Orbital Motions

7 Key excerpts on "Orbital Motions"

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  Satellite Geodesy

  Foundations, Methods, and Applications

  • Günter Seeber(Author)
  • 2008(Publication Date)
  • De Gruyter

    (Publisher)

  66 3 Satellite Orbital Motion In mathematical formulation this means for different planets P i with periods of revo-lution U i , mean motions: n i = 2 π/U i , (3.9) and semi-major axes a i : a 3 i U 2 i = C 2 4 π 2 . (3.10) C is a constant for the planetary system. Inserting (3.9) into (3.10) gives the commonly used expression a 3 i · n 2 i = C 2 . (3.11) This law was found empirically by Kepler because it approximates very well the motion of the large planets. However, a completely different numerical value for C 2 is obtained for the system of Jovian moons. Therefore a more general relation is useful a 3 U 2 = k 2 4 π 2 (M + m), (3.12) where k is a universal constant and M, m are the respective masses. Using (3.12) it is possible to determine the masses of celestial bodies. Kepler’s laws describe the simplest form of motion of celestial bodies under the assumption that no external perturbing forces are present, and that the respective masses can be considered to be point masses or homogeneous bodies with spherical mass distribution. For the motion of an artificial Earth satellite these assumptions are only valid in a first approximation. Keplerian orbits, consequently, can only be used as a simple reference orbit and they give only qualitative information on the kind of motion. Kepler himself was convinced that his three, empirically found, laws followed a more general law. This more general law was formulated by Isaac Newton (1643–1727) in the form of the Law of Gravitation . 3.1.2 Newtonian Mechanics, Two-Body Problem 3.1.2.1 Equation of Motion In the first book of “Principia” Newton introduced his three laws of motion: 1. Every body continues in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by an external impressed force. 2. The rate of change of momentum of the body is proportional to the force im-pressed and is in the same direction in which the force acts.

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  Intermediate Dynamics

  • Patrick Hamill(Author)
  • 2022(Publication Date)
  • Cambridge University Press

    (Publisher)

  The equations of motion can be used to obtain the position of the planet as a function of time or to determine the equation of the orbit, r = r(θ). If the total energy is negative, the equation of the orbit is r = l 2 /GMm 2 1 + Al 2 GMm 2 cos(θ − θ 0 ) , which has the form of the equation for an ellipse, r = a(1 − e 2 ) 1 + e cos θ . Using these relations one can relate Kepler’s laws to physical concepts as follows. First law: Planets move in elliptical orbits. This is a consequence of the inverse square nature of the gravitational force. Second law: Planets sweep out equal areas in equal times. This is a consequence of the conservation of angular momentum in a central force field. Third law: The period squared is proportional to the semimajor axis cubed. This is a consequence of the conservation of energy and the fact that the magnitude of the total energy is inversely proportional to the semimajor axis. When a planet in a circular orbit is perturbed, it will oscillate radially around the circular path, converting the orbit into an ellipse. The orbit is stable for any force law having the form f = −cr n as long as n is greater than −3. 10.14 Problems Problem 10.1 (Bohr model of the atom.) Consider an electron in a circular orbit around a proton. Assume the angular momentum can only take on values equal to nh where n is an integer (n = 1, 2, 3, . . .) and h is a constant. Determine the possible values of the radius for the orbit and the possible values for the total energy. Make up a table of the energy for the first four energy states (that is, for the electron in the four smallest orbits). 274 10 CENTRAL FORCE MOTION: THE KEPLER PROBLEM Problem 10.2 A particle of mass m is in a Lennard-Jones force field. Assume the force center is stationary. The particle has velocity v. (a) Write the Lagrangian. (b) Obtain the equations of motion. (c) Determine the angular momentum and show that it is a constant.

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  Classical Mechanics in Geophysical Fluid Dynamics

  • Osamu Morita(Author)
  • 2022(Publication Date)
  • CRC Press

    (Publisher)

  Chapter 11 Orbital Motion of Planets

  DOI: 10.1201/9781003310068-11

  Johannes Kepler found three laws of the orbital motion of planets based on the enormous observational data of Tycho Brahe. Isaac Newton established the basis of classical mechanics and found the law of universal gravitation in which Kepler's three laws played a very important role. In this chapter, we will review the process classical mechanics was established and prove Kepler's three laws exactly. Further, we will discuss the universal gravitation exerted by bodies of finite extent, and the oceanic tides and tidal effects on the Earth–Moon system. We will discuss the general orbits due to a central force. As an application of the orbital motion, Rutherford scattering is discussed precisely.

