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Introduction to Astronautics Sissejuhatus kosmonautikasse Vladislav Pustõnski 2009-2012 Tallinn University of Technology

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2 Basics of Orbital Mechanics Central gravitational attraction – gravitational constant, 6.67·10 -11 m 3 /(kg ·s 2 ) M – mass of the central body m s – mass of the satellite r – distance from the center of the body to the satellite – gravitational parameter = M. Acceleration of the satellite The universal force of gravity is the main force influencing a satellite in a stable orbit. In most of cases, satellite moves in the vicinity of a single body (a planet or the Sun, further central body), so other celestial bodies (and fields) affect its motion (relative to the central body) in much smaller extent than the gravity of the central body. Thus, it makes sense to examine first the motion due to gravitational attraction of the central body and regard other forces as perturbations of higher orders of magnitude. Since most celestial bodies are nearly spherical, we may use the Newtons law of gravity and consider non-sphericity as a perturbation. Lets write main equations to define physical quantities (considering the satellite as pointlike mass). Newtons law of gravity Earth 4.0·10 14 m 3 /s 2, Sun 1.3·10 20 m 3 /s 2.

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3 It is obvious that the acceleration is the same if r is the same. In particular, on the surface of the central body (further we will consider a planet, although in particular cases it may be the Sun, a natural satellite or even an asteroid) R 0 – radius of the planet, g 0 – surface gravity Gravitational acceleration may be represented as

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4 To place a satellite to a circular orbit with a radius r, one should provide its centripetal acceleration to be equal to the gravitational acceleration: Thus the circular velocity may be found as For the zero-height orbit (r=R 0 ) For the Earth, For the period we get So P 2 ~ r 3. According to the Keplers Third Law, this is true not only for circular orbits, but also for elliptical orbits. It is seen that for larger orbital radii, circular velocities are smaller and periods are larger. For instance, for the Moon (orbital radius 380 000 km) V circ 1 km/s, P 27 d. However, it does not mean that it is simpler to put a satellite into a higher orbit, as we will see later. Circular velocity

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5 Escape velocity Imagine a satellite (regarded as point mass m s ) at an initial distance r from the center of the planet. Let it has a velocity V. At this distance the total energy of the satellite is The first term is the kinetic energy and the second term is the potential energy. Lets found the minimum velocity V esc that the satellite requires to move apart from the planet (to an infinite distance). Consider the fact that in infinity the potential energy tends to zero; the kinetic energy should also become zero since we are seeking for a minimum V for a fixed r, thus providing a minimum total energy. Thus the total energy in the infinity is zero, and from the energy conservation law The velocity V esc is known as escape velocity. For the surface of the Earth, If a satellite has got the escape velocity, it will apart from the planet to infinity, irrespectively of the direction of its velocity vector. Its trajectory relative to the planet will be parabolic, with the center of the planet being in the focus of the parabola. However, it would be wrong to think that if a real satellite of a planet is provided with the escape velocity, that will be sufficient to go to infinity. The satellite will still remain under influence of other celestial bodies (mainly of the Sun), so it will continue in a heliocentric orbit. Later we will examine conditions which should be met for missions to distant objects.

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6 Elliptical orbits Imagine a satellite (regarded as point mass m s ) situated at an initial distance r per from the center of the planet and having a velocity V directed tangentially. If V=V circ, the trajectory will be circular; if V=V esc, the satellite will move to infinity, so that r per will be its minimal distance to the center. We will proceed from this distance, since it is often used as a starting orbit for orbital maneuvers (by the reasons that will be examined later). From mechanics, for V circ < V < V esc the satellite will never go away to infinity, its trajectory will be an ellipse with the planet in one of its focii (Keplers First Law). Equation of ellipse in polar coordinates has a form r – distance from the center – true anomaly p – parameter e – eccentricity, e<1 r has a minimum at for = 0, r has a maximum for = 180 0. The respective values are called apsides: periapsis r per and apoapsis r ap. For some celestial objects they have proper names: Earth: perigee, apogee Moon: periselene, aposelene Sun: perihelion, apohelion Star: periastron, apoastron Pegiapsis and apoapsis lie at the line which is known as line of apsides. r per + r ap is the major axis of the ellipse; its half a is the semi-major axis of the orbit. In the case of parabolic and hyperbolic orbits, only periapsis is present.

