacity and a reduction capacity. Double base propellants usually consist of nitrocellulose and nitroglycerine, to which a plasticiser is added. Homogeneous propellants do not usually have specific impulses greater than about 210 seconds under normal conditions. Their main asset is that they do not produce traceable fumes and are, therefore, commonly used in tactical weapons. They are also often used to perform subsidiary functions such as jettisoning spent parts or separating one stage from another.

Modern composite propellants are heterogeneous powders (mixtures) which use a crystallized or finely ground mineral salt as an oxidizer, often ammonium perchlorate, which constitutes between 60% and 90% of the mass of the propellant. The fuel itself is generally aluminum. The propellant is held together by a polymeric binder, usually polyurethane or polybutadienes, which is also consumed as fuel. Additional compounds are sometimes included, such as a catalyst to help increase the burning rate, or other agents to make the powder easier to manufacture. The final product is rubberlike substance with the consistency of a hard rubber eraser.

Composite propellants are often identified by the type of polymeric binder used. The two most common binders are polybutadiene acrylic acid acrylonitrile (PBAN) and hydroxy-terminator polybutadiene (HTPB). PBAN formulations give a slightly higher specific impulse, density, and burn rate than equivalent formulations using HTPB. However, PBAN propellant is the more difficult to mix and process and requires an elevated curing temperature. HTPB binder is stronger and more flexible than PBAN binder. Both PBAN and HTPB formulations result in propellants that deliver excellent performance, have good mechanical properties, and offer potentially long burn times.

Solid propellant motors have a variety of uses. Small solids often power the final stage of a launch vehicle, or attach to payloads to boost them to higher orbits. Medium solids such as the Payload Assist Module (PAM) and the Inertial Upper Stage (IUS) provide the added boost to place satellites into geosynchronous orbit or on planetary trajectories.

The Titan, Delta, and Space Shuttle launch vehicles use strap-on solid propellant rockets to provide added thrust at liftoff. The Space Shuttle uses the largest solid rocket motors ever built and flown. Each booster contains 500,000 kg (1,100,000 pounds) of propellant and can produce up to 14,680,000 Newtons (3,300,000 pounds) of thrust.

Hybrid PropellantsHybrid propellant engines represent an intermediate group between solid and liquid propellant engines. One of the substances is solid, usually the fuel, while the other, usually the oxidizer, is liquid. The liquid is injected into the solid, whose fuel reservoir also serves as the combustion chamber. The main advantage of such engines is that they have high performance, similar to that of solid propellants, but the combustion can be moderated, stopped, or even restarted. It is difficult to make use of this concept for vary large thrusts, and thus, hybrid propellant engines are rarely built.

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SpaceShipOne, which won the Ansari X-Prize, was powered by a hybrid engine burning nitrous oxide as the liquid oxidizer and HTPB rubber as the solid fuel.

PROPERTIES OF ROCKET PROPELLANTS 

Compound ChemicalFormula

MolecularWeight Density Melting

PointBoilingPoint

Liquid Oxygen O2 32.00 1.14 g/ml -218.8oC -183.0oCLiquid Fluorine F2 38.00 1.50 g/ml -219.6oC -188.1oC

Nitrogen Tetroxide N2O4 92.01 1.45 g/ml -9.3oC 21.15oCNitric Acid HNO3 63.01 1.55 g/ml -41.6oC 83oC

Hydrogen Peroxide H2O2 34.02 1.44 g/ml -0.4oC 150.2oCNitrous Oxide N2O 44.01 1.22 g/ml -90.8oC -88.5oC

Chlorine Pentafluoride ClF5 130.45 1.9 g/ml -103oC -13.1oCAmmonium Perchlorate ClH4NO4 117.49 1.95 g/ml 240oC N/A

Liquid Hydrogen H2 2.016 0.071 g/ml -259.3oC -252.9oCLiquid Methane CH4 16.04 0.423 g/ml -182.5oC -161.6oCEthyl Alcohol C2H5OH 46.07 0.789 g/ml -114.1oC 78.2oC

n-Dodecane (Kerosene) C12H26 170.34 0.749 g/ml -9.6oC 216.3oCRP-1 CnH1.953n ≈175 0.820 g/ml N/A 177-274oC

Hydrazine N2H4 32.05 1.004 g/ml 1.4oC 113.5oCMethyl Hydrazine CH3NHNH2 46.07 0.866 g/ml -52.4oC 87.5oC

Dimethyl Hydrazine (CH3)2NNH2 60.10 0.791 g/ml -58oC 63.9oCAluminum Al 26.98 2.70 g/ml 660.4oC 2467oC

Polybutadiene (C4H6)n ≈3000 ≈0.9 g/ml N/A N/ANOTES: Chemically, kerosene is a mixture of hydrocarbons; the chemical composition depends on its source, but it usually consists of about ten different hydrocarbons, each containing from 10 to 16 carbon atoms per molecule; the constituents include n-dodecane, alkyl benzenes, and naphthalene and its derivatives. Kerosene is usually represented by the single compound n-dodecane. RP-1 is a special type of kerosene covered by Military Specification MIL-R-25576. In Russia, similar specifications were developed under specifications T-1 and RG-1. Nitrogen tetroxide and nitric acid are hypergolic with hydrazine, MMH and UDMH. Oxygen is not hypergolic with any commonly used fuel. Ammonium perchlorate decomposes, rather than melts, at a temperature of about 240 oC.

ROCKET PROPELLANT PERFORMANCE 

Combustion chamber pressure, Pc = 68 atm (1000 PSI) ... Nozzle exit pressure, Pe = 1 atm

Oxidizer Fuel Hypergolic Mixture Ratio

Specific Impulse

(s, sea level)

Density Impulse

(kg-s/l, S.L.)Liquid Oxygen Liquid Hydrogen No 5.00 381 124

Liquid Methane No 2.77 299 235Ethanol + 25% water No 1.29 269 264

Kerosene No 2.29 289 294Hydrazine No 0.74 303 321

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MMH No 1.15 300 298UDMH No 1.38 297 28650-50 No 1.06 300 300

Liquid FluorineLiquid Hydrogen Yes 6.00 400 155

Hydrazine Yes 1.82 338 432FLOX-70 Kerosene Yes 3.80 320 385

Nitrogen Tetroxide

Kerosene No 3.53 267 330Hydrazine Yes 1.08 286 342

MMH Yes 1.73 280 325UDMH Yes 2.10 277 31650-50 Yes 1.59 280 326

Red-Fuming Nitric Acid

(14% N2O4)

Kerosene No 4.42 256 335Hydrazine Yes 1.28 276 341

MMH Yes 2.13 269 328UDMH Yes 2.60 266 32150-50 Yes 1.94 270 329

Hydrogen Peroxide(85% concentration)

Kerosene No 7.84 258 324Hydrazine Yes 2.15 269 328

Nitrous Oxide HTPB (solid) No 6.48 248 290Chlorine Pentafluoride Hydrazine Yes 2.12 297 439

Ammonium Perchlorate(solid)

Aluminum + HTPB (a) No 2.12 266 469Aluminum + PBAN

(b) No 2.33 267 472

NOTES: Specific impulses are theoretical maximum assuming 100% efficiency; actual performance will be less. All mixture ratios are optimum for the operating pressures indicated, unless otherwise noted. LO2/LH2 and LF2/LH2 mixture ratios are higher than optimum to improve density impulse. FLOX-70 is a mixture of 70% liquid fluorine and 30% liquid oxygen. Where kerosene is indicated, the calculations are based on n-dodecane. Solid propellant formulation (a): 68% AP + 18% Al + 14% HTPB. Solid propellant formulation (b): 70% AP + 16% Al + 12% PBAN + 2% epoxy curing agent.

SELECTED ROCKETS AND THEIR PROPELLANTS 

Rocket Stage Engines Propellant Specific Impulse

Atlas/Centaur (1962)012

Rocketdyne YLR89-NA7 (x2)Rocketdyne YLR105-NA7

P&W RL-10A-3-3 (x2)

LOX/RP-1LOX/RP-1LOX/LH2

259s sl / 292s vac220s sl / 309s vac

444s vacuum

Titan II (1964) 12

Aerojet LR-87-AJ-5 (x2)Aerojet LR-91-AJ-5

NTO/Aerozine 50NTO/Aerozine 50

259s sl / 285s vac312s vacuum

Saturn V (1967)123

Rocketdyne F-1 (x5)Rocketdyne J-2 (x5)

Rocketdyne J-2

LOX/RP-1LOX/LH2LOX/LH2

265s sl / 304s vac424s vacuum424s vacuum

Space Shuttle (1981)

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OMSRCS

Thiokol SRB (x2) Rocketdyne SSME (x3)

Aerojet OMS (x2)Kaiser Marquardt R-40 & R-1E

PBAN SolidLOX/LH2

NTO/MMHNTO/MMH

242s sl / 268s vac 363s sl / 453s vac

313s vacuum280s vacuum

Delta II (1989)012

Castor 4A (x9)Rocketdyne RS-27Aerojet AJ10-118K

HTPB SolidLOX/RP-1

NTO/Aerozine 50

238s sl / 266s vac264s sl / 295s vac

320s vacuum

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ROCKET PROPULSION

Thrust Conservation of Momentum Impulse & Momentum Combustion & Exhaust Velocity Specific Impulse Rocket Engines Power Cycles Engine Cooling Solid Rocket Motors Monopropellant Engines Staging

Isaac Newton stated in his third law of motion that "for every action there is an equal and opposite reaction." It is upon this principle that a rocket operates. Propellants are combined in a combustion chamber where they chemically react to form hot gases which are then accelerated and ejected at high velocity through a nozzle, thereby imparting momentum to the engine. The thrust force of a rocket motor is the reaction experienced by the motor structure due to ejection of the high velocity matter. This is the same phenomenon which pushes a garden hose backward as water flows from the nozzle, or makes a gun recoil when fired.

ThrustThrust is the force that propels a rocket or spacecraft and is measured in pounds, kilograms or Newtons. Physically speaking, it is the result of pressure which is exerted on the wall of the combustion chamber.

Figure 1.1 shows a combustion chamber with an opening, the nozzle, through which gas can escape. The pressure distribution within the chamber is asymmetric; that is, inside the chamber the pressure varies little, but near the nozzle it decreases somewhat. The force due to gas pressure on the bottom of the chamber is not compensated for from the outside. The resultant force F due to the internal and external pressure difference, the thrust, is opposite to the direction of the gas jet. It pushes the chamber upwards.

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To create high speed exhaust gases, the necessary high temperatures and pressures of combustion are obtained by using a very energetic fuel and by having the molecular weight of the exhaust gases as low as possible. It is also necessary to reduce the pressure of the gas as much as possible inside the nozzle by creating a large section ratio. The section ratio, or expansion ratio, is defined as the area of the exit Ae divided by the area of the throat At.

The thrust F is the resultant of the forces due to the pressures exerted on the inner and outer walls by the combustion gases and the surrounding atmosphere, taking the boundary between the inner and outer surfaces as the cross section of the exit of the nozzle. As we shall see in the next section, applying the principle of the conservation of momentum gives

where q is the rate of the ejected mass flow, Pa the pressure of the ambient atmosphere, Pe the pressure of the exhaust gases and Ve their ejection speed. Thrust is specified either at sea level or in a vacuum.

Conservation of MomentumThe linear momentum (p), or simply momentum, of a particle is the product of its mass and its velocity. That is,

Newton expressed his second law of motion in terms of momentum, which can be stated as "the resultant of the forces acting on a particle is equal to the rate of change of the linear momentum of the particle". In symbolic form this becomes

which is equivalent to the expression F=ma.

If we have a system of particles, the total momentum P of the system is the sum of the momenta of the individual particles. When the resultant external force acting on a system is zero, the total linear momentum of the system remains constant. This is called the principle of conservation of linear momentum. Let's now see how this principle is applied to rocket mechanics.

Consider a rocket drifting in gravity free space. The rocket's engine is fired for time t and, during this period, ejects gases at a constant rate and at a constant speed relative to the rocket (exhaust velocity). Assume there are no external forces, such as gravity or air resistance.

Figure 1.2(a) shows the situation at time t. The rocket and fuel have a total mass M and the combination is moving with velocity v as seen from a particular frame of reference. At a time t later the configuration has changed to that shown in Figure 1.2(b). A mass

M has been ejected from the rocket and is moving with velocity u as seen by the observer. The rocket is reduced to mass M- M and the velocity v of the rocket is changed to v+ v.

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Because there are no external forces, dP/dt=0. We can write, for the time interval t

where P2 is the final system momentum, Figure 1.2(b), and P1 is the initial system momentum, Figure 1.2(a). We write

If we let t approach zero, v/ t approaches dv/dt, the acceleration of the body. The quantity M is the mass ejected in t; this leads to a decrease in the mass M of the original body. Since dM/dt, the change in mass of the body with time, is negative in this case, in the limit the quantity M/ t is replaced by -dM/dt. The quantity u-(v+ v) is Vrel, the relative velocity of the ejected mass with respect to the rocket. With these changes, equation (1.4) can be written as

The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it. The rocket designer can make the thrust as large as possible by designing the rocket to eject mass as rapidly as possible (dM/dt large) and with the highest possible relative speed (Vrel large).

In rocketry, the basic thrust equation is written as

where q is the rate of the ejected mass flow, Ve is the exhaust gas ejection speed, Pe is the pressure of the exhaust gases at the nozzle exit, Pa is the pressure of the ambient atmosphere, and Ae is the area of the nozzle exit. The product qVe, which we derived above (Vrel × dM/dt), is called the momentum, or velocity, thrust. The product (Pe-Pa)Ae, called the pressure thrust, is the result of unbalanced pressure forces at the nozzle exit. As we shall see latter, maximum thrust occurs when Pe=Pa.

Click here for example problem #1.1 (use your browser's "back" function to return)

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Equation (1.6) may be simplified by the definition of an effective exhaust gas velocity, C, defined as

Equation (1.6) then reduces to

Impulse & MomentumIn the preceding section we saw that Newton's second law may be expressed in the form

Multiplying both sides by dt and integrating from a time t1 to a time t2, we write

The integral is a vector known as the linear impulse, or simply the impulse, of the force F during the time interval considered. The equation expresses that, when a particle is acted upon by a force F during a given time interval, the final momentum p2 of the particle may be obtained by adding its initial momentum p1 and the impulse of the force F during the interval of time.

When several forces act on a particle, the impulse of each of the forces must be considered. When a problem involves a system of particles, we may add vectorially the momenta of all the particles and the impulses of all the forces involved. When can then write

For a time interval t, we may write equation (1.10) in the form

Let us now see how we can apply the principle of impulse and momentum to rocket mechanics.

