Central Force Motion Kepler’s Laws-Classical Mechanics-Lecture Handout, Exercises of Classical Mechanics

Download Central Force Motion Kepler’s Laws-Classical Mechanics-Lecture Handout and more Exercises Classical Mechanics in PDF only on Docsity! Lecture L15 - Central Force Motion: Kepler’s Laws When the only force acting on a particle is always directed to­ wards a fixed point, the motion is called central force motion. This type of motion is particularly relevant when studying the orbital movement of planets and satellites. The laws which gov­ ern this motion were first postulated by Kepler and deduced from observation. In this lecture, we will see that these laws are a con­ sequence of Newton’s second law. An understanding of central force motion is necessary for the design of satellites and space vehicles. Kepler’s Problem We consider the motion of a particle of mass m, in an inertial reference frame, under the influence of a force, F , directed towards the origin. We will be particularly interested in the case when the force is inversely proportional to the square of the distance between the particle and the origin, such as the gravitational force. In this case, µ F = − r2 mer, where µ is the gravitational parameter, r is the modulus of the position vector, r, and er = r/r. It can be shown that, in general, Kepler’s problem is equivalent to the two-body problem, in which two masses, M and m, move solely due to the influence of their mutual gravitational attraction. This equivalence is obvious when M � m, since, in this case, the center of mass of the system can be taken to be at M . 1 docsity.com However, even in the more general case when the two masses are of similar size, we shall show that the problem can be reduced to a ”Kepler” problem. Although most problems in celestial mechanics involve more than two bodies, many problems of practical interest can be accurately solved by just looking at two bodies at a time. When more than two bodies are involved, the problem is considerably more complicated, and, in this case, no general solutions are known. The two body problem was studied by Kepler (1571-1630) who lived before Newton was born. His interest was in describing the motion of planets around the sun. He postulated the following laws: 1.- The orbits of the planets are ellipses with the Sun at one focus 2.- The line joining a planet to the Sun sweeps out equal areas in equal intervals of time 3.- The square of the period of a planet is proportional to the cube of the major axis of its elliptical orbit In this lecture, we will start from Newton’s laws and verify that the above three laws can indeed be derived from Newtonian mechanics. Equivalence between the two-body problem and Kepler’s problem Here we consider the problem of two isolated bodies of masses M and m which interact though gravitational attraction. Let rM and rm denote the position vectors of the two bodies relative to a fixed origin O. Since the only force acting on the bodies is the force of mutual gravitational attraction, the motion is governed by Newton’s law with an equal and opposite force acting on each body. Mm M r̈M = G r2 er, (1) Mm mr̈m = −G r2 er , (2) where r = |r|, er = r/r, and G is the gravitational constant. The position of the center of gravity, G, of the two bodies will be rG = MrM + mrm . (3) M + m 2 docsity.com � � � � � � � � orthogonal to the angular momentum vector, and, as a consequence, the motion will be planar. Using cylindrical coordinates, with ez being parallel to the angular momentum vector, we have, − r µ 2 er = (r̈ − rθ̇2)er + (rθ ̈+ 2ṙθ̇)eθ. Now, we consider the radial and circumferential components of this equation separately. Circumferential component We have, ¨ ˙0 = rθ + 2ṙθ . Using the following identity, � � 1 d (r 2θ̇) = rθ ̈+ 2ṙθ, ˙ r dt the above equation implies that r 2θ̇ = h ≡ constant. (8) We note that the constant of integration, h, that will be determined by the initial conditions, is precisely the magnitude of the specific angular momentum vector, i.e. h = |r × v|. In a time dt, the area, dA, swept by r will be dA = r rdθ/2. Therefore, dA 1 h = r 2θ̇ = dt 2 2 , which proves Kepler’s second law:The line joining a planet to the Sun sweeps out equal areas in equal intervals of time. Radial component The radial component of the equation of motion reads, − r µ 2 = r̈ − rθ̇2 . (9) 2 d 1 2Since −r � � = r, ˙ and r = h/θ̇ from equation 8, we can write dt r h d 1 d 1 ṙ = − θ̇ dt r = −h dθ r . Differentiating with respect to time, r̈ = −h d 2 1 θ̇ = − h 2 d2 1 . dθ2 r r2 dθ2 r 5 docsity.com Inserting this expression into equation 9, and using equation 8, we obtain the following differential equation for 1/r as a function of θ. � � d2 1 1 µ + = . dθ2 r r h2 This is a linear second order ordinary differential equation which has a general solution of the form, 1 µ = (1 + e cos(θ + ψ)) , r h2 where e and ψ are two constants of integration. If we choose θ to be zero when r is minimum, then e will be positive, and ψ = 0. The equation describing the trajectory will be h2/µ r = . (10) 1 + e cos θ We shall see below that this is the equation of a conic section in polar coordinates. Conic Sections Conic sections are planar curves that are defined as follows: given a line, or directrix, and a point, or focus O, a conic section is the locus of points, P , such that the ratio of the distance between the point and the focus, PO, to the distance between the point and the directrix, PA, is a constant e. That is, e = PO/PA. Since PO = r and PA = p/e − r cos θ, we have r = p . (11) 1 + e cos θ Here, p is the parameter of the conic and is equal to r when θ = ±90o . The constant e ≥ 0 is called the eccentricity, and, depending on its value, the conic surface will be either an open or closed curve. In particular, we have that when e = 0 the curve is a circle e < 1 the curve is an ellipse e = 1 the curve is a parabola e > 1 the curve is a hyperbola. 