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 The Hyperbolic Orbit

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 Sections

A 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

Satellite orbits can be any of the four conic sections. This page deals mostly with elliptical orbits, though we conclude with an examination of the hyperbolic orbit.

Orbital Elements

To 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.

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 Orbits

For 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 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 Gravitation

Newton'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.

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 Motion

In 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

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 Satellites

Through 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

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.

Click here for example problem #4.2

Click here for example problem #4.3

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.

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

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 Vehicle

The 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 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.

Click here for example problem #4.8

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

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:

The semi-major axis is, of course, equal to (Rp+Ra)/2, though it may be easier to

calculate it directly as follows:

Click here for example problem #4.11

If e is solved for directly using equation (4.27) or (4.30), and a is solved for using

equation (4.32), Rp and Ra can be solved for simply using equations (4.21) and (4.22).

Orbit Tilt, Rotation and Orientation

Above we determined the size and shape of the orbit, but to determine the orientation of the orbit in space, we must know the latitude and longitude and the heading of the space

vehicle at burnout.

Figure 4.9 above illustrates the location of a space vehicle at engine burnout, or orbit

insertion. is the azimuth heading measured in degrees clockwise from north, is the

geocentric latitude (or declination) of the burnout point, is the angular distance

between the ascending node and the burnout point measured in the equatorial plane,

and is the angular distance between the ascending node and the burnout point

measured in the orbital plane. 1 and 2 are the geographical longitudes of the ascending node and the burnout point at the instant of engine burnout. Figure 4.10

pictures the orbital elements, where i is the inclination, is the longitude at the

ascending node, is the argument of periapsis, and is the true anomaly.

If , , and 2 are given, the other values can be calculated from the following

relationships:

In equation (4.36), the value of is found using equation (4.28) or (4.31). If is

positive, periapsis is west of the burnout point (as shown in Figure 4.10); if is

negative, periapsis is east of the burnout point.

The longitude of the ascending node, , is measured in celestial longitude, while 1 is

geographical longitude. The celestial longitude of the ascending node is equal to the local

apparent sidereal time, in degrees, at longitude 1 at the time of engine burnout. Sidereal time is defined as the hour angle of the vernal equinox at a specific locality and

time; it has the same value as the right ascension of any celestial body that is crossing

the local meridian at that same instant. At the moment when the vernal equinox crosses the local meridian, the local apparent sidereal time is 00:00. See this sidereal time

calculator.

Click here for example problem #4.12

Geodetic Latitude, Geocentric Latitude, and Declination

Latitude is the angular distance of a point on Earth's

surface north or south of Earth's equator, positive

north and negative south. The geodetic latitude (or

geographical latitude), , is the angle defined by

the intersection of the reference ellipsoid normal

through the point of interest and the true equatorial

plane. The geocentric latitude, ', is the angle

between the true equatorial plane and the radius

vector to the point of intersection of the reference

ellipsoid and the reference ellipsoid normal passing

through the point of interest. Declination, , is the

angular distance of a celestial object north or south

of Earth's equator. It is the angle between the

geocentric radius vector to the object of interest and

the true equatorial plane.

R is the magnitude of the reference ellipsoid's geocentric radius vector to the point of interest on its

surface, r is the magnitude of the geocentric radius vector to the celestial object of interest, and the altitude

h is the perpendicular distance from the reference ellipsoid to the celestial object of interest. The value of

R at the equator is a, and the value of R at the poles is b. The ellipsoid's flattening, f, is the ratio of the

equatorial-polar length difference to the equatorial length. For Earth, a equals 6,378,137 meters, b equals

6,356,752 meters, and f equals 1/298.257.

When solving problems in orbital mechanics, the measurements of greatest usefulness are the magnitude

of the radius vector, r, and declination, , of the object of interest. However, we are often given, or

required to report, data in other forms. For instance, at the time of some specific event, such as "orbit

insertion", we may be given the spacecraft's altitude along with the geodetic latitude and longitude of the

sub-vehicle point. In such cases, it may be necessary to convert the given data to a form more suitable for

our calculations.

The relationship between geodetic and geocentric latitude is,

The radius of the reference ellipsoid is given by,

The length r can be solved from h, or h from r, using one of the following,

And declination is calculated using,

For spacecraft in low earth orbit, the difference between and ' is very small, typically not more than

about 0.00001 degree. Even at the distance of the Moon, the difference is not more than about 0.01 degree.

Unless very high accuracy is needed, for operations near Earth we can assume ≈ ' and r ≈ R + h.

It is important to note that the value of h is not always measured as described and illustrated above. In

some applications it is customary to express h as the perpendicular distance from a reference sphere, rather

than the reference ellipsoid. In this case, R is considered constant and is often assigned the value of Earth's

equatorial radius, hence h = r – a. This is the method typically used when a spacecraft's orbit is expressed

in a form such as "180 km × 220 km". The example problems presented in this web site also assume this

method of measurement.

Position in an Elliptical Orbit

Johannes 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 is 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

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.

Click here for example problem #4.13 Click here for example problem #4.14

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 #4.15

Orbit Perturbations

The 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,

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.

Click here for example problem #4.16

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.

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.

Click here for example problem #4.17

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 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 (4.53), taking into account the changes in atmospheric density with both altitude and solar activity.

Click here for example problem #4.18

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 Maneuvers

At 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

4.11 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 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 (4.59) and (4.60) are the same as equation (4.6), and equations

(4.61) and (4.62) are the same as equation

(4.45).

Click here for example problem #4.19

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

4.12 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.

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 (4.59) through (4.63) and

(4.65) above, and the following equations:

Click here for example problem #4.20

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 #4.21

From equation (4.73) 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 (4.74) is in the same form as equation (4.69).

Click here for example problem #4.22

As can be seen from equation (4.74), 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

Click here for example problem #4.23

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 #4.24

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

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.