Consider a rocket of initial mass M which it launched vertically at time t=0. The fuel is consumed at a constant rate q and is expelled at a constant speed Ve relative to the rocket. At time t, the mass of the rocket shell and remaining fuel is M-qt, and the velocity is v. During the time interval t, a mass of fuel q t is expelled. Denoting by u the absolute velocity of the expelled fuel, we apply the principle of impulse and momentum between time t and time t+ t. Please note, this derivation neglects the effect of air resistance.

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We write

We divide through by t and replace u-(v+ v) with Ve, the velocity of the expelled mass relative to the rocket. As t approaches zero, we obtain

Separating variables and integrating from t=0, v=0 to t=t, v=v, we obtain

which equals

The term -gt in equation (1.15) is the result of Earth's gravity pulling on the rocket. For a rocket drifting in space, -gt is not applicable and can be omitted. Furthermore, it is more appropriate to express the resulting velocity as a change in velocity, or V. Equation (1.15) thus becomes

Click here for example problem #1.2

Note that M represents the initial mass of the rocket and M-qt the final mass. Therefore, equation (1.16) is often written as

where mo/mf is called the mass ratio. Equation (1.17) is also known as Tsiolkovsky's rocket equation, named after Russian rocket pioneer Konstantin E. Tsiolkovsky (1857-1935) who first derived it.

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In practical application, the variable Ve is usually replaced by the effective exhaust gas velocity, C. Equation (1.17) therefore becomes

Alternatively, we can write

where e is a mathematical constant approximately equal to 2.71828.

Click here for example problem #1.3

For many spacecraft maneuvers it is necessary to calculate the duration of an engine burn required to achieve a specific change in velocity. Rearranging variables, we have

Click here for example problem #1.4

Combustion & Exhaust VelocityThe combustion process involves the oxidation of constituents in the fuel that are capable of being oxidized, and can therefore be represented by a chemical equation. During a combustion process the mass of each element remains the same. Consider the reaction of methane with oxygen

This equation states that one mole of methane reacts with two moles of oxygen to form one mole of carbon dioxide and two moles of water. This also means that 16 g of methane react with 64 g of oxygen to form 44 g of carbon dioxide and 36 g of water. All the initial substances that undergo the combustion process are called the reactants, and the substances that result from the combustion process are called the products.

The above combustion reaction is an example of a stoichiometric mixture, that is, there is just enough oxygen present to chemically react with all the fuel. The highest flame temperature is achieved under these conditions, however it is often desirable to operate a rocket engine at a "fuel-rich" mixture ratio. Mixture ratio is defined as the mass flow of oxidizer divided by the mass flow of fuel.

Consider the following reaction of kerosene(1) with oxygen,

Given the molecular weight of C12H26 is 170 and that of O2 is 32, we have a mixture ratio of

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which is typical of many rocket engines using kerosene, or RP-1, fuel.

The optimum mixture ratio is typically that which will deliver the highest engine performance (measured by specific impulse), however in some situations a different O/F ratio results in a better overall system. For a volume-constrained vehicle with a low-density fuel such as liquid hydrogen, significant reductions in vehicle size can be achieved by shifting to a higher O/F ratio. In that case, the losses in performance are more than compensated for by the reduced fuel tankage requirement. Also consider the example of bipropellant systems using NTO/MMH, where a mixture ratio of 1.67 results in fuel and oxidizer tanks of equal size. Equal sizing simplifies tank manufacturing, system packaging, and integration.

As we have seen previously, impulse thrust is equal to the product of the propellant mass flow rate and the exhaust gas ejection speed. The ideal exhaust velocity is given by

where k is the specific heat ratio, R' is the universal gas constant (8,314.51 N-m/kg mol-K in SI units, or 49,720 ft-lb/slug mol-oR in U.S. units), Tc is the combustion temperature, M is the average molecular weight of the exhaust gases, Pc is the combustion chamber pressure, and Pe is the pressure at the nozzle exit.

Specific heat ratio(2) varies depending on the composition and temperature of the exhaust gases, but it is usually about 1.2. The thermodynamics involved in calculating combustion temperatures are quite complicated, however, flame temperatures generally range from about 2,500 to 3,600 oC (4,500-6,500 oF). Chamber pressures can range from about about 7 to 250 atmospheres. Pe should be equal to the ambient pressure at which the engine will operate, more on this later. Click Here for charts providing optimum mixture ratio, adiabatic flame temperature, gas molecular weight, and specific heat ratio for some common rocket propellants.

From equation (1.22) we see that high chamber temperature and pressure, and low exhaust gas molecular weight results in high ejection velocity, thus high thrust. Based on this criterion, we can see why liquid hydrogen is very desirable as a rocket fuel.

Click here for example problem #1.5

It should be pointed out that in the combustion process there will be a dissociation of molecules among the products. That is, the high heat of combustion causes the separation of molecules into simpler constituents that are then capable of recombining. Consider the reaction of kerosene with oxygen. The true products of combustion will be an equilibrium mixture of atoms and molecules consisting of C, CO, CO2, H, H2, H2O, HO, O, and O2. Dissociation has a significant effect on flame temperature.

(1) In dealing with combustion of liquid hydrocarbon fuels it is convenient to express the composition in terms of a single hydrocarbon, even though it is a mixture of many hydrocarbons. Thus gasoline is usually considered to be octane, C8H18, and kerosene is considered to be dodecane, C12H26.

(2) Specific heat, or heat capacity, represents the amount of heat necessary to raise the temperature of one gram of a substance one degree C. Specific heat is measured at constant-pressure, CP, or at constant-volume, CV. The ratio CP/CV is called the specific heat ratio, represented by k.

Specific Impulse

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The specific impulse of a rocket, Isp, is the ratio of the thrust to the flow rate of the weight ejected, that is

where F is thrust, q is the rate of mass flow, and go is standard gravity (9.80665 m/s2).

Specific impulse is expressed in seconds. When the thrust and the flow rate remain constant throughout the burning of the propellant, the specific impulse is the time for which the rocket engine provides a thrust equal to the weight of the propellant consumed.

For a given engine, the specific impulse has different values on the ground and in the vacuum of space because the ambient pressure is involved in the expression for the thrust. It is therefore important to state whether specific impulse is the value at sea level or in a vacuum.

There are a number of losses within a rocket engine, the main ones being related to the inefficiency of the chemical reaction (combustion) process, losses due to the nozzle, and losses due to the pumps. Overall, the losses affect the efficiency of the specific impulse. This is the ratio of the real specific impulse (at sea level, or in a vacuum) and the theoretical specific impulse obtained with an ideal nozzle from gases coming from a complete chemical reaction. Calculated values of specific impulse are several percent higher than those attained in practice.

Click here for example problem #1.6

From Equation (1.8) we can substitute qC for F in Equation (1.23), thus obtaining

Equation (1.24) is very useful when solving Equations (1.18) through (1.21). It is rare we are given the value of C directly, however rocket engine specific impulse is a commonly given parameter from which we can easily calculate C.

Another important figure of merit for evaluating rocket performance is the characteristic exhaust velocity, C* (pronounced "C star"), which is a measure of the energy available from the combustion process and is given by

where Pc is the combustion chamber pressure and At is the area of the nozzle throat. Delivered values of C* range from about 1,333 m/s for monopropellant hydrazine up to about 2,360 m/s for cryogenic oxygen/hydrogen.

Rocket Engines

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A typical rocket engine consists of the nozzle, the combustion chamber, and the injector, as shown in Figure 1.4. The combustion chamber is where the burning of propellants takes place at high pressure. The chamber must be strong enough to contain the high pressure generated by, and the high temperature resulting from, the combustion process. Because of the high temperature and heat transfer, the chamber and nozzle are usually cooled. The chamber must also be of sufficient length to ensure complete combustion before the gases enter the nozzle.

Nozzle

The function of the nozzle is to convert the chemical-thermal energy generated in the combustion chamber into kinetic energy. The nozzle converts the slow moving, high pressure, high temperature gas in the combustion chamber into high velocity gas of lower pressure and temperature. Since thrust is the product of mass and velocity, a very high gas velocity is desirable. Nozzles consist of a convergent and divergent section. The minimum flow area between the convergent and divergent section is called the nozzle throat. The flow area at the end of the divergent section is called the nozzle exit area. The nozzle is usually made long enough (or the exit area is great enough) such that the pressure in the combustion chamber is reduced at the nozzle exit to the pressure existing outside the nozzle. It is under this condition, Pe=Pa where Pe is the pressure at the nozzle exit and Pa is the outside ambient pressure, that thrust is maximum and the nozzle is said to be adapted, also called optimum or correct expansion. When Pe is greater than Pa, the nozzle is under-extended. When the opposite is true, it is over-extended.

We see therefore, a nozzle is designed for the altitude at which it has to operate. At the Earth's surface, at the atmospheric pressure of sea level (0.1 MPa or 14.7 psi), the discharge of the exhaust gases is limited by the separation of the jet from the nozzle wall. In the cosmic vacuum, this physical limitation does not exist. Therefore, there have to be two different types of engines and nozzles, those which propel the first stage of the launch vehicle through the atmosphere, and those which propel subsequent stages or control the orientation of the spacecraft in the vacuum of space.

The nozzle throat area, At, can be found if the total propellant flow rate is known and the propellants and operating conditions have been selected. Assuming perfect gas law theory, we have

where q is the propellant mass flow rate, Pt is the gas pressure at the nozzle throat, Tt is the gas temperature at the nozzle throat, R' is the universal gas constant, and k is the specific heat ratio. Pt and Tt are given by

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where Pc is the combustion chamber pressure and Tc is the combustion chamber flame temperature.

Click here for example problem #1.7

The hot gases must be expanded in the diverging section of the nozzle to obtain maximum thrust. The pressure of these gases will decrease as energy is used to accelerate the gas. We must find that area of the nozzle where the gas pressure is equal to the outside atmospheric pressure. This area will then be the nozzle exit area.

Mach number Nm is the ratio of the gas velocity to the local speed of sound. The Mach number at the nozzle exit is given by the perfect gas expansion expression

where Pa is the pressure of the ambient atmosphere.

The nozzle exit area, Ae, corresponding to the exit Mach number is given by

The section ratio, or expansion ratio, is defined as the area of the exit Ae divided by the area of the throat At.

Click here for example problem #1.8

For launch vehicles (particularly first stages) where the ambient pressure varies during the burn period, trajectory computations are performed to determine the optimum exit pressure. However, an additional constraint is the maximum allowable diameter for the nozzle exit cone, which in some cases is the limiting constraint. This is especially true on stages other than the first, where the nozzle diameter may not be larger than the outer diameter of the stage below. For space engines, where the ambient pressure is zero, thrust always increases as nozzle expansion ratio increases. On these engines, the nozzle expansion ratio is generally increased until the additional weight of the longer nozzle costs more performance than the extra thrust it generates.

(For additional information, please see Supplement #1: Optimizing Expansion for Maximum Thrust.)

Since the flow velocity of the gases in the converging section of the rocket nozzle is relatively low, any smooth and well-rounded convergent nozzle section will have very low energy loses. By contrast, the contour of the diverging nozzle section is very important to performance, because of the very high flow velocities involved. The selection of an optimum nozzle shape for a given expansion ratio is generally influenced by the following design considerations and goals: (1) uniform, parallel, axial gas flow at the nozzle exit for maximum momentum vector, (2) minimum separation and turbulence losses within the nozzle, (3) shortest possible nozzle length for minimum space envelope, weight, wall friction losses, and cooling requirements, and (4) ease of manufacturing.

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Conical nozzle: In early rocket engine applications, the conical nozzle, which proved satisfactory in most respects, was used almost exclusively. A conical nozzle allows ease of manufacture and flexibility in converting an existing design to higher or lower expansion ratio without major redesign.

The configuration of a typical conical nozzle is shown in Figure 1.4. The nozzle throat section has the contour of a circular arc with radius R, ranging from 0.25 to 0.75 times the throat diameter, Dt. The half-angle of the nozzle convergent cone section, , can range from 20 to 45 degrees. The divergent cone half-angle, , varies from approximately 12 to 18 degrees. The conical nozzle with a 15-degree divergent half-angle has become almost a standard because it is a good compromise on the basis of weight, length, and performance.

Since certain performance losses occur in a conical nozzle as a result of the nonaxial component of the exhaust gas velocity, a correction factor, , is applied in the calculation of the exit-gas momentum. This factor (thrust efficiency) is the ratio between the exit-gas momentum of the conical nozzle and that of an ideal nozzle with uniform, parallel, axial gas-flow. The value of can be expressed by the following equation:

Bell nozzle: To gain higher performance and shorter length, engineers developed the bell-shaped nozzle. It employs a fast-expansion (radial-flow) section in the initial divergent region, which leads to a uniform, axially directed flow at the nozzle exit. The wall contour is changed gradually enough to prevent oblique shocks.

An equivalent 15-degree half-angle conical nozzle is commonly used as a standard to specify bell nozzles. For instance, the length of an 80% bell nozzle (distance between throat and exit plane) is 80% of that of a 15-degree half-angle conical nozzle having the same throat area, radius below the throat, and area expansion ratio. Bell nozzle lengths beyond approximately 80% do not significantly contribute to performance, especially when weight penalties are considered. However, bell nozzle lengths up to 100% can be optimum for applications stressing very high performance.

One convenient way of designing a near optimum thrust bell nozzle contour uses the parabolic approximation procedures suggested by G.V.R. Rao. The design configuration of a parabolic approximation bell nozzle is shown in Figure 1.5. The nozzle contour immediately upstream of the throat T is a circular arc with a radius of 1.5 Rt. The divergent section nozzle contour is made up of a circular entrance section with a radius of 0.382 Rt from the throat T to the point N and parabola from there to the exit E.

Design of a specific nozzle requires the following data: throat diameter Dt, axial length of the nozzle from throat to exit plane Ln (or the desired fractional length, Lf, based on a 15-degree conical nozzle), expansion ratio , initial wall angle of the parabola n, and nozzle exit wall angle e. The wall angles n and e are shown in Figure 1.6 as a function of the expansion ratio. Optimum nozzle contours can be approximated very accurately by selecting the proper inputs. Although no allowance is made for different propellant

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combinations, experience has shown only small effect of the specific heat ratio upon the contour.

Combustion Chamber

The combustion chamber serves as an envelope to retain the propellants for a sufficient period to ensure complete mixing and combustion. The required stay time, or combustion residence time, is a function of many parameters. The theoretically required combustion chamber volume is a function of the mass flow rate of the propellants, the average density of the combustion products, and the stay time needed for efficient combustion. This relationship can be expressed by the following equation:

where Vc is the chamber volume, q is the propellant mass flow rate, V is the average specific volume, and ts is the propellant stay-time.