6 docsity.com Comparing equation(11) which deals solely with the property of a conic section, and equation(10) which provides the solution of the motion of a point mass in a gravitational field, we can identify the properties of the conic section orbits in terms of the physical parameters of the Kepler problem. In particular, we see that the trajectory of a mass under the influence of a central force will be a conic curve with parameter p = h2/µ. (12) When e < 1, the trajectory is an ellipse, thus proving Kepler’s first law:The orbits of the planets are ellipses with the Sun at one focus. The point in the trajectory which is closest to the focus is called the periapsis and is denoted by π. For elliptical orbits, the point in the trajectory which is farthest away from the focus is called the apoapsis and is denoted by α. When considering orbits around the earth, these points are called the perigee and apogee, whereas for orbits around the sun, these points are called the perihelion and aphelion, respectively. Elliptical Trajectories If a is the semi-major axis of the ellipse, then 2a = rπ + rα. (13) Using equation 11 to evaluate rπ (θ = 0) and rα (θ = π), we obtain a = p/(1 − e 2). (14) Thus from the geometric properties of an ellipse, rπ = 1 + p e = a(1 − e), rα = 1 − p e = a(1 + e). Also, the distance between O and the center of the ellipse will be a − rπ = a e. (15) 7 docsity.com major axis define an angle u, which is referred to as the eccentric anomaly. In addition, we define a third anomaly, the mean anomaly M of the point P , as 2πtP MP = . (24) τ Here, tP is the time of flight from the periapsis to the point P . Thus, if we want to determine the time of flight between two points 1 and 2 on the ellipse, we can use equation (24) and write TOF = τ (M2 − M1) = A2 − A1 τ , = t2 − t1 2π πab where A2 − A1 is the area swept out between points 1 and 2. The mean anomaly for point P can also be written as 2π ∗ AP MP = , (25) AT where AP is the area swept out up to the point P . When the area swept out equals the total area of the ellipse AT , the time t equals the period τ and the mean anomaly Mπ = 2 ∗ π. (The subscript π denotes the return to the periapsis π.) Thus the mean anomaly can be thought of as the fraction of the total angle 2π that would be swept out in a time τ by an object reaching point P . The focus is on time not on actual spatial angle. All is needed now is an expression for the mean anomaly M as a function of the orbit parameters. We start by obtaining a relation between θ and u. From simple trigonometry, we have that a cos u − r cos θ = ae (26) or noting that r = a(1 − e2)/(1 + e cos θ), cos u = e + cos θ , cos θ = cos u − e . (27) 1 + e cos θ → 1 − e cos u We now develop relationships between the various areas indicated on the figure, with the goal to find the formula for the area AP , the area swept out by the point r as it travels from the periapsis π to the point P . The area A1 is the wedge in the circle occupied by the angle u. A1 = a2u; the area of the large triangle formed by the angle u within the circle is A2 = a2cos(u)sin(u)/2. Therefore, the area of the large curved segment from O”, P �, π is A1 − A2 = (1/2) × a 2 × (u − Cos(u)Sin(u)). (28) The base of the small triangle of area A4, O,O��, P , is r × cos(θ) = a × cos(u) − ae by equation (26). The height of the small triangle is b × sin(u). Therefore, the area of the small triangle is A4 = (1/2) × (a ∗ cos(u) − e) × b × sin(u). This area plus the curved segment O”, P, π is the total area swept by the point P . The final step in identifying the area segment swept out between point π and P is to identify the curved segment from O”, P, π, which is then added to the triangle section A4 to form the complete swept area. The curved vertical segment formed from removing the large imbedded triangle A2 from the arc segment of 10 docsity.com the circle A1 –call it A3 – is geometrically similar to the curved segment formed by removing the small triangle from the area of the swept segment of the ellipse. Since the vertical height of the ellipse is b, and the vertical height of the circle is a, the area of the desired curved segment can by obtained from that of the corresponding segment of the circle by multiplying by b/a. Specifically, AP − A4 = (b/a) × (A1 − A2). (29) Therefore, the final result for the area swept out by the point r moving from point π to point P is AP = b/a × (A1 − A2) + A4 (30) And the mean anomaly for the point P is MP = 2π ∗ AP = 2π × (b/a × (A1 − A2)) + A4 (31) AT AT Thus, combining equations (24), (28) and (30), we obtain the mean anomaly for the point P , called Kepler’s equation (It took a Kepler to work this out.) u − e sin u = MP = 2πtP . τ where u is the eccentric anomaly for the point P , defined in the figure. This equation is very easy to use if we want to know the time tP at which the satellite is at position θ. The only thing required, in this case, is the calculation of the eccentric anomaly u using equation (27). On the other hand, if we need to find the position θ of the satellite at a given time t, then, we need to solve Kepler’s equation which is non-linear using an iterative numerical algorithm such as Newton’s method. ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 3/13 (except energy analysis) 11 docsity.com