  11.1 The Law of Universal Gravitation

  Newton considered that an apple falls to the ground owing to some attractive force exerted by the Earth and the same force exerts on the Moon. Orbital motion of the Moon around the Earth is the same that the Moon is continuously falling to the Earth. Newton calculated the falling distance of the Moon in 1 s, from which he obtained the acceleration due to gravity at the center of the Moon. He found that the ratio of the magnitude of the acceleration due to gravity at the center of the Moon to that of the Earth's surface is almost equal to the square ratio of the Earth's radius to the distance between the Earth and the Moon. We will review the calculation performed by Newton. Here we will use the following physical quantities,

    r = 3.84 ×

  10 8

    m

  : the distance from the Earth to the Moon,

    a = 6.37 ×

  10 6

    m

  : the radius of the Earth,

    T = 2.36 ×

  10 6

    s

  : the orbital period of the Moon.

  At first, we will calculate the distance that the Moon falls to the Earth in 1 s. Let the center of the Earth be O, the position of the Moon at time t(s) be A and the position of the Moon at

  t + 1

  (s) be B. Suppose that the Moon would move to point C at

  t + 1

  (s) if it traveled at the constant velocity (Fig. 11.1

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  Dynamics and Control of Autonomous Space Vehicles and Robotics

  • Ranjan Vepa(Author)
  • 2019(Publication Date)
  • Cambridge University Press

    (Publisher)

  2 Space Vehicle Orbit Dynamics 2.1 Orbit Dynamics: An Introduction All space vehicle flights occur under the influence of the gravitational force of a multitude of planetary objects in the Solar system. In particular, the motion of an artificial Earth satellite is primarily influenced by the gravitational pull of the Earth and the gravitational forces exerted on it by the Moon and the Sun. Given that the Moon is at distance of 384,440 km (semi-major axis), orbits the Earth in a near circular orbit (eccentricity, e = 0.0549), and has a mass of only 0.01226 of the Earth’s mass, it is conceivable that for artificial Earth satellites orbiting the Earth at a distance of less than approximately 38,444 km the influence of the Moon can be neglected. Furthermore, the Sun is at a distance of 149, 599  10 3 km, which is 400 times further away than the Moon, although it is 332,952 times heavier than the Earth. For an artificial satellite orbiting the Earth at a distance of less than approximately 38,440 km, the influence of the Sun can be assumed to be no different than its influence on the Earth. 2.2 Planetary Motion: The Two-Body Problem As far as motion around the Earth is concerned, any orbiting artificial satellite and the Earth may be considered in isolation as two interacting independent celestial bodies as long as the artificial satellite is sufficiently close to Earth. This leads to the classical two-body approximation problem that serves as a valuable paradigm for developing the theory of planetary motion. The key question is: How close is sufficiently close? That is a question we shall not seek to answer yet, although it is indeed a fundamental one. 2.2.1 Kepler’s Laws The German astronomer Johann Kepler (1571–1630) formulated three empirical laws of planetary motion based on astronomical data provided to him by the Danish astronomer Tycho Brahe in the late 1590s.

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  Principles of Dynamics

  • Rodney Hill(Author)
  • 2016(Publication Date)
  • Pergamon

    (Publisher)

  1.7 Motion of Two Gravitating Bodies The prototype problem in celestial mechanics is that of the periodic relative motion of two bodies idealized as mass-points whose accelerations are affected equally, at least in first approxima-tion, by other matter. It should be realized that the additional 12 THEORY OF P L A N E T A R Y SYSTEMS [1.7 F IG. I By deñnition G is such that μι^ + μ2 ^2 = Ö (1) where and rg are the position vectors measured from G towards the bodies (Fig. 1 shows two of the possible configurations). Since the latter are necessarily always coUinear with G, the orbits relative to G are similar, with linear dimensions in inverse ratio to the masses (Principia: Book I, Prop. LVII). The gravitational equations of motion, (5) or (8) of §1.6, reduce here to = μΐμ2 (^2 -Γι)/ 1 ^2 - ΓΐΙ' = -/ ^ 2 Ϊ 2 . (2) only one being independent in view of (1). With its help they become (Ml + μ2) = -μΙ^Κ, (μ^ + μ^)% = -μ1^2ΐ4. (3) both of which evidently define inverse-square central orbits with t Thus, the acceleration of the moon by the earth is only about one half the accelerations of the moon and earth separately by the sun, but is about 100 χ their difference. accelerations of each body due to external influences need not be small in themselves, but that their difference must be small compared with the additional accelerations produced by the bodies in each other.f Examples are a planet and its major satellite; the sun and any planet; and an isolated binary (interacting double stars). Then, exactly as in the transition from (1) to (5) in §1.6 we may show that the mass-centre (G say) of the pair moves like an isolated mass-point under the external ñeld; and that, in a sidereal frame, the acceleration of each body relative to G is just that due to the presence of the other. 1 .7] MOTION OF T W O G R A V I T A T I N G BODIES 13 respect to G.