A useful parameter relative to chamber volume and residence time is the characteristic length, L* (pronounced "L star"), the chamber volume divided by the nozzle sonic throat area:

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The L* concept is much easier to visualize than the more elusive "combustion residence time", expressed in small fractions of a second. Since the value of At is in nearly direct proportion to the product of q and V, L* is essentially a function of ts.

The customary method of establishing the L* of a new thrust chamber design largely relies on past experience with similar propellants and engine size. Under a given set of operating conditions, such as type of propellant, mixture ratio, chamber pressure, injector design, and chamber geometry, the value of the minimum required L* can only be evaluated by actual firings of experimental thrust chambers. Typical L* values for various propellants are shown in the table below. With throat area and minimum required L* established, the chamber volume can be calculated by equation (1.33).

Table 1: Chamber Characteristic Length, L* Propellant Combination L*, cm

Nitric acid/hydrazine-base fuel 76-89

Nitrogen tetroxide/hydrazine-base fuel 76-89

Hydrogen peroxide/RP-1 (including catalyst bed)   152-178

Liquid oxygen/RP-1 102-127

Liquid oxygen/ammonia 76-102

Liquid oxygen/liquid hydrogen (GH2 injection) 56-71

Liquid oxygen/liquid hydrogen (LH2 injection) 76-102

Liquid fluorine/liquid hydrogen (GH2 injection) 56-66

Liquid fluorine/liquid hydrogen (LH2 injection) 64-76

Liquid fluorine/hydrazine 61-71

Chlorine trifluoride/hydrazine-base fuel 51-89

Three geometrical shapes have been used in combustion chamber design - spherical, near-spherical, and cylindrical - with the cylindrical chamber being employed most frequently in the United States. Compared to a cylindrical chamber of the same volume, a spherical or near-spherical chamber offers the advantage of less cooling surface and weight; however, the spherical chamber is more difficult to manufacture and has provided poorer performance in other respects.

The total combustion process, from injection of the reactants until completion of the chemical reactions and conversion of the products into hot gases, requires finite amounts of time and volume, as expressed by the characteristic length L*. The value of this factor is significantly greater than the linear length between injector face and throat plane. The contraction ratio is defined as the major cross-sectional area of the combuster divided by the throat area. Typically, large engines are constructed with a low contraction ratio and a comparatively long length; and smaller chambers employ a large contraction ratio with a shorter length, while still providing sufficient L* for adequate vaporization and combustion dwell-time.

As a good place to start, the process of sizing a new combustion chamber examines the dimensions of previously successful designs in the same size class and plotting such data in a rational manner. The throat size of a new engine can be generated with a fair degree of confidence, so it makes sense to plot the data from historical sources in relation to throat diameter. Figure 1.7 plots chamber length as a function of throat diameter (with approximating equation). It is important that the output of any modeling program not be slavishly applied, but be considered a logical starting point for specific engine sizing.

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The basic elements of a cylindrical thrust-chamber are identified in Figure 1.4. In design practice, it has been arbitrarily defined that the combustion chamber volume includes the space between the injector face and the nozzle throat plane. The approximate volume of the combustion chamber can be expressed by the following equation:

Rearranging equation (1.34) we get the following, which can be solved for the chamber diameter via iteration:

Click here for example problem #1.9

Injector

The injector, as the name implies, injects the propellants into the combustion chamber in the right proportions and the right conditions to yield an efficient, stable combustion process. Placed at the forward, or upper, end of the combustor, the injector also performs the structural task of closing off the top of the combustion chamber against the high pressure and temperature it contains. The injector has been compared to the carburetor of an automobile engine, since it provides the fuel and oxidizer at the proper rates and in the correct proportions, this may be an appropriate comparison. However, the injector, located directly over the high-pressure combustion, performs many other functions

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related to the combustion and cooling processes and is much more important to the function of the rocket engine than the carburetor is for an automobile engine.

No other component of a rocket engine has as great an impact upon engine performance as the injector. In various and different applications, well-designed injectors may have a fairly wide spread in combustion efficiency, and it is not uncommon for an injector with C* efficiency as low as 92% to be considered acceptable. Small engines designed for special purposes, such as attitude control, may be optimized for response and light weight at the expense of combustion efficiency, and may be deemed very satisfactory even if efficiency falls below 90%. In general, however, recently well-designed injection systems have demonstrated C* efficiencies so close to 100% of theoretical that the ability to measure this parameter is the limiting factor in its determination. High levels of combustion efficiency derive from uniform distribution of the desired mixture ratio and fine atomization of the liquid propellants. Local mixing within the injection-element spray pattern must take place at virtually a microscopic level to ensure combustion efficiencies approaching 100%.

Combustion stability is also a very important requirement for a satisfactory injector design. Under certain conditions, shock and detonation waves are generated by local disturbances in the chamber, possibly caused by fluctuations in mixing or propellant flow. These may trigger pressure oscillations that are amplified and maintained by the combustion processes. Such high-amplitude waves - referred to as combustion instability - produce high levels of vibration and heat flux that can be very destructive. A major portion of the design and development effort therefore concerns stable combustion. High performance can become secondary if the injector is easily triggered into destructive instability, and many of the injector parameters that provide high performance appear to reduce the stability margin.

Power CyclesLiquid bipropellant rocket engines can be categorized according to their power cycles, that is, how power is derived to feed propellants to the main combustion chamber. Described below are some of the more common types.

Gas-generator cycle: The gas-generator cycle, also called open cycle, taps off a small amount of fuel and oxidizer from the main flow (typically 3 to 7 percent) to feed a burner called a gas generator. The hot gas from this generator passes through a turbine to generate power for the pumps that send propellants to the combustion chamber. The hot gas is then either dumped overboard or sent into the main nozzle downstream. Increasing the flow of propellants into the gas generator increases the speed of the turbine, which increases the flow of propellants into the main combustion chamber, and hence, the amount of thrust produced. The gas generator must burn propellants at a less-than-optimal mixture ratio to keep the temperature low for the turbine blades. Thus, the cycle is appropriate for moderate power requirements but not high-power systems, which would have to divert a large portion of the main flow to the less efficient gas-generator flow.

As in most rocket engines, some of the propellant in a gas generator cycle is used to cool the nozzle and combustion chamber, increasing efficiency and allowing higher engine temperature.

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Staged combustion cycle: In a staged combustion cycle, also called closed cycle, the propellants are burned in stages. Like the gas-generator cycle, this cycle also has a burner, called a preburner, to generate gas for a turbine. The preburner taps off and burns a small amount of one propellant and a large amount of the other, producing an oxidizer-rich or fuel-rich hot gas mixture that is mostly unburned vaporized propellant. This hot gas is then passed through the turbine, injected into the main chamber, and burned again with the remaining propellants. The advantage over the gas-generator cycle is that all of the propellants are burned at the optimal mixture ratio in the main chamber and no flow is dumped overboard. The staged combustion cycle is often used for high-power applications. The higher the chamber pressure, the smaller and lighter the engine can be to produce the same thrust. Development cost for this cycle is higher because the high pressures complicate the development process. Further disadvantages are harsh turbine conditions, high temperature piping required to carry hot gases, and a very complicated feedback and control design.

Staged combustion was invented by Soviet engineers and first appeared in 1960. In the West, the first laboratory staged combustion test engine was built in Germany in 1963.

Expander cycle: The expander cycle is similar to the staged combustion cycle but has no preburner. Heat in the cooling jacket of the main combustion chamber serves to vaporize the fuel. The fuel vapor is then passed through the turbine and injected into the main chamber to burn with the oxidizer. This cycle works with fuels such as hydrogen or methane, which have a low boiling point and can be vaporized easily. As with the staged combustion cycle, all of the propellants are burned at the optimal mixture ratio in the main chamber, and typically no flow is dumped overboard; however, the heat transfer to the fuel limits the power available to the turbine, making this cycle appropriate for small to midsize engines. A variation of the system is the open, or bleed, expander cycle, which uses only a portion of the fuel to drive the turbine. In this variation, the turbine exhaust is dumped overboard to ambient pressure to increase the turbine pressure ratio and power output. This can achieve higher chamber pressures than the closed expander cycle although at lower efficiency because of the overboard flow.

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Pressure-fed cycle: The simplest system, the pressure-fed cycle, does not have pumps or turbines but instead relies on tank pressure to feed the propellants into the main chamber. In practice, the cycle is limited to relatively low chamber pressures because higher pressures make the vehicle tanks too heavy. The cycle can be reliable, given its reduced part count and complexity compared with other systems.

Engine CoolingThe heat created during combustion in a rocket engine is contained within the exhaust gases. Most of this heat is expelled along with the gas that contains it; however, heat is transferred to the thrust chamber walls in quantities sufficient to require attention.

Thrust chamber designs are generally categorized or identified by the hot gas wall cooling method or the configuration of the coolant passages, where the coolant pressure inside may be as high as 500 atmospheres. The high combustion temperatures (2,500 to 3,600o K) and the high heat transfer rates (up to 16 kJ/cm2-s) encountered in a combustion chamber present a formidable challenge to the designer. To meet this challenge, several chamber cooling techniques have been utilized successfully. Selection of the optimum cooling method for a thrust chamber depends on many considerations, such as type of propellant, chamber pressure, available coolant pressure, combustion chamber configuration, and combustion chamber material.

Regenerative cooling is the most widely used method of cooling a thrust chamber and is accomplished by flowing high-velocity coolant over the back side of the chamber hot gas wall to convectively cool the hot gas liner. The coolant with the heat input from cooling the liner is then discharged into the injector and utilized as a propellant.

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Earlier thrust chamber designs, such as the V-2 and Redstone, had low chamber pressure, low heat flux and low coolant pressure requirements, which could be satisfied by a simplified "double wall chamber" design with regenerative and film cooling. For subsequent rocket engine applications, however, chamber pressures were increased and the cooling requirements became more difficult to satisfy. It became necessary to design new coolant configurations that were more efficient structurally and had improved heat transfer characteristics.

This led to the design of "tubular wall" thrust chambers, by far the most widely used design approach for the vast majority of large rocket engine applications. These chamber designs have been successfully used for the Thor, Jupiter, Atlas, H-1, J-2, F-1, RS-27 and several other Air Force and NASA rocket engine applications. The primary advantage of the design is its light weight and the large experience base that has accrued. But as chamber pressures and hot gas wall heat fluxes have continued to increase (>100 atm), still more effective methods have been needed.

One solution has been "channel wall" thrust chambers, so named because the hot gas wall cooling is accomplished by flowing coolant through rectangular channels, which are machined or formed into a hot gas liner fabricated from a high-conductivity material, such as copper or a copper alloy. A prime example of a channel wall combustion chamber is the SSME, which operates at 204 atmospheres nominal chamber pressure at 3,600 K for a duration of 520 seconds. Heat transfer and structural characteristics are excellent.

In addition to the regeneratively cooled designs mentioned above, other thrust chamber designs have been fabricated for rocket engines using dump cooling, film cooling, transpiration cooling, ablative liners and radiation cooling. Although regeneratively cooled combustion chambers have proven to be the best approach for cooling large liquid rocket engines, other methods of cooling have also been successfully used for cooling thrust chamber assemblies. Examples include:

Dump cooling, which is similar to regenerative cooling because the coolant flows through small passages over the back side of the thrust chamber wall. The difference, however, is that after cooling the thrust chamber, the coolant is discharged overboard through openings at the aft end of the divergent nozzle. This method has limited application because of the performance loss resulting from dumping the coolant overboard. To date, dump cooling has not been used in an actual application.

Film cooling provides protection from excessive heat by introducing a thin film of coolant or propellant through orifices around the injector periphery or through manifolded orifices in the chamber wall near the injector or chamber throat region. This method is typically used in high heat flux regions and in combination with regenerative cooling.

Transpiration cooling provides coolant (either gaseous or liquid propellant) through a porous chamber wall at a rate sufficient to maintain the chamber hot gas wall to the desired temperature. The technique is really a special case of film cooling.

With ablative cooling, combustion gas-side wall material is sacrificed by melting, vaporization and chemical changes to dissipate heat. As a result, relatively cool gases flow over the wall surface, thus lowering the boundary-layer temperature and assisting the cooling process.

With radiation cooling, heat is radiated from the outer surface of the combustion chamber or nozzle extension wall. Radiation cooling is typically used for small thrust chambers with a high-temperature wall material (refractory) and in low-heat flux regions, such as a nozzle extension.

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Solid Rocket MotorsSolid rockets motors store propellants in solid form. The fuel is typically powdered aluminum and the oxidizer is ammonium perchlorate. A synthetic rubber binder such as polybutadiene holds the fuel and oxidizer powders together. Though lower performing than liquid propellant rockets, the operational simplicity of a solid rocket motor often makes it the propulsion system of choice.

Solid Fuel Geometry

A solid fuel's geometry determines the area and contours of its exposed surfaces, and thus its burn pattern. There are two main types of solid fuel blocks used in the space industry. These are cylindrical blocks, with combustion at a front, or surface, and cylindrical blocks with internal combustion. In the first case, the front of the flame travels in layers from the nozzle end of the block towards the top of the casing. This so-called end burner produces constant thrust throughout the burn. In the second, more usual case, the combustion surface develops along the length of a central channel. Sometimes the channel has a star shaped, or other, geometry to moderate the growth of this surface.

The shape of the fuel block for a rocket is chosen for the particular type of mission it will perform. Since the combustion of the block progresses from its free surface, as this surface grows, geometrical considerations determine whether the thrust increases, decreases or stays constant.

Fuel blocks with a cylindrical channel (1) develop their thrust progressively. Those with a channel and also a central cylinder of fuel (2) produce a relatively constant thrust, which reduces to zero very quickly when the fuel is used up. The five pointed star profile (3) develops a relatively constant thrust which decreases slowly to zero as the last of the fuel is consumed. The 'cruciform' profile (4) produces progressively less thrust. Fuel in a

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block with a 'double anchor' profile (5) produces a decreasing thrust which drops off quickly near the end of the burn. The 'cog' profile (6) produces a strong inital thrust, followed by an almost constant lower thrust.

Burn Rate

The burning surface of a rocket propellant grain recedes in a direction perpendicular to this burning surface. The rate of regression, typically measured in millimeters per second (or inches per second), is termed burn rate. This rate can differ significantly for different propellants, or for one particular propellant, depending on various operating conditions as well as formulation. Knowing quantitatively the burning rate of a propellant, and how it changes under various conditions, is of fundamental importance in the successful design of a solid rocket motor.

Propellant burning rate is influenced by certain factors, the most significant being: combustion chamber pressure, initial temperature of the propellant grain, velocity of the combustion gases flowing parallel to the burning surface, local static pressure, and motor acceleration and spin. These factors are discussed below.