The Hyperbolic Orbit

The discussion thus far has focused on the elliptical orbit, which will result whenever a spacecraft has insufficient velocity to escape the gravity of its primary. There is a

velocity, called the escape velocity, Vesc, such that if the spacecraft is launched with an

initial velocity greater than Vesc, it will travel away from the planet and never return. To

achieve escape velocity we must give the spacecraft enough kinetic energy to overcome all of the negative gravitational potential energy. Thus, if m is the mass of the

spacecraft, M is the mass of the planet, and r is the radial distance between the

spacecraft and planet, the potential energy is -GmM /r. The kinetic energy of the

spacecraft, when it is launched, is mv2/2. We thus have

which is independent of the mass of the spacecraft.

Click here for example problem #4.25

A space vehicle that has

exceeded the escape velocity

of a planet will travel a hyperbolic path relative to the

planet. The hyperbola is an

unusual and interesting conic

section because it has two branches. The arms of a

hyperbola are asymptotic to

two intersecting straight line

(the asymptotes). If we consider the left-hand focus, f,

as the prime focus (where the

center of our gravitating body

is located), then only the left branch of the hyperbola

represents the possible orbit.

If, instead, we assume a force

of repulsion between our

satellite and the body located at f (such as the force between

two like-charged electric

particles), then the right-hand

branch represents the orbit. The parameters a, b and c are

labeled in Figure 4.14. We can

see that c2 = a2+ b2 for the

hyperbola. The eccentricity is,

The angle between the

asymptotes, which represents

the angle through which the

path of a space vehicle is turned by its encounter with a

planet, is labeled . This

turning angle is related to the geometry of the hyperbola as

follows:

If we let equal the angle between the periapsis vector and the departure asymptote, i.e. the true anomaly at infinity, we have

If we know the radius, r, velocity, v, and flight path angle, , of a point on the orbit (see Figure 4.15), we can calculate the eccentricity and semi-major axis using equations

(4.30) and (4.32) as previously presented. Note that the semi-major axis of a hyperbola

is negative.

The true anomaly corresponding to known valves of r, v and can be calculated using equation (4.31), however special care must be taken to assure the angle is placed in the

correct quadrant. It may be easier to first calculate e and a, and then calculate true

anomaly using equation (4.43), rearranged as follows:

Whenever is positive, should be taken as positive; whenever is negative, should be taken as negative.

The impact parameter, b, is the distance of closest approach that would result between a

spacecraft and planet if the spacecraft trajectory was undeflected by gravity. The impact

parameter is,

Closet approach occurs at periapsis, where the radius distance, ro, is equal to

p is a geometrical constant of the conic called the parameter or semi-latus rectum, and

is equal to

Click here for example problem #4.26

At any known true anomaly, the magnitude of a spacecraft's radius vector, its flight-path

angle, and its velocity can be calculated using equations (4.43), (4.44) and (4.45).

Click here for example problem #4.27

Early we introduced the variable eccentric anomaly and its use in deriving the time of

flight in an elliptical orbit. In a similar manner, the analytical derivation of the hyperbolic

time of flight, using the hyperbolic eccentric anomaly, F, can be derived as follows:

where,

Whenever is positive, F should be taken as positive; whenever is negative, F should be taken as negative.

Click here for example problem #4.28

Hyperbolic Excess Velocity

If you give a space vehicle exactly

escape velocity, it will just barely

escape the gravitational field, which means that its velocity will

be approaching zero as its distance

from the force center approaches

infinity. If, on the other hand, we give our vehicle more than escape

velocity at a point near Earth, we would expect the velocity at a great distance from

Earth to be approaching some finite constant value. This residual velocity the vehicle

would have left over even at infinity is called hyperbolic excess velocity. We can calculate this velocity from the energy equation written for two points on the hyperbolic

escape trajectory – a point near Earth called the burnout point and a point an infinite

distance from Earth where the velocity will be the hyperbolic excess velocity, v∞. Solving

for v∞ we obtain

Note that if v∞ = 0 (as it is on a parabolic trajectory), the burnout velocity, vbo, becomes

simply the escape velocity.

Click here for example problem #4.29

It is, or course, absurd to talk about a space vehicle "reaching infinity" and in this sense

it is meaningless to talk about escaping a gravitational field completely. It is a fact, however, that once a space vehicle is a great distance from Earth, for all practical

purposes it has escaped. In other words, it has already slowed down to very nearly its

hyperbolic excess velocity. It is convenient to define a sphere around every gravitational

body and say that when a probe crosses the edge of this sphere of influence it has escaped. Although it is difficult to get agreement on exactly where the sphere of

influence should be drawn, the concept is convenient and is widely used, especially in

lunar and interplanetary trajectories. For most purposes, the radius of the sphere of

influence for a planet can be calculated as follows:

where Dsp is the distance between the Sun and the planet, Mp is the mass of the planet,

and Ms is the mass of the Sun. Equation (4.89) is also valid for calculating a moon's

sphere of influence, where the moon is substituted for the planet and the planet for the

Sun.

Click here for example problem #4.30

Compiled, edited and written in part by Robert A. Braeunig, 1997, 2005, 2007, 2008,

2011, 2012, 2013.

Bibliography

Back Home Next

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

change in 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

PROBLEM 1.3

A spacecraft's dry mass is 75,000 kg and the effective exhaust gas velocity

of its main engine is 3,100 m/s. How much propellant must be carried if

the

propulsion system is 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 escape trajectory. The engine expels mass at a rate of 10 kg/s and

an

effective velocity of 3,000 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

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

expanded to 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)) × (8,314.46 × 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

flow rate 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)

PROBLEM 1.7

A rocket engine uses the same propellant, mixture ratio, and combustion

chamber

pressure as that in problem 1.5. If the propellant flow rate is 500 kg/s,

calculate the 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 = 2.839×106 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 / 2.839×106) × SQRT[ (8,314.46 × 3,125) / (21.40 × 1.221) ]

At = 0.1756 m2

</sup.o<>

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,

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 1.9

For the rocket engine in problem 1.7, calculate the volume and dimensions

of a

possible combustion 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)

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 1.70 g/ml. Calculate

the

burn rate and the product generation rate when the chamber pressure is 5.0

MPa.