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7 From the energy conservation law and from the angular momentum conservation law Joint solution of these two equations enables one to compute the apoapsis r ap and the apoapsis velocity V ap from the known periapsis r per and the periapsis velocity V per. The semi-major axis is deduced as From energy considerations one may find velocity of the satellite V at an arbitrary distance r from the center. The energy conservation lay yields vis viva equation It follows from the angular momentum conservation law that the orbital velocity decreases as the satellite moves along the ascending branch of the trajectory from the periapasis to the apoapsis, achieving minimum value in the apoapsis, and it rises again on the descending branch, achieving maximum value in the periapsis. The same is true for parabolic and hyperbolic orbits: in the periapsis the satellite has the highest velocity. If V per < V circ, periapsis and apoapsis change places: the distance r per corresponds to apoapsis and vice versa. Eccentricity is the measure of oblateness of orbit. If e = 0, the orbit is circular, if e = 1, the orbit is parabolic. Using conservation laws, one may find the relation between velocities in apsides and distances to the center in apsides.

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8 To find the position of the satellite on an elliptical orbit (i.e. its true anomaly and the distance r) as a function of time t, the following algorithm may be used. First, compute the mean anomaly M. Than, solve Keplers equation to find the eccentric anomaly E. The true anomaly is deduced from the eccentric anomaly and the distance is deduced from the ellipse equation. Mean anomaly equation Keplers Equation Relation between the mean anomaly and the eccentric anomaly. According to the Keplers Third Law, a satellite on an elliptical orbit with the semimajor axis a has the same period as the satellite on a circular orbit with the radius r = a, thus The total energy of the satellite on an elliptical orbit is negative.

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9 Hyperbolic velocity If the velocity V of a satellite is higher than the escape velocity V esc for the corresponding distance r from the center of the planet, i.e. V > V esc the total energy of the satellite is positive. Thus, the satellite is able (and will) move away to infinity. However, even in infinity, where its potential energy will reach its highest value E inf = 0, it still will retain a positive (kinetic) energy surplus, thus having a positive velocity V exc. This excess is known as hyperbolic excess velocity. The trajectory of such satellite will be hyperbolic, so the corresponding velocity is called hyperbolic excess velocity. The distance from the center of the planet is found from the following equation in polar coordinates r – distance from the center – true anomaly p – parameter e – eccentricity, e>1 The center of the planet coincides with the inner focus of the hyperbola. The range of the true anomaly is The semi-major axis of the hyperbolic orbit is found as The hyperbolic excess velocity may be found from a as Velocity in the arbitrary point of the trajectory is To find the position of the satellite on an hyperbolic orbit, the same procedure is used as for elliptical orbits, but the formulae for hyperbolic anomaly and true anomaly have the following form

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10 Types of orbits in a central gravitational field and real orbits e = 0, V = V circ : circular orbit e < 1, V circ < V

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11 Energy considerations & characteristic velocity As we have seen earlier, a satellite should be provided a certain velocity in order to be placed on the desired orbit. For instance, at the orbit with the height of h = 300 km above the Earth surface this velocity should be So the total specific energy (energy per unit mass) is the sum of the specific kinetic energy and the specific potential energy Accelerating a satellite to that speed means providing it with corresponding kinetic energy. However, for a real satellite, it is not enough for the rocket to provide it only with that of energy. The required amount of energy depends on the current position and velocity of the satellite. If the rocket with the satellite in our example are resting on the surface of the Earth, the satellite should not only be accelerated to V circ : it should also be lifted to the height of 300 km. Thus the corresponding potential energy should be provided as well, and it is

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12 Calculation gives This is the minimal energy that should be provided to a unit mass to be transferred from the surface of the Earth to the selected circular orbit. The rocket provides a satellite with energy by accelerating it. As we will see later, one of the most important characteristics of a rocket is the maximal velocity to which it is able to accelerate a payload of a given mass in void space (so called ideal velocity). To compare the rocket ability to accelerate the satellite and the energy that is required to place the satellite in the chosen orbit, it is often convenient to express the energy (required for the corresponding orbital transfer) in terms of velocity. To deliver the specific energy E spec to a unit mass, we should accelerate it to the velocity This velocity is called the mission characteristic velocity. For the example analyzed We see that it is about 350 m/s higher than the circular velocity. The difference arises from the need to lift the satellite to the altitude of 300 km. However, for any real rocket ideal velocity of 8080 m/s would not be sufficient to reach the orbit. Gravity drag and aerodynamic drag add to this value up to 1500 m/s or even more (steering losses should also be added and the linear velocity of the launch site due to the Earths rotation should be subtracted), so launch vehicles should have ideal velocity of about 9 500 m/s to place a satellite to a low Earth orbit.