Burn rate is profoundly affected by chamber pressure. The usual representation of the pressure dependence on burn rate is the Saint-Robert's Law,

where r is the burn rate, a is the burn rate coefficient, n is the pressure exponent, and Pc is the combustion chamber pressure. The values of a and n are determined empirically for a particular propellant formulation and cannot be theoretically predicted. It is important to realize that a single set of a, n values are typically valid over a distinct pressure range. More than one set may be necessary to accurately represent the full pressure regime of interest.

Example a, n values are 5.6059* (pressure in MPa, burn rate in mm/s) and 0.35 respectively for the Space Shuttle SRBs, which gives a burn rate of 9.34 mm/s at the average chamber pressure of 4.3 MPa.

* NASA publications gives a burn rate coefficient of 0.0386625 (pressure in PSI, burn rate in inch/s).

Temperature affects the rate of chemical reactions and thus the initial temperature of the propellant grain influences burning rate. If a particular propellant shows significant sensitivity to initial grain temperature, operation at temperature extremes will affect the time-thrust profile of the motor. This is a factor to consider for winter launches, for example, when the grain temperature may be lower than "normal" launch conditions.

For most propellants, certain levels of local combustion gas velocity (or mass flux) flowing parallel to the burning surface leads to an increased burning rate. This "augmentation" of burn rate is referred to as erosive burning, with the extent varying with propellant type and chamber pressure. For many propellants, a threshold flow velocity exists. Below this flow level, either no augmentation occurs, or a decrease in burn rate is experienced (negative erosive burning).

The effects of erosive burning can be minimized by designing the motor with a sufficiently large port-to-throat area ratio (Aport/At). The port area is the cross-section area of the flow channel in a motor. For a hollow-cylindrical grain, this is the cross-section area of the core. As a rule of thumb, the ratio should be a minimum of 2 for a grain L/D ratio of 6. A greater Aport/At ratio should be used for grains with larger L/D ratios.

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In an operating rocket motor, there is a pressure drop along the axis of the combustion chamber, a drop that is physically necessary to accelerate the increasing mass flow of combustion products toward the nozzle. The static pressure is greatest where gas flow is zero, that is, at the front of the motor. Since burn rate is dependant upon the local pressure, the rate should be greatest at this location. However, this effect is relatively minor and is usually offset by the counter-effect of erosive burning.

Burning rate is enhanced by acceleration of the motor. Whether the acceleration is a result of longitudinal force (e.g. thrust) or spin, burning surfaces that form an angle of about 60-90o with the acceleration vector are prone to increased burn rate.

It is sometimes desirable to modify the burning rate such that it is more suitable to a certain grain configuration. For example, if one wished to design an end burner grain, which has a relatively small burning area, it is necessary to have a fast burning propellant. In other circumstances, a reduced burning rate may be sought after. For example, a motor may have a large L/D ratio to generate sufficiently high thrust, or it may be necessary for a particular design to restrict the diameter of the motor. The web would be consequently thin, resulting in short burn duration. Reducing the burning rate would be beneficial.

There are a number of ways of modifying the burning rate: decrease the oxidizer particle size, increase or reduce the percentage of oxidizer, adding a burn rate catalyst or suppressant, and operate the motor at a lower or higher chamber pressure. These factors are discussed below.

The effect of the oxidizer particle size on burn rate seems to be influenced by the type of oxidizer. Propellants that use ammonium perchlorate (AP) as the oxidizer have a burn rate that is significantly affected by AP particle size. This most likely results from the decomposition of AP being the rate-determining step in the combustion process.

The burn rate of most propellants is strongly influenced by the oxidizer/fuel ratio. Unfortunately, modifying the burn rate by this means is quite restrictive, as the performance of the propellant, as well as mechanical properties, are also greatly affected by the O/F ratio.

Certainly the best and most effective means of increasing the burn rate is the addition of a catalyst to the propellant mixture. A catalyst is a chemical compound that is added in small quantities for the sole purpose of tailoring the burning rate. A burn rate suppressant is an additive that has the opposite effect to that of a catalyst – it is used to decrease the burn rate.

For a propellant that follows the Saint-Robert's burn rate law, designing a rocket motor to operate at a lower chamber pressure will provide for a lower burning rate. Due to the nonlinearity of the pressure-burn rate relationship, it may be necessary to significantly reduce the operating pressure to get the desired burning rate. The obvious drawback is reduced motor performance, as specific impulse similarly decays with reducing chamber pressure.

Product Generation Rate

The rate at which combustion products are generated is expressed in terms of the regression speed of the grain. The product generation rate integrated over the port surface area is

where q is the combustion product generation rate at the propellant surface, p is the solid propellant density, Ab is the area of the burning surface, and r is the propellant burn rate.

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It is important to note that the combustion products may consist of both gaseous and condensed-phase mass. The condensed-phase, which manifests itself as smoke, may be either solid or liquid particles. Only the gaseous products contribute to pressure development. The condensed-phase certainly does, however, contribute to the thrust of the rocket motor, due to its mass and velocity.

Click here for example problem #1.10

Chamber Pressure

The pressure curve of a rocket motor exhibits transient and steady state behavior. The transient phases are when the pressure varies substantially with time – during the ignition and start-up phase, and following complete (or nearly complete) grain consumption when the pressure falls down to ambient level during the tail-off phase. The variation of chamber pressure during the steady state burning phase is due mainly to variation of grain geometry with associated burn rate variation. Other factors may play a role, however, such as nozzle throat erosion and erosive burn rate augmentation.

Monopropellant EnginesBy far the most widely used type of propulsion for spacecraft attitude and velocity control is monopropellant hydrazine. Its excellent handling characteristics, relative stability under normal storage conditions, and clean decomposition products have made it the standard. The general sequence of operations in a hydrazine thruster is:

When the attitude control system signals for thruster operation, an electric solenoid valve opens allowing hydrazine to flow. The action may be pulsed (as short as 5 ms) or long duration (steady state).

The pressure in the propellant tank forces liquid hydrazine into the injector. It enters as a spray into the thrust chamber and contacts the catalyst beds.

The catalyst bed consists of alumina pellets impregnated with iridium. Incoming hydrazine heats to its vaporizing point by contact with the catalyst bed and with the hot gases leaving the catalyst particles. The temperature of the hydrazine rises to a point where the rate of its decomposition becomes so high that the chemical reactions are self-sustaining.

By controlling the flow variables and the geometry of the catalyst chamber, a designer can tailor the proportion of chemical products, the exhaust temperature, the molecular weight, and thus the enthalpy for a given application. For a thruster application where specific impulse is paramount, the designer attempts to provide 30-40% ammonia dissociation, which is about the lowest percentage that can be maintained reliably. For gas-generator application, where lower temperature gases are usually desired, the designer provides for higher levels of ammonia dissociation.

Finally, in a space thruster, the hydrazine decomposition products leave the catalyst bed and exit from the chamber through a high expansion ratio exhaust nozzle to produce thrust.

Monopropellant hydrazine thrusters typically produce a specific impulse of about 230 to 240 seconds.

Other suitable propellants for catalytic decomposition engines are hydrogen peroxide and nitrous oxide, however the performance is considerably lower than that obtained with hydrazine - specific impulse of about 150 s with H2O2 and about 170 s with N2O.

Monopropellant systems have successfully provided orbit maintenance and attitude control functions, but lack the performance to provide weight-efficient large V maneuvers required for orbit insertion. Bipropellant systems are attractive because they

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can provide all three functions with one higher performance system, but they are more complex than the common solid rocket and monopropellant combined systems. A third alternative are dual mode systems. These systems are hybrid designs that use hydrazine both as a fuel for high performance bipropellant engines and as a monopropellant with conventional low-thrust catalytic thrusters. The hydrazine is fed to both the bipropellant engines and the monopropellant thrusters from a common fuel tank.

Cold gas propulsion is just a controlled, pressurized gas source and a nozzle. It represents the simplest form of rocket engine. Cold gas has many applications where simplicity and/or the need to avoid hot gases are more important than high performance. The Manned Maneuvering Unit used by astronauts is an example of such a system.

StagingMultistage rockets allow improved payload capability for vehicles with a high V requirement such as launch vehicles or interplanetary spacecraft. In a multistage rocket, propellant is stored in smaller, separate tanks rather than a larger single tank as in a single-stage rocket. Since each tank is discarded when empty, energy is not expended to accelerate the empty tanks, so a higher total V is obtained. Alternatively, a larger payload mass can be accelerated to the same total V. For convenience, the separate tanks are usually bundled with their own engines, with each discardable unit called a stage.

Multistage rocket performance is described by the same rocket equation as single-stage rockets, but must be determined on a stage-by-stage basis. The velocity increment, Vi, for each stage is calculated as before,

where moi represents the total vehicle mass when stage i is ignited, and mfi is the total vehicle mass when stage i is burned out but not yet discarded. It is important to realize that the payload mass for any stage consists of the mass of all subsequent stages plus the ultimate payload itself. The velocity increment for the vehicle is then the sum of those for the individual stages where n is the total number of stages.

Click here for example problem #1.11

We define the payload fraction as the ratio of payload mass to initial mass, or mpl/mo.

For a multistage vehicle with dissimilar stages, the overall vehicle payload fraction depends on how the V requirement is partitioned among stages. Payload fractions will be reduced if the V is partitioned suboptimally. The optimal distribution may be determined by trial and error. A V distribution is postulated and the resulting payload fraction calculated. The V distribution is varied until the payload fraction is maximized. Once the V distribution is selected, vehicle sizing is accomplished by starting with the uppermost or final stage (whose payload is the actual deliverable payload) and calculating the initial mass of this assembly. This assembly then forms the payload for the previous stage and the process repeats until all stages are sized. Results reveal that to maximize payload fraction for a given V requirement:

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1. Stages with higher Isp should be above stages with lower Isp. 2. More V should be provided by the stages with the higher Isp. 3. Each succeeding stage should be smaller than its predecessor. 4. Similar stages should provide the same V.

ORBITAL MECHANICS

Conic Sections Orbital Elements Types of Orbits Newton's Laws of Motion and Universal Gravitation Uniform Circular Motion Motions of Planets and Satellites Launch of a Space Vehicle Position in an Elliptical Orbit Orbit Perturbations Orbit Maneuvers Escape Velocity

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Orbital mechanics, also called flight mechanics, is the study of the motions of artificial satellites and space vehicles moving under the influence of forces such as gravity, atmospheric drag, thrust, etc. Orbital mechanics is a modern offshoot of celestial mechanics which is the study of the motions of natural celestial bodies such as the moon and planets. The root of orbital mechanics can be traced back to the 17th century when mathematician Isaac Newton (1642-1727) put forward his laws of motion and formulated his law of universal gravitation. The engineering applications of orbital mechanics include ascent trajectories, reentry and landing, rendezvous computations, and lunar and interplanetary trajectories.

Conic SectionsA conic section, or just conic, is a curve formed by passing a plane through a right circular cone. As shown in Figure 4.1, the angular orientation of the plane relative to the cone determines whether the conic section is a circle, ellipse, parabola, or hyerbola. The circle and the ellipse arise when the intersection of cone and plane is a bounded curve. The circle is a special case of the ellipse in which the plane is perpendicular to the axis of the cone. If the plane is parallel to a generator line of the cone, the conic is called a parabola. Finally, if the intersection is an unbounded curve and the plane is not parallel to a generator line of the cone, the figure is a hyperbola. In the latter case the plane will intersect both halves of the cone, producing two separate curves.

We can define all conic sections in terms of the eccentricity. The type of conic section is also related to the semi-major axis and the energy. The table below shows the relationships between eccentricity, semi-major axis, and energy and the type of conic section.

Conic Section Eccentricity, e Semi-major axis Energy

Circle 0 = radius < 0

Ellipse 0 < e < 1 > 0 < 0

Parabola 1 infinity 0

Hyperbola > 1 < 0 > 0

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Satellite orbits can be any of the four conic sections. In this section we will discuss bounded conic orbits, i.e. circles and ellipses.

Orbital ElementsTo mathematically describe an orbit one must define six quantities, called orbital elements. They are

Semi-Major Axis, a Eccentricity, e Inclination, i Argument of Periapsis, Time of Periapsis Passage, T Longitude of Ascending Node,

An orbiting satellite follows an oval shaped path known as an ellipse with the body being orbited, called the primary, located at one of two points called foci. An ellipse is defined to be a curve with the following property: for each point on an ellipse, the sum of its distances from two fixed points, called foci, is constant (see Figure 4.2). The longest and shortest lines that can be drawn through the center of an ellipse are called the major axis and minor axis, respectively. The semi-major axis is one-half of the major axis and represents a satellite's mean distance from its primary. Eccentricity is the distance between the foci divided by the length of the major axis and is a number between zero and one. An eccentricity of zero indicates a circle.

Inclination is the angular distance between a satellite's orbital plane and the equator of its primary (or the ecliptic plane in the case of heliocentric, or sun centered, orbits). An inclination of zero degrees indicates an orbit about the primary's equator in the same direction as the primary's rotation, a direction called prograde (or direct). An inclination of 90 degrees indicates a polar orbit. An inclination of 180 degrees indicates a retrograde equatorial orbit. A retrograde orbit is one in which a satellite moves in a direction opposite to the rotation of its primary.

Periapsis is the point in an orbit closest to the primary. The opposite of periapsis, the farthest point in an orbit, is called apoapsis. Periapsis and apoapsis are usually modified to apply to the body being orbited, such as perihelion and aphelion for the Sun, perigee and apogee for Earth, perijove and apojove for Jupiter, perilune and apolune for the Moon, etc. The argument of periapsis is the angular distance between the ascending node and the point of periapsis (see Figure 4.3). The time of periapsis passage is the time in which a satellite moves through its point of periapsis.

Nodes are the points where an orbit crosses a plane, such as a satellite crossing the Earth's equatorial plane. If the satellite crosses the plane going from south to north, the node is the ascending node; if moving from north to south, it is the descending node. The longitude of the ascending node is the node's celestial longitude. Celestial longitude is analogous to longitude on Earth and is measured in degrees counter-clockwise from zero with zero longitude being in the direction of the vernal equinox.

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In general, three observations of an object in orbit are required to calculate the six orbital elements. Two other quantities often used to describe orbits are period and true anomaly. Period, P, is the length of time required for a satellite to complete one orbit. True anomaly, , is the angular distance of a point in an orbit past the point of periapsis, measured in degrees.