SOLUTION,

Given: a = 5.5

n = 0.4

Pc = 5.0 MPa

p = 1.70 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 = 1.70 × 31.416 × 10.47 = 559 kg/s

PROBLEM 1.11

Calculate the ideal density of a solid rocket propellant consisting of 68%

ammonium perchlorate, 18% aluminum, and 14% HTPB by mass.

SOLUTION,

Given: wAP = 0.68

wAl = 0.18

wHTPB = 0.14

From Properties of Rocket Propellants we have,

AP = 1.95 g/ml

Al = 2.70 g/ml

HTPB = ≈0.93 g/ml

Equation (1.38),

p = 1 / i (w / )i

p = 1 / [(0.68 / 1.95) + (0.18 / 2.70) + (0.14 / 0.93)]

p = 1.767

PROBLEM 1.12

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 stages are 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),

C1 = Isp1g

C1 = 260 × 9.80665 = 2,550 m/s

C2 = Isp2g

C2 = 320 × 9.80665 = 3,138 m/s

Equation (1.39),

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 (1.40),

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 orbit at an altitude of 200 km above the Earth's surface.

SOLUTION,

From Basics Constants,

Radius of Earth = 6,378.14 km

GM of Earth = 3.986005×1014 m3/s2

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

Equation (4.6),

v = SQRT[ GM / r ]

v = SQRT[ 3.986005×1014 / 6,578,140 ]

v = 7,784 m/s

PROBLEM 4.2

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

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 / 3.986005×1014 ]

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 × 3.986005×1014 / (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 of the 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

Equations (4.16) and (4.17),

Vp = SQRT[ 2 × GM × Ra / (Rp × (Ra + Rp)) ]

Vp = SQRT[ 2 × 3.986005×1014 × 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 × 3.986005×1014 × 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 km above the Earth's surface and at a velocity of 7,850 m/s. Calculate

the

apogee altitude of 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 × 3.986005×1014 / (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 / 3.986005×1014 - 1

e = 0.01696

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

an altitude 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)

C = 2 × 3.986005×1014 / (6,628,140 × 7,9002)

C = 1.927179

(Rp / r1)1,2 = ( -1.927179 ± SQRT[ 1.9271792 - 4 × -0.927179 × -

sin2(89) ]) / (2 × -0.927179)

(Rp / r1)1,2 = 0.996019 and 1.082521

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 × 1.082521

Ra = 7,175,100 m

Altitude @ agogee = 7,175,100 / 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 / 3.986005×1014 - 1)2 × sin2(89) +

cos2(89) ]

e = 0.0416170

PROBLEM 4.10

Calculate the angle from perigee point to launch point for the satellite

in problem 4.8.

SOLUTION,

Given: r1 = 6,628,140 m

v1 = 7,900 m/s

= 89o

Equation (4.28),

tan = (r1 × v12 / GM) × sin × cos / [(r1 × v12 / GM) × sin2

- 1]

tan = (6,628,140 × 7,9002 / 3.986005×1014) × sin(89) × cos(89)

/ [(6,628,140 × 7,9002 / 3.986005×1014) × sin2(89) - 1]

tan = 0.48329

= arctan(0.48329)

= 25.794o

PROBLEM 4.11

Calculate the semi-major axis of the orbit for the satellite in problem

4.8.

SOLUTION,

Given: r1 = 6,628,140 m

v1 = 7,900 m/s

Equation (4.32),

a = 1 / ( 2 / r1 - v12 / GM )

a = 1 / ( 2 / 6,628,140 - 7,9002 / 3.986005×1014) )

a = 6,888,430 m

PROBLEM 4.12

For the satellite in problem 4.8, burnout occurs 2000-10-20, 15:00 UT. The

geocentric coordinates at burnout are 32o N latitude, 60o W longitude, and

the

azimuth heading is 86o. Calculate the orbit's inclination, argument of

perigee,

and longitude of ascending node.

SOLUTION,

Given: = 86o

= 32o

2 = -60o

From problem 4.10,

= 25.794o

Equation (4.33),

cos(i) = cos( ) × sin( )

cos(i) = cos(32) × sin(86)

i = 32.223o

Equations (4.34) and (4.36),

tan( ) = tan( ) / cos( )

tan( ) = tan(32) / cos(86)

= 83.630o

= -

= 83.630 - 25.794

= 57.836o

Equations (4.35) and (4.37),

tan( ) = sin( ) × tan( )

tan( ) = sin(32) × tan(86)

= 82.483o

1 = 2 -

1 = -60 - 82.483

1 = -142.483o

= Sidereal time at -142.483 longitude, 2000-10-20, 15:00 UT

= 7h 27' 34" = 111.892o

PROBLEM 4.13

A satellite is in an orbit with a semi-major axis of 7,500 km and an

eccentricity

of 0.1. Calculate the time it takes to move from a position 30 degrees

past

perigee to 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 (4.40),

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 (4.41),

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 (4.39),

n = SQRT[ GM / a3 ]

n = SQRT[ 3.986005×1014 / 7,500,000

3 ]

n = 0.00097202 rad/s

Equation (4.38),

M - Mo = n × (t - tO)

t = tO + (M - Mo) / n

t = 0 + (1.37113 - 0.42978) / 0.00097202

t = 968.4 s

PROBLEM 4.14

The satellite in problem 4.13 has a true anomaly of 90 degrees. What will

be the

satellite'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 4.13,

Mo = 1.37113 rad

n = 0.00097202 rad/s

Equation (4.38),

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 (4.42),

~ 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 (4.41),

M = E - e × sin E

2.53755 = E - 0.1 × sin E

By iteration, E = 2.58996 radians

Equation (4.40),

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 4.15

For the satellite in problems 4.13 and 4.14, calculate the length of its

position

vector, its flight-path angle, and its velocity when the satellite's true

anomaly

is 225 degrees.

