Baixe Mecâni ca Orbital e outras Notas de estudo em PDF para Engenharia Aeronáutica, somente na Docsity!
ORBITAL MECHANICS
- Conic Sections
- Orbital Elements
- Types of Orbits
- (^) Newton's Laws of Motion and Universal Gravitation
- Uniform Circular Motion
- Motions of Planets and Satellites
- Launch of a Space Vehicle
- Position in an Elliptical Orbit
- Orbit Perturbations
- Orbit Maneuvers
- Escape Velocity
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 the figure to the right, 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. In this section we will discuss bounded conic orbits, i.e. circles and ellipses.
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 to right). 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 below). 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.
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- m 2 /kg 2 (3.4389x10 -8^ lb-ft^2 /slug^2 ).
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 /r^2. 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 /r^2 )/m , or
At the surface of the Earth this acceleration has the valve 9.80665 m/s 2 (32.174 ft/s 2 ).
Many of the upcoming computations will be somewhat simplified if we express the product GM as a constant, which for Earth has the value 3.986005x10 14 m^3 /s^2 (1.408x10^16 ft 3 /s 2 ). 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 = v 2 /r. From Newton's law of universal gravitation we know that g = GM /r 2. Therefore, by setting these equations equal to one another we find that, for a circular orbit,
Click here for example problem #3. (use your browser's "back" function to return)
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:
- All planets move in elliptical orbits with the sun at one focus.
- A line joining any planet to the sun sweeps out equal areas in equal times.
- 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 2 r must equal M 2 R. 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
Click here for example problem #3. Click here for example problem #3. The eccentricity e of an orbit is given by
Click here for example problem #3.
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 #3.
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 to right). If we let r1, v1 , and be the initial (launch) values of r, v , and , then we may consider these as given quantities. If we let point P represent the perigee, then equation (3.13) becomes
Substituting equation (3.23) into (3.15), we can obtain an equation for the perigee radius Rp.
Multiplying through by -Rp 2 /(r1^2 v1^2 ) and rearranging, we get
Note that this is a simple quadratic equation in the ratio (Rp/r1) and that 2GM /(r1 x v1 2 ) 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 #3.
Equation (3.26) 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 #3.
To pin down a satellite's orbit in space, we need to know the angle from the periapsis point to the launch point. This angle is given by
Click here for example problem #3.
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 the figure to the right. When flight-path angle is used, equations (3.26) through (3.28) are rewritten as follows:
Position in an Elliptical Orbit
Johannes Kepler was able to solve the problem of relating position in an orbit to the elapsed time, t-t (^) O , 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 M (^) O in the mean anomaly at time t (^) O 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 v 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 e 3 ):
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 #3. Click here for example problem #3.
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 #3.
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
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 F (^) D on a body acts in the opposite direction of the velocity vector and is given by the equation
where C (^) D 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/(C (^) DA) , 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 (3.47), taking into account the changes in atmospheric density with both altitude and solar activity.
Click here for example problem #3.
Perturbations from Solar Radiation
Solar radiation pressure causes periodic variations in all of the orbital elements. The magnitude of the acceleration in m/s 2 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. The figure to the right 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, r (^) A and
rB , we can calculate the total velocity change using the following equations:
Note that equations (3.53) and (3.54) are the same as equation (3.6), and equations (3.55) and (3.56) are variations of equations (3.16) and (3.17) respectively.
Click here for example problem #3.
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. The figure to the right 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 (3.53) through (3.57) and (3.59) above, and the following equations:
Click here for example problem #3.
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
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.
Escape Velocity
We know that if we throw a ball up from the surface of the Earth, it will rise for a while and then return. If we give it a larger initial velocity, it will rise higher and then return. There is a velocity, called the escape velocity, Vesc , such that if the ball is launched with an initial velocity greater than Vesc , it will rise and never return. We must give the particle enough kinetic energy to overcome all of the negative gravitational potential energy. Thus, if m is the mass of the ball, M is the mass of the Earth, and R is the radius of the Earth, the potential energy is -GmM /R. The kinetic energy of the ball, when it is launched, is mv 2 /2. We thus have
which is independent of the mass of the ball. For a spacecraft launched to escape velocity from a parking orbit, R is the radius of the orbit. Click here for example problem #3.
http://www.braeunig.us/space/
BASIC CONSTANTS
Mathematical Constants
e 2.
Physical Constants Speed of light (c) 299,792,458 m/s
Constant of gravitation (G) 6.67259x10-11^ N-m^2 /kg^2
Acceleration of gravity (g) 9.80665 (^) m/s^2
Universal gas constant (R) 8,314.510 N-m/kg-K
Standard atmosphere, sea level 101,325 Pa
Astronomical Constants Astronomical unit (AU) 149,597,870 km
Light year (ly) 9.460530x10^12 km
Parsec (pc) 3.261633 ly
Sidereal year 365.256366 days
Mass of Sun 1.9891x10^30 kg
Radius of Sun 696,000 km
Mass of Earth 5.9737x10^24 kg
Equatorial radius of Earth 6,378.140 km
Earth oblateness 1/298.