Types Of OrbitsFor a spacecraft to achieve Earth orbit, it must be launched to an elevation above the Earth's atmosphere and accelerated to orbital velocity. The most energy efficient orbit, that is one that requires the least amount of propellant, is a direct low inclination orbit. To achieve such an orbit, a spacecraft is launched in an eastward direction from a site near the Earth's equator. The advantage being that the rotational speed of the Earth contributes to the spacecraft's final orbital speed. At the United States' launch site in Cape Canaveral (28.5 degrees north latitude) a due east launch results in a "free ride" of 1,471 km/h (914 mph). Launching a spacecraft in a direction other than east, or from a site far from the equator, results in an orbit of higher inclination. High inclination orbits are less able to take advantage of the initial speed provided by the Earth's rotation, thus the launch vehicle must provide a greater part, or all, of the energy required to attain orbital velocity. Although high inclination orbits are less energy efficient, they do have advantages over equatorial orbits for certain applications. Below we describe several types of orbits and the advantages of each:

Geosynchronous orbits (GEO) are circular orbits around the Earth having a period of 24 hours. A geosynchronous orbit with an inclination of zero degrees is called a geostationary orbit. A spacecraft in a geostationary orbit appears to hang motionless above one position on the Earth's equator. For this reason, they are ideal for some types of communication and meteorological satellites. A spacecraft in an inclined geosynchronous orbit will appear to follow a regular figure-8 pattern in the sky once every orbit. To attain geosynchronous orbit, a spacecraft is first launched into an elliptical orbit with an apogee of 35,786 km (22,236 miles) called a geosynchronous transfer orbit (GTO). The orbit is then circularized by firing the spacecraft's engine at apogee.

Polar orbits (PO) are orbits with an inclination of 90 degrees. Polar orbits are useful for satellites that carry out mapping and/or surveillance operations because as the planet rotates the spacecraft has access to virtually every point on the planet's surface.

Walking orbits: An orbiting satellite is subjected to a great many gravitational influences. First, planets are not perfectly spherical and they have slightly uneven mass distribution. These fluctuations have an effect on a spacecraft's trajectory. Also, the sun, moon, and planets contribute a gravitational influence on an orbiting satellite. With

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proper planning it is possible to design an orbit which takes advantage of these influences to induce a precession in the satellite's orbital plane. The resulting orbit is called a walking orbit, or precessing orbit.

Sun synchronous orbits (SSO) are walking orbits whose orbital plane precesses with the same period as the planet's solar orbit period. In such an orbit, a satellite crosses periapsis at about the same local time every orbit. This is useful if a satellite is carrying instruments which depend on a certain angle of solar illumination on the planet's surface. In order to maintain an exact synchronous timing, it may be necessary to conduct occasional propulsive maneuvers to adjust the orbit.

Molniya orbits are highly eccentric Earth orbits with periods of approximately 12 hours (2 revolutions per day). The orbital inclination is chosen so the rate of change of perigee is zero, thus both apogee and perigee can be maintained over fixed latitudes. This condition occurs at inclinations of 63.4 degrees and 116.6 degrees. For these orbits the argument of perigee is typically placed in the southern hemisphere, so the satellite remains above the northern hemisphere near apogee for approximately 11 hours per orbit. This orientation can provide good ground coverage at high northern latitudes.

Hohmann transfer orbits are interplanetary trajectories whose advantage is that they consume the least possible amount of propellant. A Hohmann transfer orbit to an outer planet, such as Mars, is achieved by launching a spacecraft and accelerating it in the direction of Earth's revolution around the sun until it breaks free of the Earth's gravity and reaches a velocity which places it in a sun orbit with an aphelion equal to the orbit of the outer planet. Upon reaching its destination, the spacecraft must decelerate so that the planet's gravity can capture it into a planetary orbit.

To send a spacecraft to an inner planet, such as Venus, the spacecraft is launched and accelerated in the direction opposite of Earth's revolution around the sun (i.e. decelerated) until it achieves a sun orbit with a perihelion equal to the orbit of the inner planet. It should be noted that the spacecraft continues to move in the same direction as Earth, only more slowly.

To reach a planet requires that the spacecraft be inserted into an interplanetary trajectory at the correct time so that the spacecraft arrives at the planet's orbit when the planet will be at the point where the spacecraft will intercept it. This task is comparable to a quarterback "leading" his receiver so that the football and receiver arrive at the same point at the same time. The interval of time in which a spacecraft must be launched in order to complete its mission is called a launch window.

Newton's Laws of Motion and Universal GravitationNewton's laws of motion describe the relationship between the motion of a particle and the forces acting on it.

The first law states that if no forces are acting, a body at rest will remain at rest, and a body in motion will remain in motion in a straight line. Thus, if no forces are acting, the velocity (both magnitude and direction) will remain constant.

The second law tells us that if a force is applied there will be a change in velocity, i.e. an acceleration, proportional to the magnitude of the force and in the direction in which the force is applied. This law may be summarized by the equation

where F is the force, m is the mass of the particle, and a is the acceleration.

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The third law states that if body 1 exerts a force on body 2, then body 2 will exert a force of equal strength, but opposite in direction, on body 1. This law is commonly stated, "for every action there is an equal and opposite reaction".

In his law of universal gravitation, Newton states that two particles having masses m1 and m2 and separated by a distance r are attracted to each other with equal and opposite forces directed along the line joining the particles. The common magnitude F of the two forces is

where G is an universal constant, called the constant of gravitation, and has the value 6.67259x10-11 N-m2/kg2 (3.4389x10-8 lb-ft2/slug2).

Let's now look at the force that the Earth exerts on an object. If the object has a mass m, and the Earth has mass M, and the object's distance from the center of the Earth is r, then the force that the Earth exerts on the object is GmM /r2 . If we drop the object, the Earth's gravity will cause it to accelerate toward the center of the Earth. By Newton's second law (F = ma), this acceleration g must equal (GmM /r2)/m, or

At the surface of the Earth this acceleration has the valve 9.80665 m/s2 (32.174 ft/s2).

Many of the upcoming computations will be somewhat simplified if we express the product GM as a constant, which for Earth has the value 3.986005x1014 m3/s2 (1.408x1016

ft3/s2). The product GM is often represented by the Greek letter .

For additional useful constants please see the appendix Basic Constants.

For a refresher on SI versus U.S. units see the appendix Weights & Measures.

Uniform Circular MotionIn the simple case of free fall, a particle accelerates toward the center of the Earth while moving in a straight line. The velocity of the particle changes in magnitude, but not in direction. In the case of uniform circular motion a particle moves in a circle with constant speed. The velocity of the particle changes continuously in direction, but not in magnitude. From Newton's laws we see that since the direction of the velocity is changing, there is an acceleration. This acceleration, called centripetal acceleration is directed inward toward the center of the circle and is given by

where v is the speed of the particle and r is the radius of the circle. Every accelerating particle must have a force acting on it, defined by Newton's second law (F = ma). Thus, a particle undergoing uniform circular motion is under the influence of a force, called centripetal force, whose magnitude is given by

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The direction of F at any instant must be in the direction of a at the same instant, that is radially inward.

A satellite in orbit is acted on only by the forces of gravity. The inward acceleration which causes the satellite to move in a circular orbit is the gravitational acceleration caused by the body around which the satellite orbits. Hence, the satellite's centripetal acceleration is g, that is g = v2/r. From Newton's law of universal gravitation we know that g = GM /r2. Therefore, by setting these equations equal to one another we find that, for a circular orbit,

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Motions of Planets and SatellitesThrough a lifelong study of the motions of bodies in the solar system, Johannes Kepler (1571-1630) was able to derive three basic laws known as Kepler's laws of planetary motion. Using the data compiled by his mentor Tycho Brahe (1546-1601), Kepler found the following regularities after years of laborious calculations:

1.  All planets move in elliptical orbits with the sun at one focus. 2.  A line joining any planet to the sun sweeps out equal areas in equal times. 3.  The square of the period of any planet about the sun is proportional to the cube of the planet's mean distance from the sun.

These laws can be deduced from Newton's laws of motion and law of universal gravitation. Indeed, Newton used Kepler's work as basic information in the formulation of his gravitational theory.

As Kepler pointed out, all planets move in elliptical orbits, however, we can learn much about planetary motion by considering the special case of circular orbits. We shall neglect the forces between planets, considering only a planet's interaction with the sun. These considerations apply equally well to the motion of a satellite about a planet.

Let's examine the case of two bodies of masses M and m moving in circular orbits under the influence of each other's gravitational attraction. The center of mass of this system of two bodies lies along the line joining them at a point C such that mr = MR. The large body of mass M moves in an orbit of constant radius R and the small body of mass m in an orbit of constant radius r, both having the same angular velocity . For this to happen, the gravitational force acting on each body must provide the necessary centripetal acceleration. Since these gravitational forces are a simple action-reaction pair, the centripetal forces must be equal but opposite in direction. That is, m 2r must equal M 2R. The specific requirement, then, is that the gravitational force acting on either body must equal the centripetal force needed to keep it moving in its circular orbit, that is

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If one body has a much greater mass than the other, as is the case of the sun and a planet or the Earth and a satellite, its distance from the center of mass is much smaller than that of the other body. If we assume that m is negligible compared to M, then R is negligible compared to r. Thus, equation (4.7) then becomes

If we express the angular velocity in terms of the period of revolution, = 2 /P, we obtain

where P is the period of revolution. This is a basic equation of planetary and satellite motion. It also holds for elliptical orbits if we define r to be the semi-major axis (a) of the orbit.

A significant consequence of this equation is that it predicts Kepler's third law of planetary motion, that is P2~r3.

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In celestial mechanics where we are dealing with planetary or stellar sized bodies, it is often the case that the mass of the secondary body is significant in relation to the mass of the primary, as with the Moon and Earth. In this case the size of the secondary cannot be ignored. The distance R is no longer negligible compared to r and, therefore, must be carried through the derivation. Equation (4.9) becomes

More commonly the equation is written in the equivalent form

where a is the semi-major axis. The semi-major axis used in astronomy is always the primary-to-secondary distance, or the geocentric semi-major axis. For example, the Moon's mean geocentric distance from Earth (a) is 384,403 kilometers. On the other hand, the Moon's distance from the barycenter (r) is 379,732 km, with Earth's counter-orbit (R) taking up the difference of 4,671 km.

Kepler's second law of planetary motion must, of course, hold true for circular orbits. In such orbits both and r are constant so that equal areas are swept out in equal times by the line joining a planet and the sun. For elliptical orbits, however, both and r will vary with time. Let's now consider this case.

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Figure 4.5 shows a particle revolving around C along some arbitrary path. The area swept out by the radius vector in a short time interval t is shown shaded. This area, neglecting the small triangular region at the end, is one-half the base times the height or approximately r(r t)/2. This expression becomes more exact as t approaches zero, i.e. the small triangle goes to zero more rapidly than the large one. The rate at which area is being swept out instantaneously is therefore

For any given body moving under the influence of a central force, the value r2 is constant.

Let's now consider two points P1 and P2 in an orbit with radii r1 and r2, and velocities v1 and v2. Since the velocity is always tangent to the path, it can be seen that if is the angle between r and v, then

where vsin is the transverse component of v. Multiplying through by r, we have

or, for two points P1 and P2 on the orbital path

Note that at periapsis and apoapsis, = 90 degrees. Thus, letting P1 and P2 be these two points we get

Let's now look at the energy of the above particle at points P1 and P2. Conservation of energy states that the sum of the kinetic energy and the potential energy of a particle remains constant. The kinetic energy T of a particle is given by mv2/2 while the potential energy of gravity V is calculated by the equation -GMm/r. Applying conservation of energy we have

From equations (4.14) and (4.15) we obtain

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Rearranging terms we get

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The eccentricity e of an orbit is given by

Click here for example problem #4.6

If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

Click here for example problem #4.7

Launch of a Space VehicleThe launch of a satellite or space vehicle consists of a period of powered flight during which the vehicle is lifted above the Earth's atmosphere and accelerated to orbital velocity by a rocket, or launch vehicle. Powered flight concludes at burnout of the rocket's last stage at which time the vehicle begins its free flight. During free flight the space vehicle is assumed to be subjected only to the gravitational pull of the Earth. If the vehicle moves far from the Earth, its trajectory may be affected by the gravitational influence of the sun, moon, or another planet.

A space vehicle's orbit may be determined from the position and the velocity of the vehicle at the

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beginning of its free flight. A vehicle's position and velocity can be described by the variables r, v, and , where r is the vehicle's distance from the center of the Earth, v is its velocity, and is the angle between the position and the velocity vectors, called the zenith angle (see Figure 4.7). If we let  r1, v1, and 1 be the initial (launch) values of  r, v, and , then we may consider these as given quantities. If we let point P2 represent the perigee, then equation (4.13) becomes

Substituting equation (4.23) into (4.15), we can obtain an equation for the perigee radius Rp.

Multiplying through by -Rp2/(r1

2v12) and rearranging, we get

Note that this is a simple quadratic equation in the ratio (Rp/r1) and that 2GM /(r1 × v12) is

a nondimensional parameter of the orbit.

Solving for (Rp/r1) gives

Like any quadratic, the above equation yields two answers. The smaller of the two answers corresponds to Rp, the periapsis radius. The other root corresponds to the apoapsis radius, Ra.

Please note that in practice spacecraft launches are usually terminated at either perigee or apogee, i.e. =90. This condition results in the minimum use of propellant.

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Equation (4.26) gives the values of Rp and Ra from which the eccentricity of the orbit can be calculated, however, it may be simpler to calculate the eccentricity e directly from the equation

Click here for example problem #4.9

To pin down a satellite's orbit in space, we need to know the angle , the true anomaly, from the periapsis point to the launch point. This angle is given by

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Click here for example problem #4.10

In most calculations, the complement of the zenith angle is used, denoted by . This angle is called the flight-path angle, and is positive when the velocity vector is directed away from the primary as shown in Figure 4.8. When flight-path angle is used, equations (4.26) through (4.28) are rewritten as follows:

Position in an Elliptical OrbitJohannes Kepler was able to solve the problem of relating position in an orbit to the elapsed time, t-to, or conversely, how long it takes to go from one point in an orbit to another. To solve this, Kepler introduced the quantity M, called the mean anomaly, which is the fraction of an orbit period that has elapsed since perigee.  The mean anomaly equals the true anomaly for a circular orbit. By definition,

where Mo in the mean anomaly at time to and n is the mean motion, or the average angular velocity, determined from the semi-major axis of the orbit as follows:

This solution will give the average position and velocity, but satellite orbits are elliptical with a radius constantly varying in orbit. Because the satellite's velocity depends on this varying radius, it changes as well. To resolve this problem we can define an intermediate variable E, called the eccentric anomaly, for elliptical orbits, which is given by

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where is the true anomaly. Mean anomaly is a function of eccentric anomaly by the formula

For small eccentricities a good approximation of true anomaly can be obtained by the following formula (the error is of the order e3):

The preceding five equations can be used to (1) find the time it takes to go from one position in an orbit to another, or (2) find the position in an orbit after a specific period of time. When solving these equations it is important to work in radians rather than degrees, where 2 radians equals 360 degrees.