SOLUTION,

Given: a = 7,500,000 m

e = 0.1

= 225 degrees

Equations (4.43) and (4.44),

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 (4.45),

v = SQRT[ GM × (2 / r - 1 / a)]

v = SQRT[ 3.986005×1014 × (2 / 7,989,977 - 1 / 7,500,000)]

v = 6,828 m/s

PROBLEM 4.16

Calculate the perturbations in longitude of the ascending node and argument

of

perigee caused by the Moon and Sun for the International Space Station

orbiting

at 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 (4.46) through (4.49),

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 4.17

A 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 (4.50) and (4.51),

J2 = -2.06474×1014 × a-7/2 × (cos i) × (1 - e2)-2

J2 = -2.06474×1014 × (7,500)-7/2 × (cos 28.5) × (1 - (0.1)2)-2

J2 = -5.067 deg/day

J2 = 1.03237×1014 × a-7/2 × (4 - 5 × sin2 i) × (1 - e2)-2

J2 = 1.03237×1014 × (7,500)-7/2 × (4 - 5 × sin2 28.5) × (1 - (0.1)2)-2

J2 = 8.250 deg/day

PROBLEM 4.18

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.

The satellite 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 drag and 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,

= 2.62×10-12 kg/m3

H = 58.2 km

Equation (4.6),

V = SQRT[ GM / a ]

V = SQRT[ 3.986005×1014 / 6,778,140 ]

V = 7,669 m/s

Equations (4.53) through (4.55),

arev = (-2 × × CD × A × × a2) / m

arev = (-2 × × 2.67 × 8 × 2.62×10-12 × 6,778,1402) / 1,000

arev = -16.2 m

Prev = (-6 × 2 × CD × A × × a2) / (m × V)

Prev = (-6 × 2 × 2.67 × 8 × 2.62×10-12 × 6,778,1402) / (1,000 ×

7,669)

Prev = -0.0199 s

Vrev = ( × CD × A × × a × V) / m

Vrev = ( × 2.67 × 8 × 2.62×10-12 × 6,778,140 × 7,669) / 1,000

Vrev = 0.00914 m/s

Equation (4.56),

L ~ -H / arev

L ~ -(58.2 × 1,000) / -16.2

L ~ 3,600 revolutions

PROBLEM 4.19

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 at geosynchronous altitude.

SOLUTION,

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

From problem 4.3,

rB = 42,164,170 m

Equations (4.58) through (4.65),

atx = (rA + rB) / 2

atx = (6,578,140 + 42,164,170) / 2

atx = 24,371,155 m

ViA = SQRT[ GM / rA ]

ViA = SQRT[ 3.986005×1014 / 6,578,140 ]

ViA = 7,784 m/s

VfB = SQRT[ GM / rB ]

VfB = SQRT[ 3.986005×1014 / 42,164,170 ]

VfB = 3,075 m/s

VtxA = SQRT[ GM × (2 / rA - 1 / atx)]

VtxA = SQRT[ 3.986005×1014 × (2 / 6,578,140 - 1 / 24,371,155)]

VtxA = 10,239 m/s

VtxB = SQRT[ GM × (2 / rB - 1 / atx)]

VtxB = SQRT[ 3.986005×1014 × (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

PROBLEM 4.20

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 a

transfer ellipse with a semi-major axis of 30,000 km. Calculate the total

required velocity change 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 (4.66) through (4.68),

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 (4.59) through (4.63),

ViA = SQRT[ GM / rA ]

ViA = SQRT[ 3.986005×1014 / 6,578,140 ]

ViA = 7,784 m/s

VfB = SQRT[ GM / rB ]

VfB = SQRT[ 3.986005×1014 / 42,164,170 ]

VfB = 3,075 m/s

VtxA = SQRT[ GM × (2 / rA - 1 / atx)]

VtxA = SQRT[ 3.986005×1014 × (2 / 6,578,140 - 1 / 30,000,000)]

VtxA = 10,388 m/s

VtxB = SQRT[ GM × (2 / rB - 1 / atx)]

VtxB = SQRT[ 3.986005×1014 × (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 (4.69),

VB = SQRT[ VtxB2 + VfB

2 - 2 × VtxB × VfB × cos ]

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

VB = 2,260 m/s

Equation (4.65),

VT = VA + VB

VT = 2,604 + 2,260

VT = 4,864 m/s

Equations (4.70) and (4.71),

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 /

3.986005×1014 ]

TOF = 11,931 s = 3.314 hours

PROBLEM 4.21

Calculate the velocity change required to transfer a satellite from a

circular

600 km orbit with an inclination of 28 degrees to an orbit of equal size

with

an inclination 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[ 3.986005×1014 / 6,978,140 ]

Vi = 7,558 m/s

Equation (4.73),

V = 2 × Vi × sin( /2)

V = 2 × 7,558 × sin(8/2)

V = 1,054 m/s

PROBLEM 4.22

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

satellite to 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 4.19,

VfB = 3,075 m/s

VtxB = 1,597 m/s

VA = 2,455 m/s

Equation (4.74),

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 (4.65),

VT = VA + VB

VT = 2,455 + 1,826

VT = 4,281 m/s

PROBLEM 4.23

A 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 (4.75),

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

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 4.24

Calculate the latitude and longitude of the intersection nodes between the

initial and final orbits for the spacecraft in problem 4.23.

SOLUTION,

From problem 4.21,

a1 = 0.129410

a2 = 0.482963

a3 = 0.866025

b1 = 0.0920195

b2 = 0.521869

b3 = 0.848048

Equations (4.76) and (4.77),

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

PROBLEM 4.25

Calculate the escape velocity of a spacecraft launched from an Earth orbit

with

an altitude of 200 km.

SOLUTION,

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

Equation (4.78),

Vesc = SQRT[ 2 × GM / r ]

Vesc = SQRT[ 2 × 3.986005×1014 / 6,578,140 ]

Vesc = 11,009 m/s

PROBLEM 4.26

A space probe is approaching Mars on a hyperbolic flyby trajectory. When

at

a distance of 100,000 km, its velocity relative to Mars is 5,140.0 m/s and

its flight path angle is -85.300 degrees. Calculate the probe's

eccentricity,

semi-major axis, turning angle, angle , true anomaly, impact parameter,

periapsis radius, and parameter p.