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13 In the general case, the mission consists of several operations (for instance, orbital transfers, landings, launches etc.). Characteristic velocity for such complex mission should be calculated taking into account all operations. It should be emphasized that such calculations should be made from energy considerations, it is not always enough to simply add characteristic velocities of individual phases! That means that the characteristic velocity of the whole mission is not necessarily equal to the sum of characteristic velocities of its phases. In our example, to lift a unit of mass to the altitude of 300 km energy of 2.8 ·10 6 J/kg is required. So, the characteristic velocity of this operation is 2370 m/s: that is the velocity to which a rocket should accelerate vertically the satellite so that it could reach the altitude of 300 km. Thus, if we divide the mission to two stages: 1) vertical accent to 300 km; and 2) horizontal acceleration to the orbital velocity of V circ = 7730 m/s; - the characteristic velocity of such mission will be V char = 7730 + 2370 = 10 100 m/s, i.e. by 2000 m/s larger than the minimal characteristic velocity. Note that this is only the minimum value of the mission characteristic velocity (and it also does not take into account the Earths rotation). As the matter of fact, a certain space mission may be realized in a number of ways, a specific value of characteristic velocity corresponds to each realization. One may make sure that for a higher orbit the characteristic velocity is always larger than for a lower orbit, although the circular velocity decreases with height. The corresponding formula is

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14 It should be always kept in mind that from the point of view of space flight, characteristic velocity should be regarded as exponential parameter (due to the ideal rocket equation) defining availability of a mission. Increment of characteristic velocity by ~2300 m/s means the need to increase the mass of the vehicle twice. This number is approximately valid for non-cryogenic propellants that are usually used onboard of spacecraft, and also for LOX/hydrocarbon propellants used by launch vehicles. For LH2/LOX propellant doubling of the vehicle mass increases the ideal velocity of the vehicle by ~3000 m/s, but this cryogenic propellant cannot be used in long missions since hydrogen boils out. Today it is used only in launch vehicles, in the nearest future it may become available also in short (up to about two weeks) deep space missions, for instance for Moon landers. The above-mentioned factors mean that a Moon soft-lander should be about two times heavier than a Moon fly-by probe (since characteristic velocity of about ~2500 m/s is required to soft- land the Moon from the fly-by trajectory). From the same considerations we may conclude that since the total characteristic velocity of launch to Low Earth Orbit (LEO) is about ~9500 m/s (including losses), the weight of payload would make up ~ (1/2 3.2 – 1/2 4 ) 6 % – 10 % of the launch mass. In practice, these are correct numbers for total mass delivered to LEO, but useful payload is only half of this mass, i.e. ~ 3 % – 5 %; the second half is the expendable upper stage of the launch vehicle. If we consider Mars with its escape velocity of ~5000 m/s, we should conclude that the mass of the object sent to an escape trajectory from the Martian surface would be ~ ¼ of its launch mass (on the Martian surface).

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15 At the present level of technology, high ideal velocities of rockets are quite expensive, costs dependence on mass is approximately exponential. So, we should organize a real mission in a way providing as small characteristic velocity as possible. Although the theoretical minimum value is unachievable due to many reasons (gravity and aerodynamic drag, steering losses, finite time of acceleration etc.), smart mission planning is indispensable. For instance, we should launch to the East and not to the West when possible in order to take advantage of the Earth rotation. If the launch site is on the equator and we launch directly to the East, that may add up to 465 m/s to the rocket velocity, thus decreasing the required mission characteristic velocity by the same value (however, other considerations may sometimes oblige to launch to the West; for instant, Israel should do so to avoid debris to fall onto the neighboring countries). Launches to polar orbits also do not take advantage of the additive due to rotation of the Earth. Let us compute characteristic velocities of some important missions, neglecting the Earths rotation.

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16 Example calculation of characteristic velocities If we wish to send a spacecraft out of the Solar system, we should 1) provide it with enough energy to leave the Earth gravity field; 2) provide it with the escape velocity relative to the Sun. The first specific energy corresponds to the Earth escape velocity of V esc 11 100 m/s, i.e. V esc 2 /2 6.2 ·10 7 J/kg. After leaving the Earth, the satellite continues to orbit the Sun at the distance of R E 1.5 ·10 11 m, that is the radius of the Earth orbit. The Sun-relative orbital velocity at this distance is The escape velocity is The satellite, already heaving the orbital velocity, should be provided with the velocity increment of V Sun,esc - V sun,circ 12 200 m/s to leave the gravity field of the Sun. This corresponds to the specific energy of 12 200 2 /2 7.4 ·10 7 J/kg. Total specific energy is (6.2 ·10 7 + 7.4 ·10 7 ) 13.6 ·10 7 J/kg, which can be expressed in terms of specific velocity as (2·1.36 ·10 8 ) 1/2 16 500 m/s. This is a very high value, but it is possible to decrease it using gravity assist of the planets (mostly the Jupiter). Note that this velocity should be given to the satellite at once, by one impulse near the Earth.