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At any time in its orbit, the magnitude of a spacecraft's position vector, i.e. its distance from the primary body, and its flight-path angle can be calculated from the following equations:

And the spacecraft's velocity is given by,

Click here for example problem #4.13

Orbit PerturbationsThe orbital elements discussed at the beginning of this section provide an excellent reference for describing orbits, however there are other forces acting on a satellite that perturb it away from the nominal orbit. These perturbations, or variations in the orbital elements, can be classified based on how they affect the Keplerian elements. Secular variations represent a linear variation in the element, short-period variations are periodic in the element with a period less than the orbital period, and long-period variations are those with a period greater than the orbital period. Because secular variations have long-term effects on orbit prediction (the orbital elements affected continue to increase or decrease), they will be discussed here for Earth-orbiting satellites. Precise orbit determination requires that the periodic variations be included as well.

Third-Body Perturbations

The gravitational forces of the Sun and the Moon cause periodic variations in all of the orbital elements, but only the longitude of the ascending node, argument of perigee, and mean anomaly experience secular variations. These secular variations arise from a gyroscopic precession of the orbit about the ecliptic pole. The secular variation in mean anomaly is much smaller than the mean motion and has little effect on the orbit,

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however the secular variations in longitude of the ascending node and argument of perigee are important, especially for high-altitude orbits.

For nearly circular orbits the equations for the secular rates of change resulting from the Sun and Moon are

Longitude of the ascending node:

Argument of perigee:

where i is the orbit inclination, n is the number of orbit revolutions per day, and and are in degrees per day. These equations are only approximate; they neglect the variation caused by the changing orientation of the orbital plane with respect to both the Moon's orbital plane and the ecliptic plane.

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Perturbations due to Non-spherical Earth

When developing the two-body equations of motion, we assumed the Earth was a spherically symmetrical, homogeneous mass. In fact, the Earth is neither homogeneous nor spherical. The most dominant features are a bulge at the equator, a slight pear shape, and flattening at the poles. For a potential function of the Earth, we can find a satellite's acceleration by taking the gradient of the potential function. The most widely used form of the geopotential function depends on latitude and geopotential coefficients, Jn, called the zonal coefficients.

The potential generated by the non-spherical Earth causes periodic variations in all the orbital elements. The dominant effects, however, are secular variations in longitude of the ascending node and argument of perigee because of the Earth's oblateness, represented by the J2 term in the geopotential expansion. The rates of change of and due to J2 are

where n is the mean motion in degrees/day, J2 has the value 0.00108263, RE is the Earth's equatorial radius, a is the semi-major axis in kilometers, i is the inclination, e is the eccentricity, and and are in degrees/day. For satellites in GEO and below, the J2 perturbations dominate; for satellites above GEO the Sun and Moon perturbations dominate.

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Molniya orbits are designed so that the perturbations in argument of perigee are zero. This conditions occurs when the term 4-5sin2i is equal to zero or, that is, when the inclination is either 63.4 or 116.6 degrees.

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Perturbations from Atmospheric Drag

Drag is the resistance offered by a gas or liquid to a body moving through it. A spacecraft is subjected to drag forces when moving through a planet's atmosphere. This drag is greatest during launch and reentry, however, even a space vehicle in low Earth orbit experiences some drag as it moves through the Earth's thin upper atmosphere. In time, the action of drag on a space vehicle will cause it to spiral back into the atmosphere, eventually to disintegrate or burn up. If a space vehicle comes within 120 to 160 km of the Earth's surface, atmospheric drag will bring it down in a few days, with final disintegration occurring at an altitude of about 80 km. Above approximately 600 km, on the other hand, drag is so weak that orbits usually last more than 10 years - beyond a satellite's operational lifetime. The deterioration of a spacecraft's orbit due to drag is called decay.

The drag force FD on a body acts in the opposite direction of the velocity vector and is given by the equation

where CD is the drag coefficient, is the air density, v is the body's velocity, and A is the area of the body normal to the flow. The drag coefficient is dependent on the geometric form of the body and is generally determined by experiment. Earth orbiting satellites typically have very high drag coefficients in the range of about 2 to 4. Air density is given by the appendix Atmosphere Properties.

The region above 90 km is the Earth's thermosphere where the absorption of extreme ultraviolet radiation from the Sun results in a very rapid increase in temperature with altitude. At approximately 200-250 km this temperature approaches a limiting value, the average value of which ranges between about 600 and 1,200 K over a typical solar cycle. Solar activity also has a significant affect on atmospheric density, with high solar activity resulting in high density. Below about 150 km the density is not strongly affected by solar activity; however, at satellite altitudes in the range of 500 to 800 km, the density variations between solar maximum and solar minimum are approximately two orders of magnitude. The large variations imply that satellites will decay more rapidly during periods of solar maxima and much more slowly during solar minima.

For circular orbits we can approximate the changes in semi-major axis, period, and velocity per revolution using the following equations:

where a is the semi-major axis, P is the orbit period, and V, A and m are the satellite's velocity, area, and mass respectively. The term m/(CDA), called the ballistic coefficient, is

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given as a constant for most satellites. Drag effects are strongest for satellites with low ballistic coefficients, this is, light vehicles with large frontal areas.

A rough estimate of a satellite's lifetime, L, due to drag can be computed from

where H is the atmospheric density scale height. A substantially more accurate estimate (although still very approximate) can be obtained by integrating equation (4.47), taking into account the changes in atmospheric density with both altitude and solar activity.

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Perturbations from Solar Radiation

Solar radiation pressure causes periodic variations in all of the orbital elements. The magnitude of the acceleration in m/s2 arising from solar radiation pressure is

where A is the cross-sectional area of the satellite exposed to the Sun and m is the mass of the satellite in kilograms. For satellites below 800 km altitude, acceleration from atmospheric drag is greater than that from solar radiation pressure; above 800 km, acceleration from solar radiation pressure is greater.

Orbit ManeuversAt some point during the lifetime of most space vehicles or satellites, we must change one or more of the orbital elements. For example, we may need to transfer from an initial parking orbit to the final mission orbit, rendezvous with or intercept another spacecraft, or correct the orbital elements to adjust for the perturbations discussed in the previous section. Most frequently, we must change the orbit altitude, plane, or both. To change the orbit of a space vehicle, we have to change its velocity vector in magnitude or direction. Most propulsion systems operate for only a short time compared to the orbital period, thus we can treat the maneuver as an impulsive change in velocity while the position remains fixed. For this reason, any maneuver changing the orbit of a space vehicle must occur at a point where the old orbit intersects the new orbit. If the orbits do not intersect, we must use an intermediate orbit that intersects both. In this case, the total maneuver will require at least two propulsive burns.

Orbit Altitude Changes

The most common type of in-plane maneuver changes the size and energy of an orbit, usually from a low-altitude parking orbit to a higher-altitude mission orbit such as a geosynchronous orbit. Because the initial and final orbits do not intersect, the maneuver requires a transfer orbit. Figure 4.9 represents a Hohmann transfer orbit. In this case, the transfer orbit's ellipse is tangent to both the initial and final orbits at the transfer orbit's perigee and apogee respectively. The orbits are tangential, so the velocity vectors

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are collinear, and the Hohmann transfer represents the most fuel-efficient transfer between two circular, coplanar orbits. When transferring from a smaller orbit to a larger orbit, the change in velocity is applied in the direction of motion; when transferring from a larger orbit to a smaller, the change of velocity is opposite to the direction of motion.

The total change in velocity required for the orbit transfer is the sum of the velocity changes at perigee and apogee of the transfer ellipse. Since the velocity vectors are collinear, the velocity changes are just the differences in magnitudes of the velocities in each orbit. If we know the initial and final orbits, rA and rB, we can calculate the total velocity change using the following equations:

Note that equations (4.53) and (4.54) are the same as equation (4.6), and equations (4.55) and (4.56) are variations of equations (4.16) and (4.17) respectively.

Click here for example problem #4.17

Ordinarily we want to transfer a space vehicle using the smallest amount of energy, which usually leads to using a Hohmann transfer orbit. However, sometimes we may need to transfer a satellite between orbits in less time than that required to complete the Hohmann transfer. Figure 4.10 shows a faster transfer called the One-Tangent Burn. In this instance the transfer orbit is tangential to the initial orbit. It intersects the final orbit at an angle equal to the flight path angle of the transfer orbit at the point of intersection. An infinite number of transfer orbits are tangential to the initial orbit and intersect the final orbit at some angle. Thus, we may choose the transfer orbit by specifying the size of the transfer orbit, the angular change of the transfer, or the time required to complete the transfer. We can then define the transfer orbit and calculate the required velocities.

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For example, we may specify the size of the transfer orbit, choosing any semi-major axis that is greater than the semi-major axis of the Hohmann transfer ellipse. Once we know the semi-major axis of the ellipse, atx, we can calculate the eccentricity, angular distance traveled in the transfer, the velocity change required for the transfer, and the time required to complete the transfer. We do this using equations (4.53) through (4.57) and (4.59) above, and the following equations:

Click here for example problem #4.18

Another option for changing the size of an orbit is to use electric propulsion to produce a constant low-thrust burn, which results in a spiral transfer. We can approximate the velocity change for this type of orbit transfer by

where the velocities are the circular velocities of the two orbits.

Orbit Plane Changes

To change the orientation of a satellite's orbital plane, typically the inclination, we must change the direction of the velocity vector. This maneuver requires a component of V to be perpendicular to the orbital plane and, therefore, perpendicular to the initial velocity vector. If the size of the orbit remains constant, the maneuver is called a simple plane change. We can find the required change in velocity by using the law of cosines. For the case in which Vf is equal to Vi, this expression reduces to

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

Click here for example problem #4.19

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From equation (4.67) we see that if the angular change is equal to 60 degrees, the required change in velocity is equal to the current velocity. Plane changes are very expensive in terms of the required change in velocity and resulting propellant 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 orbit. In some cases, it may even be cheaper to boost the satellite into a higher orbit, change the orbit plane at apogee, and return the satellite to its original orbit.

Typically, orbital transfers require changes in both the size and the plane of the orbit, such as transferring from an inclined parking orbit at low altitude to a zero-inclination orbit at geosynchronous altitude. We can do this transfer in two steps: a Hohmann transfer to change the size of the orbit 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. 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, Vf is the final velocity, and is the angle change required. Note that equation (4.68) is in the same form as equation (4.63).

Click here for example problem #4.20

As can be seen from equation (4.68), a small plane change can be combined with an altitude change for almost no cost in V or propellant. Consequently, in practice, geosynchronous transfer is done with a small plane change at perigee and most of the plane change at apogee.

Another option is to complete the maneuver using three burns. The first burn 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, a combined plane change maneuver is done. This places the satellite in a second transfer orbit that 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 orbit. This three-burn maneuver may save propellant, but the propellant savings comes at the expense of the total time required to complete the maneuver.

When a plane change is used to modify inclination only, the magnitude of the angle change is simply the difference between the initial and final inclinations. In this case, the initial and final orbits share the same ascending and descending nodes. The plane change maneuver takes places when the space vehicle passes through one of these two nodes.

In some instances, however, a plane change is used to alter an orbit's longitude of ascending node in addition to the inclination. An example might be a maneuver to correct out-of-plane errors to make the orbits of two space vehicles coplanar in preparation for a rendezvous. If the orbital elements of the initial and final orbits are known, the plane change angle is determined by the vector dot product. If ii and i are the inclination and longitude of ascending node of the initial orbit, and if and f are the inclination and longitude of ascending node of the final orbit, then the angle between the orbital planes, , is given by

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Click here for example problem #4.21

The plane change maneuver takes place at one of two nodes where the initial and final orbits intersect. The latitude and longitude of these nodes are determined by the vector cross product. The position of one of the two nodes is given by

Knowing the position of one node, the second node is simply

Click here for example problem #4.22

Orbit Rendezvous

Orbital transfer becomes more complicated when the object is to rendezvous with or intercept another object in space: both the interceptor and the target must arrive at the rendezvous point at the same time. This precision demands a phasing orbit to accomplish the maneuver. A phasing orbit is any orbit that 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. 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 would inject the interceptor into a Hohmann transfer orbit.

Launch Windows

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. The time of the launch depends on the launch site's latitude and longitude and the satellite orbit's inclination and longitude of ascending node.

Orbit Maintenance

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Once in their mission orbits, many satellites need no additional orbit adjustment. On the other hand, mission requirements may demand that we maneuver the satellite to correct the orbital elements when perturbing forces have changed them. Two particular cases of note are satellites with repeating ground tracks and geostationary satellites.

After the mission of a satellite is complete, several options exist, depending on the orbit. We may allow low-altitude orbits to decay and reenter the atmosphere or use a velocity change to speed up the process. We may also boost satellites at all altitudes into benign orbits to reduce the probability of collision with active payloads, especially at synchronous altitudes.

V Budget

To an orbit designer, a space mission is a series of different orbits. For example, a satellite might be released in a low-Earth parking orbit, transferred to some mission orbit, go through a series of resphasings or alternate mission orbits, and then move to some final orbit at the end of its useful life. Each of these orbit changes requires energy. The V budget is traditionally used to account for this energy. It sums all the velocity changes required throughout the space mission life. In a broad sense the V budget represents the cost for each mission orbit scenario.

Escape VelocityWe know that if we throw a ball up from the surface of the Earth, it will rise for a while and then return. If we give it a larger initial velocity, it will rise higher and then return. There is a velocity, called the escape velocity, Vesc, such that if the ball is launched with an initial velocity greater than Vesc, it will rise and never return. We must give the particle enough kinetic energy to overcome all of the negative gravitational potential energy. Thus, if m is the mass of the ball, M is the mass of the Earth, and R is the radius of the Earth, the potential energy is -GmM /R. The kinetic energy of the ball, when it is launched, is mv2/2. We thus have

which is independent of the mass of the ball.

For a spacecraft launched to escape velocity from a parking orbit, R is the radius of the orbit.

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EXAMPLE PROBLEMS

PROBLEM 1.1

A spacecraft's engine ejects mass at a rate of 30 kg/s with an exhaust velocity of 3,100 m/s. The pressure at the nozzle exit is 5 kPa and the exit area is 0.7 m2. What is the thrust of the engine in a vacuum?

SOLUTION, Given: q = 30 kg/s Ve = 3,100 m/s Ae = 0.7 m2

Pe = 5 kPa = 5,000 N/m2

Pa = 0 Equation (1.6),

F = q × Ve + (Pe - Pa) × Ae F = 30 × 3,100 + (5,000 - 0) × 0.7 F = 96,500 N

PROBLEM 1.2

The spacecraft in problem 1.1 has an initial mass of 30,000 kg. What is the changein velocity if the spacecraft burns its engine for one minute?