SOLUTION,

From Basics Constants,

GM of Mars = 4.282831×1013 m3/s2

Given: r = 100,000 × 1,000 = 100,000,000 m

v = 5,140.0 m/s

= -85.300o

Equations (4.30) and (4.32),

e = SQRT[ (r × v2 / GM - 1)2 × cos2 + sin2 ]

e = SQRT[ (100,000,000 × 5,1402 / 4.282831×1013 - 1)2 × cos2(-85.3) +

sin2(-85.3) ]

e = 5.0715

a = 1 / ( 2 / r - v2 / GM )

a = 1 / ( 2 / 100,000,000 - 5,1402 / 4.282831×1013 )

a = -1,675,400 m

Equations (4.80) through (4.85),

sin( /2) = 1 / e

= 2 × arcsin( 1 / 5.0715 )

= 22.744o

cos = -1 / e

= arccos( -1 / 5.0715 )

= 101.37o

= arccos[ (a × (1 - e2) - r) / (e × r) ]

= arccos[ (-1,675,400 × (1 - 5.07152) - 100,000,000) / (5.0715 ×

100,000,000) ]

= -96.633o

b = -a / tan( /2)

b = 1,675.4 / tan(22.744/2)

b = 8,330.0 km

ro = a × (1 - e)

ro = -1,675.4 × (1 - 5.0715)

ro = 6,821.4 km

p = a × (1 - e2)

p = -1,675.4 × (1 - 5.07152)

p = 41,416 km

PROBLEM 4.27

The space probe in problem 4.26 has moved to a true anomaly of 75 degrees.

Calculate the radius vector, flight path angle, and velocity.

SOLUTION,

Given: a = -1,675,400 m

e = 5.0715

= 75o

Equations (4.43) through (4.45),

r = a × (1 - e2) / (1 + e × cos )

r = -1,675,400 × (1 - 5.07152) / (1 + 5.0715 × cos(75))

r = 17,909,000 m

= arctan[ e × sin / (1 + e × cos )]

= arctan[ 5.0715 × sin(75) / (1 + 5.0715 × cos(75))]

= 64.729o

v = SQRT[ GM × (2 / r - 1 / a)]

v = SQRT[ 4.282831×1013 × (2 / 17,909,000 - 1 / -1,675,400)]

v = 5,508.7 m/s

PROBLEM 4.28

A spacecraft is launched from Earth on a hyperbolic trajectory with a semi-

major

axis of -36,000 km and an eccentricity of 1.1823. How long does it take to

move

from a true anomaly of 15 degrees to a true anomaly of 120 degrees?

SOLUTION,

Given: a = -36,000 × 1,000 = -36,000,000 m

e = 1.1823

O = 15o

= 120o

Equation (4.87),

cosh F = (e + cos ) / (1 + e cos )

Fo = arccosh[(1.1823 + cos(15)) / (1 + 1.1823 × cos(15))]

Fo = 0.07614

F = arccosh[(1.1823 + cos(120)) / (1 + 1.1823 × cos(120))]

F = 1.10023

Equation (4.86),

t - tO = SQRT[(-a)3 / GM ] × [(e × sinh F - F) - (e × sinh Fo - Fo)]

t - tO = SQRT[(36,000,000)3 / 3.986005×1014 ] × [(1.1823 ×

sinh(1.10023) - 1.10023)

- (1.1823 × sinh(0.07614) - 0.07614)]

t - tO = 5,035 s = 1.399 hours

PROBLEM 4.29

A spacecraft launched from Earth has a burnout velocity of 11,500 m/s at an

altitude of 200 km. What is the hyperbolic excess velocity?

SOLUTION,

Given: Vbo = 11,500 m/s

From problem 4.25,

Vesc = 11,009 m/s

Equation (4.88),

V 2 = Vbo2 - Vesc

2

V = SQRT[ 11,5002 - 11,0092 ]

V = 3,325 m/s

PROBLEM 4.30

Calculate the radius of Earth's sphere of influence.

SOLUTION,

From Basics Constants,

Dsp = 149,597,870 km

MP = 5.9737×1024 kg

MS = 1.9891×1030 kg

Equation (4.89),

REarth = Dsp × (MP / MS)0.4

REarth = 149,597,870 × (5.9737×1024 / 1.9891×1030)0.4

REarth = 925,000 km

PROBLEM 5.1

Using a one-tangent burn, calculate the change in true anomaly and the

time-of-flight for a transfer from Earth to Mars. The radius vector of

Earth at

departure is 1.000 AU and that of Mars at arrival is 1.524 AU. The semi-

major

axis of the transfer orbit is 1.300 AU.

SOLUTION,

Given: rA = 1.000 AU

rB = 1.524 AU

atx = 1.300 AU × 149.597870×109 m/AU = 194.48×109 m

From Basics Constants,

GM of Sun = 1.327124×1020 m3/s2

Equations (4.66) and (4.67),

e = 1 - rA / atx

e = 1 - 1.0 / 1.3

e = 0.230769

= arccos[(atx × (1 - e2) / rB - 1) / e ]

= arccos[(1.3 × (1 - 0.2307692) / 1.524 - 1) / 0.230769 ]

= 146.488 degrees

Equations (4.70) and (4.71),

E = arctan[(1 - e2)1/2 × sin / (e + cos )]

E = arctan[(1 - 0.2307692)1/2 × sin(146.488) / (0.230769 +

cos(146.588))]

E = 2.41383 radians

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

TOF = (2.41383 - 0.230769 × sin(2.41383)) × SQRT[ (194.48×109)3 /

1.327124×1020 ]

TOF = 16,827,800 s = 194.77 days

PROBLEM 5.2

For the transfer orbit in problem 5.1, calculate the departure phase angle,

given

that the angular velocity of Mars is 0.5240 degrees/day.