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17 If we wish to soft-land on the Moon, we should 1) provide it with enough energy to reach the Moon orbit; 2) to cancel the velocity near the Moon. In our estimations lets ignore the gravitational influence of the Sun (since its distance from the system Earth-Moon is much larger than the distance between these celestial bodies). Lets neglect also the influence of the lunar gravity field near the Earth, as well as the influence of the Earth gravity field near the Moon. Lets consider also that the satellite is launched from the 300-km orbit. However, we can organize the mission of the rocket with ideal velocity of V ideal = 16 500 m/s in other way. First, lets accelerate the satellite up to the Earth escape velocity of V esc 11 100 m/s, and when the object moves far away from the Earth (and slows down nearly to zero speed), use the rest of the fuel to accelerate it further more. This second impulse will give the satellite only 16 500 – 11 100 = 5400 m/s. It is clearly not enough to leave the Solar system, and the mission will not be completed. As we see, applying the same ideal velocity may produce quite different effects: effectiveness strongly depends on the place where the propellant is burnt. Generally, it is better to burn propellant closer to the planet. This produces greater effect and enables missions with higher characteristic velocities. This is the so-called Oberth effect which will be discussed further. We have already found that to reach a 300-km orbit, the characteristic velocity of 8080 m/s is needed. The orbital velocity at this altitude is 7730 m/s, so the escape velocity is 10 930 m/s. The distance to the Moon (i.e. the apogee of our transfer orbit) is 3.8 ·10 8 m, so we shall fly to the Moon by an ellipse with the semi-major axis of a (6.4 ·10 6 + 3.8 ·10 8 /2) 1.95 ·10 8 m. Using the formula to the right, we may found the necessary perigee velocity, which yields V per 10 840 m/s.

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18 So we should accelerate the satellite by 10 840 – 7730 3110 m/s to make the trip to the Moon orbit possible. This value should be added to 8080 m/s needed to reach a 300-km orbit. From the angular momentum conservation law, the satellite would have in the apogee the velocity of 10 830 · 6670/380 000 200 m/s. The Moon also moves along its orbit with the velocity of about 1000 m/s. Since our satellite will have in the apogee the velocity of 200 m/s, it will move with the speed of 1000 – 200 = 800 m/s relative to the Moon. So, the satellite will have in the lunar infinity a hyperbolic excess velocity of 800 m/s. It is known that the escape velocity for the surface of the Moon is V esc,Moon 2380 m/s; from energy considerations it is clear that the object falling to the surface of the Moon with zero velocity in infinity will achieve exactly the same speed. If we let the satellite to fall freely to the surface, its velocity would achieve (2380 2 + 800 2 ) 1/2 2510 m/s near the surface. So we should add 2510 m/s to the mission characteristic velocity: this velocity should be cancelled at soft-landing. In total, we get V char 8080 + 3110 + 2510 13 700 m/s. We should not forget that this estimation is quite rough, since it does not take into account all losses, imperfections, the Earth rotation, need to change inclination etc. Drag and aerodynamic losses will add about 1500 m/s or more to this figure, so the total ideal velocity of the system cannot be less than 15 000 m/s.

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19 Selected characteristic velocities Below characteristic velocities of some selected missions are given. We consider direct transfer from the Earth surface, neglecting all losses, the Earth rotation, plane changes, possible gravity assists and tricks that may help to reduce the number, but aerobraking at landing is taken into account. It is specially important for the Eath and the Mars; availability of aerobrake landing method makes the martian surface more achievable than the lunar surface. Mean figures are given. 300-km circular orbit: 8.1 km/s Geostationary transfer: 10.4 km/s Moon transfer: 11.1 km/s Earth escape: 11.2 km/s Venus transfer: 11.5 km/s Mars transfer: 11.6 km/s Geostationary orbit: 11.7 km/s Moon satellite: 11.9 km/s Mars orbit: 12.1 km/s Mars direct land: 12.5 km/s Mercury transfer: 13.4 km/s Moon direct land: 13.5 km/s

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20 Jupiter transfer: 14.2 km/s Saturn transfer: 15.2 km/s Uran transfer: 15.9 km/s Moon land and return: 15.9 km/s Neptun transfer: 16.2 km/s Pluto transfer: 16.3 km/s Direct Sun transfer: 29.2 km/s

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21 End of the Lecture 3

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Gravitational force and centripetal acceleration Circular orbit

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O – center, F 1 – focus, AP – line of apsides, A – apoapsis (apogee), P – periapsis (perigee), r – distance of the satellite from the attracting center, – true anomaly, a – semi-major axis, b – semi-minor axis, e – eccentricity, OF 1 = a·e, F 1 P = a·(1-e) Elements of elliptical orbit

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O – center, F 1 – focus, P – periapsis, AP = a – semi-major axis, r – distance of the satellite from the attracting center, – true anomaly, M = (POy) – mean anomaly, E = (POx) – eccentric anomaly Relation between true and eccentric anomaly

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C – circular orbit E 1, E 2 – elliptical orbits with different excentricities P – parabolic orbit, H – hyperbolic orbit Different paths in central field

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