SOLUTION, Given: M = 30,000 kg q = 30 kg/s Ve = 3,100 m/s t = 60 s

Equation (1.16), V = Ve × LN[ M / (M - qt) ] V = 3,100 × LN[ 30,000 / (30,000 - (30 × 60)) ] V = 192 m/s

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PROBLEM 1.3

A spacecraft's dry mass is 75,000 kg and the effective exhaust gas velocity of itsmain engine is 3,100 m/s. How much propellant must be carried if the propulsion systemis to produce a total v of 700 m/s?

SOLUTION, Given: Mf = 75,000 kg C = 3,100 m/s V = 700 m/s

Equation (1.20), Mo = Mf × e(V / C)

Mo = 75,000 × e(700 / 3,100)

Mo = 94,000 kg

Propellant mass,

Mp = Mo - Mf Mp = 94,000 - 75,000 Mp = 19,000 kg

PROBLEM 1.4

A 5,000 kg spacecraft is in Earth orbit traveling at a velocity of 7,790 m/s. Its engine is burned to accelerate it to a velocity of 12,000 m/s placing it on an escapetrajectory. The engine expels mass at a rate of 10 kg/s and an effective velocity of3,000 m/s. Calculate the duration of the burn.

SOLUTION, Given: M = 5,000 kg q = 10 kg/s C = 3,000 m/s V = 12,000 - 7,790 = 4,210 m/s

Equation (1.21),

t = M / q × [ 1 - 1 / e(V / C) ] t = 5,000 / 10 × [ 1 - 1 / e(4,210 / 3,000) ] t = 377 s

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PROBLEM 1.5

A rocket engine burning liquid oxygen and kerosene operates at a mixture ratio of 2.26 and a combustion chamber pressure of 50 atmospheres. If the nozzle is expandedto operate at sea level, calculate the exhaust gas velocity relative to the rocket.

SOLUTION, Given: O/F = 2.26 Pc = 50 atm Pe = Pa = 1 atm

From LOX/Kerosene Charts we estimate,

Tc = 3,470 K M = 21.40 k = 1.221

Equation (1.22),

Ve = SQRT[ (2 × k / (k - 1)) × (R' × Tc / M) × (1 - (Pe / Pc)(k-1)/k) ] Ve = SQRT[ (2 × 1.221 / (1.221 - 1)) × (8,314.51 × 3,470 / 21.40) × (1 - (1 / 50)(1.221-1)/1.221) ] Ve = 2,749 m/s

PROBLEM 1.6

A rocket engine produces a thrust of 1,000 kN at sea level with a propellant flowrate of 400 kg/s. Calculate the specific impulse.

SOLUTION, Given: F = 1,000,000 N q = 400 kg/s

Equation (1.23),

Isp = F / (q × g) Isp = 1,000,000 / (400 × 9.80665) Isp = 255 s (sea level)

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PROBLEM 1.7

A rocket engine uses the same propellant, mixture ratio, and combustion chamberpressure as that in problem 1.5. If the propellant flow rate is 500 kg/s, calculatethe area of the exhaust nozzle throat.

SOLUTION, Given: Pc = 50 × 0.101325 = 5.066 MPa Tc = 3,470<SUP.O< sup> K M = 21.40 k = 1.221 q = 500 kg/s

Equation (1.27),

Pt = Pc × [1 + (k - 1) / 2]-k/(k-1)

Pt = 5.066 × [1 + (1.221 - 1) / 2]-1.221/(1.221-1)

Pt = 2.839 MPa = 2.839x106 N/m2

Equation (1.28),

Tt = Tc / (1 + (k - 1) / 2) Tt = 3,470 / (1 + (1.221 - 1) / 2) Tt = 3,125 K

Equation (1.26),

At = (q / Pt) × SQRT[ (R' × Tt) / (M × k) ] At = (500 / 2.839x106) × SQRT[ (8,314.51 × 3,125) / (21.40 × 1.221) ] At = 0.1756 m2

PROBLEM 1.8

The rocket engine in problem 1.7 is optimized to operate at an elevation of 2000 meters.Calculate the area of the nozzle exit and the section ratio.

SOLUTION, Given: Pc = 5.066 MPa At = 0.1756 m2

k = 1.221

From Atmosphere Properties,

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Pa = 0.0795 MPa

Equation (1.29),

Nm2 = (2 / (k - 1)) × [(Pc / Pa)(k-1)/k - 1] Nm2 = (2 / (1.221 - 1)) × [(5.066 / 0.0795)(1.221-1)/1.221 - 1] Nm2 = 10.15 Nm = (10.15)1/2 = 3.185

Equation (1.30),

Ae = (At / Nm) × [(1 + (k - 1) / 2 × Nm2)/((k + 1) / 2)](k+1)/(2(k-1))

Ae = (0.1756 / 3.185) × [(1 + (1.221 - 1) / 2 × 10.15)/((1.221 + 1) / 2)](1.221+1)/(2(1.221-1))

Ae = 1.426 m2

Section Ratio,

Ae / At = 1.426 / 0.1756 = 8.12

PROBLEM 1.9For the rocket engine in problem 1.7, calculate the volume and dimensions of a possiblecombustion chamber. The convergent cone half-angle is 20 degrees.

SOLUTION, Given: At = 0.1756 m2 = 1,756 cm2

Dt = 2 × (1,756/ )1/2 = 47.3 cm = 20o

From Table 1,

L* = 102-127 cm for LOX/RP-1, let's use 110 cm

Equation (1.33),

Vc = At × L* Vc = 1,756 × 110 = 193,160 cm3

From Figure 1.7,

Lc = 66 cm (second-order approximation)

Equation (1.35),

Dc = SQRT[(Dt3 + 24/ × tan × Vc) / (Dc + 6 × tan × Lc)] Dc = SQRT[(47.33 + 24/ × tan(20) × 193,160) / (Dc + 6 × tan(20) × 66)] Dc = 56.6 cm (four interations)

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PROBLEM 1.10

A solid rocket motor burns along the face of a central cylindrical channel 10 meters long and 1 meter in diameter. The propellant has a burn rate coefficient of 5.5, a pressure exponent of 0.4, and a density of 1.77 g/ml. Calculate the burn rate andthe product generation rate when the chamber pressure is 5.0 MPa.

SOLUTION, Given: a = 5.5 n = 0.4 Pc = 5.0 MPa p = 1.77 g/ml Ab = × 1 × 10 = 31.416 m2

Equation (1.36),

r = a × Pcn

r = 5.5 × 5.00.4 = 10.47 mm/s

Equation (1.37),

q = p × Ab × r q = 1.77 × 31.416 × 10.47 = 582 kg/s

PROBLEM 1.11

A two-stage rocket has the following masses: 1st-stage propellant mass 120,000 kg, 1st-stage dry mass 9,000 kg, 2nd-stage propellant mass 30,000 kg, 2nd-stage dry mass 3,000 kg, and payload mass 3,000 kg. The specific impulses of the 1st and 2nd stagesare 260 s and 320 s respectively. Calculate the rocket's total V.

SOLUTION, Given: Mo1 = 120,000 + 9,000 + 30,000 + 3,000 + 3,000 = 165,000 kg Mf1 = 9,000 + 30,000 + 3,000 + 3,000 = 45,000 kg Isp1 = 260 s Mo2 = 30,000 + 3,000 + 3,000 = 36,000 kg Mf2 = 3,000 + 3,000 = 6,000 kg Isp2 = 320 s

Equation (1.24),

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C1 = Isp1g C1 = 260 × 9.80665 = 2,550 m/s

C2 = Isp2g C2 = 320 × 9.80665 = 3,138 m/s

Equation (1.38),

V1 = C1 × LN[ Mo1 / Mf1 ] V1 = 2,550 × LN[ 165,000 / 45,000 ] V1 = 3,313 m/s

V2 = C2 × LN[ Mo2 / Mf2 ] V2 = 3,138 × LN[ 36,000 / 6,000 ] V2 = 5,623 m/s

Equation (1.39),

VTotal = V1 + V2 VTotal = 3,313 + 5,623 VTotal = 8,936 m/s

PROBLEM 4.1

Calculate the velocity of an artificial satellite orbiting the Earth in a circular orbitat an altitude of 200 km above the Earth's surface.

SOLUTION, From Basics Constants,

Radius of Earth = 6,378.140 km GM of Earth = 3.986005x1014 m3/s2 Given: r = (6,378.14 + 200) × 1,000 = 6,578,140 m

Equation (4.6),

v = SQRT[ GM / r ] v = SQRT[ 3.986005x1014 / 6,578,140 ] v = 7,784 m/s

PROBLEM 4.2

Calculate the period of revolution for the satellite in problem 4.1.

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SOLUTION, Given: r = 6,578,140 m

Equation (4.9),

P2 = 4 × 2 × r3 / GM

P = SQRT[ 4 × 2 × r3 / GM ] P = SQRT[ 4 × 2 × 6,578,1403 / 3.986005x1014 ] P = 5,310 s

PROBLEM 4.3

Calculate the radius of orbit for a Earth satellite in a geosynchronous orbit, where the Earth's rotational period is 86,164.1 seconds.

SOLUTION, Given: P = 86,164.1 s

Equation (4.9),

P2 = 4 × 2 × r3 / GM

r = [ P2 × GM / (4 × 2) ]1/3

r = [ 86,164.12 × 3.986005x1014 / (4 × 2) ]1/3

r = 42,164,170 m

PROBLEM 4.4

An artificial Earth satellite is in an elliptical orbit which brings it to an altitude of 250 km at perigee and out to an altitude of 500 km at apogee. Calculate the velocity ofthe satellite at both perigee and apogee.

SOLUTION, Given: Rp = (6,378.14 + 250) × 1,000 = 6,628,140 m Ra = (6,378.14 + 500) × 1,000 = 6,878,140 m

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Equations (4.16) and (4.17),

Vp = SQRT[ 2 × GM × Ra / (Rp × (Ra + Rp)) ] Vp = SQRT[ 2 × 3.986005x1014 × 6,878,140 / (6,628,140 × (6,878,140 + 6,628,140)) ] Vp = 7,826 m/s

Va = SQRT[ 2 × GM × Rp / (Ra × (Ra + Rp)) ] Va = SQRT[ 2 × 3.986005x1014 × 6,628,140 / (6,878,140 × (6,878,140 + 6,628,140)) ] Va = 7,542 m/s

PROBLEM 4.5

A satellite in Earth orbit passes through its perigee point at an altitude of 200 kmabove the Earth's surface and at a velocity of 7,850 m/s. Calculate the apogee altitudeof the satellite.

SOLUTION, Given: Rp = (6,378.14 + 200) × 1,000 = 6,578,140 m Vp = 7,850 m/s

Equation (4.18),

Ra = Rp / [2 × GM / (Rp × Vp2) - 1] Ra = 6,578,140 / [2 × 3.986005x1014 / (6,578,140 × 7,8502) - 1] Ra = 6,805,140 m

Altitude @ apogee = 6,805,140 / 1,000 - 6,378.14 = 427.0 km

PROBLEM 4.6

Calculate the eccentricity of the orbit for the satellite in problem 4.5.

SOLUTION, Given: Rp = 6,578,140 m Vp = 7,850 m/s Equation (4.20),

e = Rp × Vp2 / GM - 1 e = 6,578,140 × 7,8502 / 3.986005x1014 - 1 e = 0.01696

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PROBLEM 4.7

A satellite in Earth orbit has a semi-major axis of 6,700 km and an eccentricity of 0.01.Calculate the satellite's altitude at both perigee and apogee.

SOLUTION, Given: a = 6,700 km e = 0.01

Equation (4.21) and (4.22),

Rp = a × (1 - e) Rp = 6,700 × (1 - .01) Rp = 6,633 km

Altitude @ perigee = 6,633 - 6,378.14 = 254.9 km

Ra = a × (1 + e) Ra = 6,700 × (1 + .01) Ra = 6,767 km

Altitude @ apogee = 6,767 - 6,378.14 = 388.9 km

PROBLEM 4.8

A satellite is launched into Earth orbit where its launch vehicle burns out at analtitude of 250 km. At burnout the satellite's velocity is 7,900 m/s with the zenith angle equal to 89 degrees. Calculate the satellite's altitude at perigee and apogee.

SOLUTION, Given: r1 = (6,378.14 + 250) × 1,000 = 6,628,140 m v1 = 7,900 m/s = 89o Equation (4.26),

(Rp / r1)1,2 = ( -C ± SQRT[ C2 - 4 × (1 - C) × -sin2 ]) / (2 × (1 - C))

where C = 2 × GM / (r1 × v12)

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C = 2 × 3.986005x1014 / (6,628,140 × 7,9002) C = 1.92718

(Rp / r1)1,2 = ( -1.92718 ± SQRT[ 1.927182 - 4 × -0.92718 × -sin2(89) ]) / (2 × -0.92718) (Rp / r1)1,2 = 0.996019 and 1.08252

Perigee Radius, Rp = Rp1 = r1 × (Rp / r1)1

Rp = 6,628,140 × 0.996019 Rp = 6,601,750 m

Altitude @ perigee = 6,601,750 / 1,000 - 6,378.14 = 223.6 km

Apogee Radius, Ra = Rp2 = r1 × (Rp / r1)2

Ra = 6,628,140 × 1.08252 Ra = 7,175,090 m

Altitude @ agogee = 7,175,090 / 1,000 - 6,378.14 = 797.0 km

PROBLEM 4.9

Calculate the eccentricity of the orbit for the satellite in problem 4.8.

SOLUTION, Given: r1 = 6,628,140 m v1 = 7,900 m/s = 89o

Equation (4.27),

e = SQRT[ (r1 × v12 / GM - 1)2 × sin2 + cos2 ] e = SQRT[ (6,628,140 × 7,9002 / 3.986005x1014 - 1)2 × sin2(89) + cos2(89) ] e = 0.04162

PROBLEM 4.10

Calculate the angle from perigee point to launch point for the satellitein problem 4.8.

SOLUTION, Given: r1 = 6,628,140 m v1 = 7,900 m/s

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= 89o

Equation (4.28),

tan = (r1 × v12 / GM) × sin × cos / [(r1 × v12 / GM) × sin2 - 1] tan = (6,628,140 × 7,9002 / 3.986005x1014) × sin(89) × cos(89)

/ [(6,628,140 × 7,9002 / 3.986005x1014) × sin2(89) - 1] tan = 0.48329

= arctan(0.48329) = 25.79o

PROBLEM 4.11A satellite is in an orbit with a semi-major axis of 7,500 km and an eccentricityof 0.1. Calculate the time it takes to move from a position 30 degrees past perigeeto 90 degrees past perigee.