Obliquity of the ecliptic, epoch 2000 23.4392911 degrees
Mean lunar distance 384,403 km
Radius of Moon 1,738 km
Mass of Moon (^) 7.348x10^22 kg
Luminosity of Sun 3.827x10^26 W
Solar constant, at 1 AU 1,358 W/m 2
Solar maxima 1990 + 11n (date)
Spaceflight Constants
GM (Sun) 1.32712438x10^20 m 3 /s 2
GM (Earth) 3.986005x10^14 m 3 /s 2
GM (Moon) (^) 4.902794x10^12 m 3 /s 2
J 2 (Earth) 0.
J 2 (Moon) 0.
Planet Data
Name Equatorial Radius (km)
Mass (10 24 kg)
Surface Gravity (m/s 2 )
GM
(10 15 m^3 /s^3 )
Semimajor Axis (10^6 km)
Eccentricity Orbit Inclination (degrees)
Sidereal Period (days)
Sidereal Rotation Period (hours) Mercury 2,439.7 0.3302 3.70 0.02203 57.91 0.2056 7.00 87.960 1,407. Venus 6,051.8 4.8685 8.87 0.3249 108.21 0.0067 3.39 224.701 -5,832. Earth 6,378.1 5.9742 9.80 0.3986 149.60 0.0167 0.000 365.256 23.
Mars 3,397.0 0.64185 3.71 0.04283 227.92 0.0935 1.850 686.980 24. Jupiter 71,492 1,898.6 24.79 126.686 778.57 0.0489 1.304 4,332.59 9. Saturn 60,268 568.46 10.44 37.931 1,433.53 0.0565 2.485 10,759.2 10. Uranus 25,559 86.832 8.87 5.794 2,872.46 0.0457 0.772 30,685.4 -17.
Neptune 24,764 102.43 11.15 6.835 4,495.06 0.0113 1.769 60,189 16. Pluto 1,195 0.0125 0.58 0.00083 5,869.66 0.2444 17.16 90,465 -153.
Pe = 5 kPa = 5,000 N/m 2 Pa = 0
Equation (1.6),
F = q x Ve + (Pe - Pa) x Ae F = 30 x 3,100 + (5,000 - 0) x 0. F = 96,500 N
PROBLEM 1.
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 x LN[ M / (M - qt) ] V = 3,100 x LN[ 30,000 / (30,000 - (30 x 60)) ] V = 192 m/s
PROBLEM 1.
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 x e^(V / C) Mo = 75,000 x e^(700 / 3,100) Mo = 94,000 kg
Propellant mass,
Mp = Mo - Mf Mp = 94,000 - 75, Mp = 19,000 kg
PROBLEM 1.
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 x [ 1 - 1 / e^(V / C) ] t = 5,000 / 10 x [ 1 - 1 / e^(4,210 / 3,000) ] t = 377 s
PROBLEM 1.
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. Pc = 50 atm Pe = Pa = 1 atm
From LOX/Kerosene Charts we estimate,
Tc = 3,470 K M = 21.
Pt = 2.839 MPa = 2.839x10 6 N/m^2
Equation (1.27),
Tt = Tc / (1 + (k - 1) / 2) Tt = 3,470 / (1 + (1.221 - 1) / 2) Tt = 3,125 K
Equation (1.25),
At = (q / Pt) x SQRT[ (R' x Tt) / (M x k) ] At = (500 / 2.839x10^6 ) x SQRT[ (8,314.51 x 3,125) / (21.40 x 1.221) ] At = 0.1756 m^2
PROBLEM 1.
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 m^2 k = 1.
From Atmosphere Properties,
Pa = 0.0795 MPa
Equation (1.28),
Nm^2 = (2 / (k - 1)) x [(Pc / Pa) (k-1)/k^ - 1] Nm^2 = (2 / (1.221 - 1)) x [(5.066 / 0.0795) (1.221-1)/1.221^ - 1] Nm^2 = 10. Nm = (10.15)1/2^ = 3.
Equation (1.29),
Ae = (At / Nm) x (1 + (k - 1) / 2 x Nm 2 )/((k + 1) / 2)/(2(k-1)) Ae = (0.1756 / 3.185) x [(1 + (1.221 - 1) / 2 x 10.15)/((1.221 + 1) / 2)] (1.221+1)/ (2(1.221-1))
Ae = 1.426 m^2
Section Ratio,
Ae / At = 1.426 / 0.1756 = 8.
PROBLEM 1.
A solid rocket motor burns along the face of a central cylindrical channel 10 meters long and 1 meter in diameter. The propellant has a burn rate coefficient of 5.5, a pressure exponent of 0.4, and a density of 1.77 g/ml. Calculate the burn rate and the product generation rate when the chamber pressure is 5.0 MPa.
SOLUTION,
Given: a = 5. n = 0. Pc = 5.0 MPa p = 1.77 g/ml Ab = x 1 x 10 = 31.416 m 3
Equation (1.30),
r = a x Pcn r = 5.5 x 5.00.4^ = 10.47 mm/s
Equation (1.31),
q = p x Ab x r q = 1.77 x 31.416 x 10.47 = 582 kg/s
PROBLEM 1.
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 Mf 1 = 9,000 + 30,000 + 3,000 + 3,000 = 45,000 kg Isp 1 = 260 s Mo 2 = 30,000 + 3,000 + 3,000 = 36,000 kg Mf2 = 3,000 + 3,000 = 6,000 kg Isp2 = 320 s
Equation (1.24),
C1 = Isp1 g C1 = 260 x 9.80665 = 2,550 m/s