SOLUTION,

Given: 2- 1 = 146.488o

t2-t1 = 194.77 days

t = 0.5240o/day

Equation (5.1),

= ( 2- 1) - t × (t2-t1)

= 146.488 - 0.5240 × 194.77

= 44.43o

PROBLEM 5.3

A flight to Mars is launched on 2020-7-20, 0:00 UT. The planned time of

flight

is 207 days. Earth's postion vector at departure is 0.473265X - 0.899215Y

AU.

Mars' postion vector at intercept is 0.066842X + 1.561256Y + 0.030948Z AU.

Calculate the parameter and semi-major axis of the transfer orbit.

SOLUTION,

Given: t = 207 days

r1 = 0.473265X - 0.899215Y AU

r2 = 0.066842X + 1.561256Y + 0.030948Z AU

GM = 1.327124×1020 m3/s2

= 1.327124×1020 / (149.597870×109)3 = 3.964016×10-14 AU3/s2

From vector magnitude,

r1 = SQRT[ 0.4732652 + (-0.899215)2 ]

r1 = 1.016153 AU

r2 = SQRT[ 0.0668422 + 1.561256

2 + 0.030948

2 ]

r2 = 1.562993 AU

From vector dot product,

= arccos[ (0.473265 × 0.066842 - 0.899215 × 1.561256) / (1.016153

× 1.562993) ]

= 149.770967o

Equations (5.9), (5.10) and (5.11),

k = r1 × r2 × (1 - cos )

k = 1.016153 × 1.562993 × (1 - cos(149.770967))

k = 2.960511 AU

= r1 + r2

= 1.016153 + 1.562993

= 2.579146 AU

m = r1 × r2 × (1 + cos )

m = 1.016153 × 1.562993 × (1 + cos(149.770967))

m = 0.215969 AU

Equations (5.18) and (5.19),

pi = k / ( + SQRT(2 × m))

pi = 2.960511 / (2.579146 + SQRT(2 × 0.215969))

pi = 0.914764 AU

pii = k / ( - SQRT(2 × m))

pii = 2.960511 / (2.579146 - SQRT(2 × 0.215969))

pii = 1.540388 AU

Since < , 0.914764 < p <

Equation (5.12),

Select trial value, p = 1.2 AU

a = m × k × p / [(2 × m - 2) × p2 + 2 × k × × p - k2]

a = 0.215969 × 2.960511 × 1.2

/ [(2 × 0.215969 - 2.5791462) × 1.22 + 2 × 2.960511 × 2.579146 ×

1.2 - 2.9605112]

a = 1.270478 AU

Equations (5.5), (5.6) and (5.7),

f = 1 - r2 / p × (1 - cos )

f = 1 - 1.562993 / 1.2 × (1 - cos(149.770967))

f = -1.427875

g = r1 × r2 × sin / SQRT[ GM × p ]

g = 1.016153 × 1.562993 × sin(149.770967) / SQRT[ 3.964016×10-14 × 1.2

]

g = 3,666,240

= SQRT[ GM / p ] × tan( /2) × [(1 - cos ) / p - 1/r1 - 1/r2 ]

= SQRT[ 3.964016×10-14 / 1.2 ] × tan(149.770967/2)

× [(1 - cos(149.770967)) / 1.2 - 1/1.016153 - 1/1.562993 ]

= -4.747601×10-8

Equation (5.13),

E = arccos[ 1 - r1 / a × (1 - f) ]

E = arccos[ 1 - 1.016153 / 1.270478 × (1 + 1.427875) ]

E = 2.798925 radians

Equation (5.16),

t = g + SQRT[ a3 / GM ] × ( E - sin E)

t = 3,666,240 + SQRT[ 1.2704783 / 3.964016×10-14 ] × (2.798925 -

sin(2.798925))

t = 21,380,951 s = 247.4647 days

Select new trial value of p and repeat above steps,

p = 1.300000 AU, a = 1.443005 AU, t = 178.9588 days

Equation (5.20),

pn+1 = pn + (t - tn) × (pn - pn-1) / (tn - tn-1)

pn+1 = 1.3 + (207 - 178.9588) × (1.3 - 1.2) / (178.9588 - 247.4647)

pn+1 = 1.259067 AU

Recalculate using new value of p,

p = 1.259067 AU, a = 1.336197 AU, t = 201.5624 days

Perform additional iterations,

p = 1.249221 AU, a = 1.318624 AU, t = 207.9408 days

p = 1.250673 AU, a = 1.321039 AU, t = 206.9733 days

p = 1.250633 AU, a = 1.320971 AU, t = 206.9999 days <-- close

enough

PROBLEM 5.4

For the Mars transfer orbit in Problem 5.3, calculate the departure and

intecept

velocity vectors.

SOLUTION,

Given: r1 = 0.473265X - 0.899215Y AU

r2 = 0.066842X + 1.561256Y + 0.030948Z AU

r1 = 1.016153 AU

r2 = 1.562993 AU

p = 1.250633 AU

a = 1.320971 AU

= 149.770967o

Equations (5.5), (5.6) and (5.7),

f = 1 - r2 / p × (1 - cos )

f = 1 - 1.562993 / 1.250633 × (1 - cos(149.770967))

f = -1.329580

g = r1 × r2 × sin / SQRT[ GM × p ]

g = 1.016153 × 1.562993 × sin(149.770967) / SQRT[ 3.964016×10-14 ×

1.250633 ]

g = 3,591,258

= SQRT[ GM / p ] × tan( /2) × [(1 - cos ) / p - 1/r1 - 1/r2 ]

= SQRT[ 3.964016×10-14 / 1.250633 ] × tan(149.770967/2)

× [(1 - cos(149.770967)) / 1.250633 - 1/1.016153 - 1/1.562993 ]

= -8.795872×10-8

= 1 - r1 / p × (1 - cos )

= 1 - 1.016153 / 1.250633 × (1 - cos(149.770967))