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  An Introduction to Mechanics

  • Daniel Kleppner, Robert Kolenkow(Authors)
  • 2013(Publication Date)
  • Cambridge University Press

    (Publisher)

  Kepler discovered his laws of planetary 394 CENTRAL FORCE MOTION Table 10.2 ∗ Planet Planet / body Mercury 0.206 Saturn 0.055 Venus 0.007 Uranus 0.051 Earth 0.017 Neptune 0.007 Mars 0.093 Pluto 0.252 Jupiter 0.048 Halley’s comet 0.967 ∗ Note 10.2 derives further geometric properties of elliptic orbits. motion by trying to fit his calculations to Brahe’s accurate observations of Mars’ orbit. Orbits Example 10.5 Geostationary Orbit For communications purposes, satellites are typically placed in a cir-cular geosynchronous orbit. If the orbit is in the equatorial plane of the Earth, it is called geostationary . A satellite’s orbital speed in a geo-stationary orbit is set to match the angular velocity Ω e of the rotating Earth, so that as seen from the Earth the satellite is stationary above a fixed point on the Equator. For a satellite of mass m in a geostationary circular orbit, Eq. ( 10.30 ) gives, with A = 2 r , μ ≈ m , C = mgR e 2 , and v = r Ω e , v 2 = ( r Ω e ) 2 = 2 C m 1 r − 1 2 r r 3 = gR e 2 Ω e 2 . With Ω e ≈ 2 π/ 86 400 rad / s, r ≈ 42 250 km so that the satellite’s altitude h above the Earth is h = (42 250 − 6400) = 35 850 km ≈ 22 280 mi . Its orbital speed in the geostationary orbit is v = r Ω e = 3070 m / s ≈ 6870 mi / hr. Orbit transfer maneuvers are frequently needed in astronautics. For example, in the Apollo flights to the Moon the vehicle was first put into near Earth orbit and then transferred to a trajectory toward the Moon. In order to transfer a spacecraft from one orbit to another, its velocity must be altered at a point where the old and new orbits intersect. The next two examples look at the physical principles of satellite launch and orbit transfer. 10.5 PLANETARY MOTION 395 Example 10.6 Satellite Orbit Transfer 1 The most energy-e ffi cient way to put a satellite into circular orbit is to launch it into an elliptical transfer orbit whose apogee is at the desired final radius.

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  Foundations of Astrophysics

  • Barbara Ryden, Bradley M. Peterson(Authors)
  • 2020(Publication Date)
  • Cambridge University Press

    (Publisher)

  The angular momentum of the planet is defined as L ≡ r × p, (3.13) where p = mv is the linear momentum. The rate of change of the angular momentum is then d L dt = d r dt × p + r × d p dt = v × mv + r × m d v dt . (3.14) 3.1 Deriving Kepler’s Laws 65 r v r Δt v t Δt M m FIGURE 3.3 The motions of a planet during a short time interval t . From Newton’s second law of motion, we know that md v/dt = F. Thus, equation (3.14) can be rewritten as d L dt = m(v × v) + r × F. (3.15) However, v × v = 0 (that’s just a vector identity), and for a central force, F is parallel to r and thus F × r ∝ r × r = 0. We conclude that for gravity or any other central force, angular momentum is conserved: d L dt = 0. (3.16) Note that the direction as well as the magnitude of L is constant; this tells us that the motion of an object moving under the influence of a central force is confined to a plane. The conservation of angular momentum is equivalent to Kepler’s second law; to demonstrate that this is true, we use equation (3.10) to write the angular momentum explicitly as L = r × mv = mrv t ˆ k = L ˆ k, (3.17) where v t is the tangential velocity. Referring to Figure 3.3, consider a planet of mass m; at a time t , it is at a distance r from the Sun, which has mass M. During a brief time interval t , the planet moves a distance v t t in the tangential direction and a distance v r t in the radial direction. The area A swept out by the planet–Sun line during this brief interval can be approximated as the sum of two triangles: A ≈ 1 2 r(v t t) + 1 2 (v r t)(v t t), (3.18) where the two terms represent the left-hand triangle and the right-hand triangle in Figure 3.3. 4 In the limit v r t r , the right-hand triangle is vanishingly small compared to the left-hand triangle, and the area swept out can be further simplified as A ≈ 1 2 r(v t t). (3.19) 4 In Figure 3.3, we are looking at the specific case v r > 0, but performing a time reversal will yield the case v r < 0.

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