SOLUTION, Given: a = 7,500 × 1,000 = 7,500,000 m e = 0.1 tO = 0 O = 30 deg × /180 = 0.52360 radians = 90 deg × /180 = 1.57080 radians Equation (4.34),

cos E = (e + cos ) / (1 + e cos )

Eo = arccos[(0.1 + cos(0.52360)) / (1 + 0.1 × cos(0.52360))] Eo = 0.47557 radians

E = arccos[(0.1 + cos(1.57080)) / (1 + 0.1 × cos(1.57080))] E = 1.47063 radians

Equation (4.35),

M = E - e × sin E

Mo = 0.47557 - 0.1 × sin(0.47557) Mo = 0.42978 radians

M = 1.47063 - 0.1 × sin(1.47063) M = 1.37113 radians

Equation (4.33),

n = SQRT[ GM / a3 ] n = SQRT[ 3.986005x1014 / 7,500,0003 ]

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n = 0.00097202 rad/s

Equation (4.32),

M - Mo = n × (t - tO)

t = tO + (M - Mo) / n t = 0 + (1.37113 - 0.42978) / 0.00097202 t = 968.4 s

PROBLEM 4.12The satellite in problem 4.11 has a true anomaly of 90 degrees. What will be thesatellite's position, i.e. it's true anomaly, 20 minutes later?

SOLUTION, Given: a = 7,500,000 m e = 0.1 tO = 0 t = 20 × 60 = 1,200 s O = 90 × /180 = 1.57080 rad

From problem 4.11,

Mo = 1.37113 rad n = 0.00097202 rad/s

Equation (4.32),

M - Mo = n × (t - tO)

M = Mo + n × (t - tO) M = 1.37113 + 0.00097202 × (1,200 - 0) M = 2.53755

METHOD #1, Low Accuracy:

Equation (4.36),

~ M + 2 × e × sin M + 1.25 × e2 × sin 2M ~ 2.53755 + 2 × 0.1 × sin(2.53755) + 1.25 × 0.12 × sin(2 × 2.53755) ~ 2.63946 = 151.2 degrees

METHOD #2, High Accuracy:

Equation (4.35),

M = E - e × sin E 2.53755 = E - 0.1 × sin E

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By iteration, E = 2.58996 radians

Equation (4.34),

cos E = (e + cos ) / (1 + e cos )

Rearranging variables gives,

cos = (cos E - e) / (1 - e cos E)

= arccos[(cos(2.58996) - 0.1) / (1 - 0.1 × cos(2.58996)] = 2.64034 = 151.3 degrees

PROBLEM 4.13For the satellite in problems 4.11 and 4.12, calculate the length of its positionvector, its flight-path angle, and its velocity when the satellite's true anomalyis 225 degrees.

SOLUTION, Given: a = 7,500,000 m e = 0.1 = 225 degrees

Equations (4.37) and (4.38),

r = a × (1 - e2) / (1 + e × cos ) r = 7,500,000 × (1 - 0.12) / (1 + 0.1 × cos(225)) r = 7,989,977 m

= arctan[ e × sin / (1 + e × cos )] = arctan[ 0.1 × sin(225) / (1 + 0.1 × cos(225))] = -4.351 degrees

Equation (4.39),

v = SQRT[ GM × a × (1 - e2)] / (r × cos ) v = SQRT[ 3.986005x1014 × 7,500,000 × (1 - 0.12)] / (7,989,977 × cos(-4.351)) v = 6,828 m/s

PROBLEM 4.14Calculate the perturbations in longitude of the ascending node and argument of

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perigee caused by the Moon and Sun for the International Space Station orbitingat an altitude of 400 km, an inclination of 51.6 degrees, and with an orbital period of 92.6 minutes.

SOLUTION, Given: i = 51.6 degrees n = 1436 / 92.6 = 15.5 revolutions/day

Equations (4.40) through (4.43),

Moon = -0.00338 × cos(i) / n Moon = -0.00338 × cos(51.6) / 15.5 Moon = -0.000135 deg/day

Sun = -0.00154 × cos(i) / n Sun = -0.00154 × cos(51.6) / 15.5 Sun = -0.0000617 deg/day

Moon = 0.00169 × (4 - 5 × sin2 i) / n Moon = 0.00169 × (4 - 5 × sin2 51.6) / 15.5 Moon = 0.000101 deg/day

Sun = 0.00077 × (4 - 5 × sin2 i) / n Sun = 0.00077 × (4 - 5 × sin2 51.6) / 15.5 Sun = 0.000046 deg/day

PROBLEM 4.15A satellite is in an orbit with a semi-major axis of 7,500 km, an inclination of 28.5 degrees, and an eccentricity of 0.1. Calculate the J2 perturbations in longitude of the ascending node and argument of perigee.

SOLUTION, Given: a = 7,500 km i = 28.5 degrees e = 0.1

Equations (4.44) and (4.45),

J2 = -2.06474x1014 × a-7/2 × (cos i) × (1 - e2)-2

J2 = -2.06474x1014 × (7,500)-7/2 × (cos 28.5) × (1 - (0.1)2)-2

J2 = -5.067 deg/day

J2 = 1.03237x1014 × a-7/2 × (4 - 5 × sin2 i) × (1 - e2)-2

J2 = 1.03237x1014 × (7,500)-7/2 × (4 - 5 × sin2 28.5) × (1 - (0.1)2)-2

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J2 = 8.250 deg/day

PROBLEM 4.16

A satellite is in a circular Earth orbit at an altitude of 400 km. The satellite has a cylindrical shape 2 m in diameter by 4 m long and has a mass of 1,000 kg. Thesatellite is traveling with its long axis perpendicular to the velocity vector and it's drag coefficient is 2.67. Calculate the perturbations due to atmospheric dragand estimate the satellite's lifetime.

SOLUTION, Given: a = (6,378.14 + 400) × 1,000 = 6,778,140 m A = 2 × 4 = 8 m2

m = 1,000 kg CD = 2.67

From Atmosphere Properties,

= 2.62x10-12 kg/m3

H = 58.2 km

Equation (4.6),

V = SQRT[ GM / a ] V = SQRT[ 3.986005x1014 / 6,778,140 ] V = 7,669 m/s

Equations (4.47) through (4.49),

arev = (-2 × × CD × A × × a2) / m arev = (-2 × × 2.67 × 8 × 2.62x10-12 × 6,778,1402) / 1,000 arev = -16.2 m

Prev = (-6 × 2 × CD × A × × a2) / (m × V) Prev = (-6 × 2 × 2.67 × 8 × 2.62x10-12 × 6,778,1402) / (1,000 × 7,669) Prev = -0.0199 s

Vrev = ( × CD × A × × a × V) / m Vrev = ( × 2.67 × 8 × 2.62x10-12 × 6,778,140 × 7,669) / 1,000 Vrev = 0.00914 m/s

Equation (4.50),

L ~ -H / arev L ~ -(58.2 × 1,000) / -16.2 L ~ 3,600 revolutions

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PROBLEM 4.17

A spacecraft is in a circular parking orbit with an altitude of 200 km. Calculate the velocity change required to perform a Hohmann transfer to a circular orbit atgeosynchronous altitude.

SOLUTION, Given: rA = (6,378.14 + 200) × 1,000 = 6,578,140 m

From problem 4.3,

rB = 42,164,170 m

Equations (4.52) through (4.59),

atx = (rA + rB) / 2 atx = (6,578,140 + 42,164,170) / 2 atx = 24,371,155 m

ViA = SQRT[ GM / rA ] ViA = SQRT[ 3.986005x1014 / 6,578,140 ] ViA = 7,784 m/s

VfB = SQRT[ GM / rB ] VfB = SQRT[ 3.986005x1014 / 42,164,170 ] VfB = 3,075 m/s

VtxA = SQRT[ GM × (2 / rA - 1 / atx)] VtxA = SQRT[ 3.986005x1014 × (2 / 6,578,140 - 1 / 24,371,155)] VtxA = 10,239 m/s

VtxB = SQRT[ GM × (2 / rB - 1 / atx)] VtxB = SQRT[ 3.986005x1014 × (2 / 42,164,170 - 1 / 24,371,155)] VtxB = 1,597 m/s

VA = VtxA - ViA

VA = 10,239 - 7,784 VA = 2,455 m/s

VB = VfB - VtxB

VB = 3,075 - 1,597 VB = 1,478 m/s

VT = VA + VB

VT = 2,455 + 1,478 VT = 3,933 m/s

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PROBLEM 4.18

A satellite is in a circular parking orbit with an altitude of 200 km. Using a one-tangent burn, it is to be transferred to geosynchronous altitude using and a transferellipse with a semi-major axis of 30,000 km. Calculate the total required velocitychange and the time required to complete the transfer.

SOLUTION, Given: rA = (6,378.14 + 200) × 1,000 = 6,578,140 m rB = 42,164,170 m atx = 30,000 × 1,000 = 30,000,000 m

Equations (4.60) through (4.62),

e = 1 - rA / atx

e = 1 - 6,578,140 / 30,000,000 e = 0.780729

= arccos[(atx × (1 - e2) / rB - 1) / e ] = arccos[(30,000,000 × (1 - 0.7807292) / 42,164,170 - 1) / 0.780729 ] = 157.670 degrees

= arctan[ e × sin / (1 + e × cos )] = arctan[ 0.780729 × sin(157.670) / (1 + 0.780729 × cos(157.670))] = 46.876 degrees

Equations (4.53) through (4.57),

ViA = SQRT[ GM / rA ] ViA = SQRT[ 3.986005x1014 / 6,578,140 ] ViA = 7,784 m/s

VfB = SQRT[ GM / rB ] VfB = SQRT[ 3.986005x1014 / 42,164,170 ] VfB = 3,075 m/s

VtxA = SQRT[ GM × (2 / rA - 1 / atx)] VtxA = SQRT[ 3.986005x1014 × (2 / 6,578,140 - 1 / 30,000,000)] VtxA = 10,388 m/s

VtxB = SQRT[ GM × (2 / rB - 1 / atx)] VtxB = SQRT[ 3.986005x1014 × (2 / 42,164,170 - 1 / 30,000,000)] VtxB = 2,371 m/s

VA = VtxA - ViA

VA = 10,388 - 7,784 VA = 2,604 m/s

Equation (4.63),

VB = SQRT[ VtxB2 + VfB

2 - 2 × VtxB × VfB × cos ]

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VB = SQRT[ 2,3712 + 3,0752 - 2 × 2,371 × 3,075 × cos(46.876)] VB = 2,260 m/s

Equation (4.59),

VT = VA + VB

VT = 2,604 + 2,260 VT = 4,864 m/s

Equations (4.64) and (4.65),

E = arctan[(1 - e2)1/2 × sin / (e + cos )] E = arctan[(1 - 0.7807292)1/2 × sin(157.670) / (0.780729 + cos(157.670))] E = 2.11688 radians

TOF = (E - e × sin E) × SQRT[ atx3 / GM ]

TOF = (2.11688 - 0.780729 × sin(2.11688)) × SQRT[ 30,000,0003 / 3.986005x1014 ] TOF = 11,931 s = 3.314 hours

PROBLEM 4.19Calculate the velocity change required to transfer a satellite from a circular600 km orbit with an inclination of 28 degrees to an orbit of equal size with aninclination of 20 degrees.

SOLUTION, Given: r = (6,378.14 + 600) × 1,000 = 6,978,140 m = 28 - 20 = 8 degrees

Equation (4.6),

Vi = SQRT[ GM / r ] Vi = SQRT[ 3.986005x1014 / 6,978,140 ] Vi = 7,558 m/s

Equation (4.67),

V = 2 × Vi × sin( /2) V = 2 × 7,558 × sin(8/2) V = 1,054 m/s

PROBLEM 4.20

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A satellite is in a parking orbit with an altitude of 200 km and an inclination of 28 degrees. Calculate the total velocity change required to transfer the satelliteto a zero-inclination geosynchronous orbit using a Hohmann transfer with a combined plane change at apogee.

Given: rA = (6,378.14 + 200) × 1,000 = 6,578,140 m rB = 42,164,170 m = 28 degrees

From problem 4.17,

VfB = 3,075 m/s VtxB = 1,597 m/s VA = 2,455 m/s

Equation (4.68),

VB = SQRT[ VtxB2 + VfB

2 - 2 × VtxB × VfB × cos ] VB = SQRT[ 1,5972 + 3,0752 - 2 × 1,597 × 3,075 × cos(28)] VB = 1,826 m/s

Equation (4.59),

VT = VA + VB

VT = 2,455 + 1,826 VT = 4,281 m/s

PROBLEM 4.21A spacecraft is in an orbit with an inclination of 30 degrees and the longitude of the ascending node is 75 degrees. Calculate the angle change required to change the inclination to 32 degrees and the longitude of the ascending node to 80 degrees.

SOLUTION, Given: ii = 30 degrees i = 75 degrees if = 32 degrees f = 80 degrees

Equation (4.69),

a1 = sin(ii)cos( i) = sin(30)cos(75) = 0.129410

a2 = sin(ii)sin( i) = sin(30)sin(75) = 0.482963

a3 = cos(ii) = cos(30) = 0.866025

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b1 = sin(if)cos( f) = sin(32)cos(80) = 0.0920195

b2 = sin(if)sin( f) = sin(32)sin(80) = 0.521869

b3 = cos(if) = cos(32) = 0.848048

= arccos(a1 × b1 + a2 × b2 + a3 × b3) = arccos(0.129410 × 0.0920195 + 0.482963 × 0.521869 + 0.866025 × 0.848048) = 3.259 degrees

PROBLEM 4.22Calculate the latitude and longitude of the intersection nodes between the initialand final orbits for the spacecraft in problem 4.22.

SOLUTION, From problem 4.21,

a1 = 0.129410 a2 = 0.482963 a3 = 0.866025 b1 = 0.0920195 b2 = 0.521869 b3 = 0.848048

Equations (4.70) and (4.71),

c1 = a2 × b3 - a3 × b2 = 0.482963 × 0.848048 - 0.866025 × 0.521869 = -0.0423757

c2 = a3 × b1 - a1 × b3 = 0.866025 × 0.0920195 - 0.129410 × 0.848048 = -0.0300543

c3 = a1 × b2 - a2 × b1 = 0.129410 × 0.521869 - 0.482963 × 0.0920195 = 0.0230928

lat1 = arctan(c3 / (c12 + c22)1/2) lat1 = arctan(0.0230928 / (-0.04237572 + -0.03005432)1/2) lat1 = 23.965 degrees

long1 = arctan(c2 / c1) + 90 long1 = arctan(-0.0300543 / -0.0423757) + 90 long1 = 125.346 degrees

lat2 = -23.965 degrees long2 = 125.346 + 180 = 305.346 degrees

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PROBLEM 4.23Calculate the escape velocity of a spacecraft launched from an Earth orbit with analtitude of 300 km.

SOLUTION, Given: r = (6,378.14 + 300) × 1,000 = 6,678,140 m

Equation (4.72),

VESC = SQRT[ 2 × GM / r ] VESC = SQRT[ 2 × 3.986005x1014 / 6,678,140 ] VESC = 10,926 m/s

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