= -0.514536

Equation (5.3),

v1 = (r2 - f × r1) / g

v1 = [(0.066842 + 1.329580 × 0.473265) / 3,591,258] X

+ [(1.561256 + 1.329580 × -0.899215) / 3,591,258] Y

+ [(0.030948 + 1.329580 × 0) / 3,591,258] Z

v1 = 0.000000193828X + 0.000000101824Y + 0.00000000861759Z AU/s ×

149.597870×109

v1 = 28996.2X + 15232.7Y + 1289.2Z m/s

Equation (5.4),

v2 = × r1 + × v1

v2 = [-8.795872×10-8 × 0.473265 - 0.514536 × 0.000000193828] X

+ [-8.795872×10-8 × -0.899215 - 0.514536 × 0.000000101824] Y

+ [-8.795872×10-8 × 0 - 0.514536 × 0.00000000861759] Z

v2 = -0.000000141359X + 0.0000000267017Y - 0.00000000443406Z AU/s ×

149.597870×109

v2 = -21147.0X + 3994.5Y - 663.3Z m/s

PROBLEM 5.5

For the Mars transfer orbit in Problems 5.3 and 5.4, calculate the orbital

elements.

SOLUTION,

Problem can be solved using either r1 & v1 or r2 & v2 – we will use r1 &

v1.

Given: r1 = (0.473265X - 0.899215Y AU) × 149.597870×109 m/AU

= 7.079944×1010X - 1.345206×1011Y m

r1 = 1.016153 × 149.597870×109 = 1.520144×1011 m

GM = 1.327124×1020 m3/s2

From problem 5.4,

v1 = 28996.2X + 15232.7Y + 1289.2Z m/s

Also,

v = SQRT[ 28996.22 + 15232.72 + 1289.22 ] = 32,779.2 m/s

Equations (5.21) and (5.22),

h = (rY vZ - rZ vY)X + (rZ vX - rX vZ)Y + (rX vY - rY vX)Z

h = (-1.345206×1011 × 1289.2 - 0 × 15232.7)X + (0 × 28996.2 -

7.079944×1010 × 1289.2)Y

+ (7.079944×1010 × 15232.7 + 1.345206×1011 × 28996.2)Z

h = -1.73424×1014X - 9.12746×1013Y + 4.97905×1015Z

n = -hY X + hX Y

n = 9.12746×1013X - 1.73424×1014Y

Also,

h = SQRT[ (-1.73424×1014)2 + (-9.12746×1013)2 + (4.97905×1015)2 ] =

4.98291×1015

n = SQRT[ (9.12746×1013)2 + (1.73424×1014)2 ] = 1.95977×1014

Equation (5.23),

e = [(v2 - GM / r) × r - (r • v) × v ] / GM

v2 - GM / r = 32779.22 - 1.327124×1020 / 1.520144×1011 = 2.01451×108

r • v = 7.079944×1010 × 28996.2 - 1.345206×1011 × 15232.7 + 0 x

1289.2 = 3.80278×1012

e = [2.01451×108 × (7.079944×1010X - 1.345206×1011Y)

- 3.80278×1012 × (28996.2X + 15232.7Y + 1289.2Z) ] / 1.327124×1020

e = 0.106639X - 0.204632Y - 0.000037Z

Equations (5.24) and (5.25),

a = 1 / ( 2 / r - v2 / GM )

a = 1 / ( 2 / 1.520144×1011 - 32779.22 / 1.327124×1020 )

a = 1.97614×1011 m

e = SQRT[ 0.1066392 + (-0.204632)2 + (-0.000037)2 ]

e = 0.230751

Equations (5.26) though (5.30),

cos i = hZ / h

cos i = 4.97905×1015 / 4.98291×1015

i = 2.255o

cos = nX / n

cos = 9.12746×1013 / 1.95977×1014

= 297.76o

cos = n • e / (n × e)

cos = (9.12746×1013 × 0.106639 - 1.73424×1014 × (-0.204632) + 0 × (-

0.000037))

/ (1.95977×1014 × 0.230751)

= 359.77o

cos o = e • r / (e × r)

cos o = (0.106639 × 7.079944×1010 - 0.204632 × (-1.345206×1011) -

0.000037 × 0)

/ (0.230751 × 1.520144×1011)

o = 0.226o

cos uo = n • r / (n × r)

uo = 0 (launch point = ascending node)

Equations (5.31) and (5.32),

= +

= 297.76 + 359.77

= 297.53o

o = + + o

o = 297.76 + 359.77 + 0.23

o = 297.76o

PROBLEM 5.6

For the spacecraft in Problems 5.3 and 5.4, calculate the hyperbolic excess

velocity at departure, the injection V, and the zenith angle of the

departure

asymptote. Injection occurs from an 200 km parking orbit. Earth's

velocity

vector at departure is 25876.6X + 13759.5Y m/s.

SOLUTION,

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

r = 0.473265X - 0.899215Y AU

VP = 25876.6X + 13759.5Y m/s

From problem 5.4,

VS = 28996.2X + 15232.7Y + 1289.2Z m/s

Equation (5.33),

VS/P = (VSX - VPX)X + (VSY - VPY)Y + (VSZ - VPZ)Z

VS/P = (28996.2 - 25876.6)X + (15232.7 - 13759.5)Y + (1289.2 - 0)Z

VS/P = 3119.6X + 1473.2Y + 1289.2Z m/s

Equation (5.34),

VS/P = SQRT[ VS/PX2 + VS/PY

2 + VS/PZ2 ]

VS/P = SQRT[ 3119.62 + 1473.22 + 1289.22 ]

VS/P = 3,683.0 m/s

V = VS/P = 3,683.0 m/s

Equations (5.35) and (5.36),

Vo = SQRT[ V2 + 2 × GM / ro ]

Vo = SQRT[ 3,683.02 + 2 × 3.986005×1014 / 6,578,140 ]

Vo = 11,608.4 m/s

V = Vo - SQRT[ GM / ro ]

V = 11,608.4 - SQRT[ 3.986005×1014 / 6,578,140 ]

V = 3,824.1 m/s

Equation (5.37),

r = SQRT[ 0.4732652 + (-0.899215)2 + 02 ]

r = 1.01615 AU

= arccos[(rX × vX + rY × vY + rZ × vZ) / (r × v)]

= arccos[ 0.473265 × 3119.6 - 0.899215 × 1473.2 + 0 × 1289.2) /

(1.01615 × 3683.0)]

= 87.677o

PROBLEM 5.7

For the spacecraft in Problems 5.3 and 5.4, given a miss distance of

+18,500 km

at arrival, calculate the hyperbolic excess velocity, impact parameter, and

semi-major axis and eccentricity of the hyperbolic approach trajectory.

Mars'

velocity vector at intercept is -23307.8X + 3112.0Y + 41.8Z m/s.

SOLUTION,

Given: d = 18,500 km / 149.597870×106 = 0.000123664 AU

r = 0.066842X + 1.561256Y + 0.030948Z AU

VP = -23307.8X + 3112.0Y + 41.8Z m/s

From Basics Constants,

GM of Mars = 4.282831×1013 m3/s2

From problem 5.4,

VS = -21147.0X + 3994.5Y - 663.3Z m/s

Equation (5.33),

VS/P = (VSX - VPX)X + (VSY - VPY)Y + (VSZ - VPZ)Z

VS/P = (-21147.0 + 23307.8)X + (3994.5 - 3112.0)Y + (-663.3 - 41.8)Z

VS/P = 2160.8X + 882.5Y - 705.1Z m/s

Equation (5.34),

VS/P = SQRT[ VS/PX2 + VS/PY

2 + VS/PZ2 ]

VS/P = SQRT[ 2160.82 + 882.52 + (-705.1)2 ]

VS/P = 2,438.2 m/s

V = VS/P = 2,438.2 m/s

Equations (5.38.A) and (5.38.B),

dx = -d × ry / SQRT[ rx2 + ry

2 ]

dx = -0.000123664 × 1.561256 / SQRT[ 0.0668422 + 1.5612562 ]

dx = -0.000123551 AU

dy = d × rx / SQRT[ rx2 + ry

2 ]

dy = 0.000123664 × 0.066842 / SQRT[ 0.0668422 + 1.5612562 ]

dy = 0.0000052896 AU

Equation (5.39),

= arccos[(dx × vx + dy × vy) / (d × v)]

= arccos[(-0.000123551 × 2160.8 + 0.0000052896 × 882.5) /

(0.000123664 × 2,438.2)]

= 150.451o

Equations (5.40) through (5.42),

b = d × sin

b = 18,500 × sin(150.451)

b = 9,123.6 km

a = -GM / V 2

a = -4.282831×1013 / 2,438.22

a = -7.2043×106 m = -7,204.3 km

e = SQRT[ 1 + b2 / a2 ]

e = SQRT[ 1 + 9,123.62 / -7,204.32 ]

e = 1.6136

PROBLEM 5.8

As a spacecraft approaches Jupiter, it has a velocity of 9,470 m/s, a

flight

path angle of 39.2 degrees, and a targeted miss distance of -2,500,000 km.

At

intercept, Jupiter's velocity is 12,740 m/s with a flight path angle of

2.40

degrees. Calculate the spacecraft's velocity and flight path angle

following

its swing-by of Jupiter.

Given: VP = 12,740 m/s

P = 2.40o

VSi = 9,470 m/s

Si = 39.2o

d = -2,500,000 km

From Basics Constants,

GM of Jupiter = 1.26686×1017 m

3/s

2

Equations (5.44) and (5.45),

VP = (VP × cos P)X + (VP × sin P)Y

VP = (12740 × cos(2.40))X + (12740 × sin(2.40))Y

VP = 12729X + 533Y m/s

VSi = (VSi × cos Si)X + (VSi × sin Si)Y

VSi = (9470 × cos(39.2))X + (9470 × sin(39.2))Y

VSi = 7339X + 5985Y m/s

Equations (5.46) and (5.47),

VS/Pi = ((VSi)X - VPX)X + ((VSi)Y - VPY)Y

VS/Pi = (7339 - 12729)X + (5985 - 533)Y

VS/Pi = -5390X + 5452Y m/s

VS/P = SQRT[ (VS/Pi)X2 + (VS/Pi)Y

2 ]

VS/P = SQRT[ (-5390)2 + 54522 ]

VS/P = 7,667 m/s

V = VS/P = 7,667 m/s

Equation (5.48),

i = arctan[ (VS/Pi)Y / (VS/Pi)X ]

i = arctan[ 5452 / -5390 ]

i = 134.67o

Equations (5.40) through (5.42),

b = d × sin

b = -2,500,000 × sin(134.67)

b = -1,777,900 km

a = -GM / V 2

a = -1.26686×1017 / 76672

a = -2.1552×109 m = -2,155,200 km

e = SQRT[ 1 + b2 / a2 ]

e = SQRT[ 1 + (-1,777,900)2 / (-2,155,200)2 ]

e = 1.2963

Equation (4.80),

sin( /2) = 1 / e

= 2 × arcsin( 1 / 1.2963 )

= 100.96o

Equation (5.49),

f = i -

f = 134.67 - 100.96

f = 33.71o

Equation (5.50),

VS/Pf = (VS/P × cos f)X + (VS/P × sin f)Y

VS/Pf = (7667 × cos(33.71))X + (7667 × sin(33.71))Y

VS/Pf = 6378X + 4255Y m/s

Equations (5.51) and (5.52),

VSf = ((VS/Pf)X + VPX)X + ((VS/Pf)Y + VPY)Y

VSf = (6378 + 12729)X + (4255 + 533)Y

VSf = 19107X + 4788Y m/s

VSf = SQRT[ (VSf)X2 + (VSf)Y

2 ]

VSf = SQRT[ 191072 + 47882 ]

VSf = 19,698 m/s

Equation (5.53),

Sf = arctan[ (VSf)Y / (VSf)X ]

Sf = arctan[ 4788 / 19107 ]

Sf = 14.07o

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