* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Word
Relativistic quantum mechanics wikipedia , lookup
Four-vector wikipedia , lookup
Fictitious force wikipedia , lookup
N-body problem wikipedia , lookup
Center of mass wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Relativistic angular momentum wikipedia , lookup
Classical mechanics wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Mass versus weight wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Equations of motion wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Centripetal force wikipedia , lookup
AST 121S: The origin and evolution of the Universe Introduction to Mathematical Handout 1 Please read this first.... This is an unusually long hand-out and one which uses in places mathematics that you may not be familiar with. Section D is not part of the course and will not be examined - it is provided only for your interest. I have included Section D in the handout because it represents one of the most important breakthroughs in science. Newton's demonstration that all of Kepler's three Laws of Planetary Motion could be explained by a simple force f = GMm/r2 and that this force was the same phenomenon as familiar terrestrial gravity was important because: (a) It was the first time that the underlying cause of diverse complex natural phenomena were explained in terms of a single very simple process, i.e. inverse square law gravity. (b) Gravity was the first of the fundamental forces of Nature to be identified and described (even if we now know that Newtonian Gravity is only a very good approximation to gravity as described by Einstein's General Relativity). (c) It was the first time that processes in the remote Universe were understood in terms of the same phenomena seen on Earth. Newton's Gravity was a Universal Force. Thus the science of astrophysics was born. The fact that the Universe follows the same physical laws as are seen in terrestrial laboratories makes cosmology possible. In this handout, Sections A and B will be familiar from the lectures. Everyone should be able to cope with the mathematics in Section C in which we derive the force required for a body to move in a circular path. The proof, in Section D, that Newtonian Gravity produces orbits which satisfy all of Kepler's Three Laws of Planetary motion, is enormously easier if we use vectors and vector calculus. Many of you will not be familiar with these tools. As stated above, I do not expect students to be able to reproduce Section D in an exam or elsewhere and you should read it for interest only. I have provided a short introduction to vectors and vector calculus for those who will want to work through this derivation. I hope many of you will try this, and "experience the breakthrough" first-hand. In Section E, I state without proof a couple of simple results that Newton proved that enabled him to calculate the gravitational field near to extended objects like the Earth. What you need to know from this handout: (a) You should know Sections A and B, be able to apply the results of Section E and be able to reproduce Section C. (b) In Section D, you need to know only (i) that the angular momentum, J = mr2 d/dt, is conserved around an orbit. (ii) that the total energy, E, is also conserved around an orbit 2 2 GMm 1 dr 1 J 1 2 GMm E mv m 2 r r 2 2 dt 2 mr In the first assignment, we will use these two results, (i) and (ii), and we will obtain the same energy equation (ii) when we come, later in the course, to construct the equation to describe the evolution of an expanding Universe. AST 121S: The origin and evolution of the Universe Mathematical Handout 1: Orbits in a simple gravitational field. A. Kepler's Three Laws of planetary motion From careful analysis of Tycho Brahe's voluminous observations of the apparent positions of the planets as seen from Earth, Kepler formulated his three "laws" of planetary motion that described how the planets, including the Earth, moved in space. Kepler's 1st Law: Each planets moves on an orbit around the Sun that is an ellipse with the Sun located at one focus of the ellipse Kepler's 2nd Law: A line drawn from the Sun to a given planet sweeps out equal areas in equal time intervals as the planet moves around the orbit. Kepler's 3rd Law: For the various planets in the Solar System, there is a relation between the period of the orbit T (the time taken to complete one orbit) and the semi-major axis of the ellipse, A: T 2 A3 It should be noted that Kepler did not understand why the planets moved in this way though he suspected that it was related to some terrestrial phenomenon and in fact was the first to coin the term astrophysics. B. Newton's three laws of motion Newton formulated three laws that described the motions of objects, based in part on many of Galileo's ideas: Newton's 1st Law: Bodies move in a straight line at constant speed (i.e. at constant velocity) unless acted upon by a force. Newton's 2nd Law: The magnitude of a force is given by the mass and the acceleration the force produces. dv F ma where a dt Newton's 3rd Law: For every force there is an equal and opposite reaction. C. Acceleration required for circular motion The velocity of a particle is continually changing as it moves around a circular path (remember that velocity is a vector quantity incorporating directional information, as distinct from speed, which does not) and it must therefore, according to Newton's First and Second Laws, be accelerating and experiencing a force. Consider a particle moving with speed v around a circular orbit of radius r. Let us look at the change in velocity while the particle moves through a very small angle around the orbit. Suppose that the particle is initially moving in the y direction. In time t it moves a distance y corresponding to the angle . motion in absence of force circular orbit x y r Now, in this same time interval, the particle must stop moving purely in y and must acquire a small velocity in the x direction, i.e. towards the centre of the orbit (remember we can make arbitrarily small). On the figure above, it moves x x y r2 with a velocity vx x 2 r t t Therefore the acceleration required is a v x r t t 2 Now we can write (t) as d v t dt r where w is the angular velocity of the orbit (measured in radians per second). Thus the acceleration required to produce a circular orbit is (1.1) v2 a r 2 r Thus in order to produce a circular orbit we must continuously apply a force, directed towards the centre of the orbit, of mr2 or mv2/r. A number of people, including Newton's great rival Hooke as well as Newton himself, realized that a force field in the Solar System that fell off with the square of the distance r (and was proportional to the mass of the planet) would reproduce Kepler's 3rd Law. This is because the period, T, of a planet in a circular orbit is related to by (1.2) T 2 Now, Kepler's 3rd Law states that T 2 r3 since A r for a circular orbit Thus, from (1.1) and (1.2) (1.3) a r 2 1 r2 Thus in the case of idealized circular orbits, the force on a given planet must be proportional to the mass of the planet and inversely proportional to the square of the distance between the Sun and the planet. But Kepler's 1st Law states that planetary orbits are not circles, but rather are ellipses. Newton's great achievement was to demonstrate that such an inverse square force field would also produce elliptical orbits, with the Sun at one focus, and with T2 A3, thereby satisfying all of Kepler's three Laws describing planetary motions. The circular orbits considered above are just a special case of more general circular orbits. Furthermore, he then showed that the force that causes the Moon to orbit the Earth was the same as causes apples to drop from apple-trees. It is these two to which we now turn. D: Elliptical orbits in a simple gravitational field (Section D is for interest only and is not a formal part of the course) The derivation of the general orbit in a inverse-square law gravitational field is most easily seen using vectors and vector calculus: Box 1: Vectors and vector calculus For those students who (a) are not familiar with vector calculus and (b) wish to go though this next section, the following is a brief overview of vectors and the calculus of them. Recall that whereas many quantities, such as mass, speed, temperature, are scalars (i.e. simply have a magnitude), others are vectors and have both magnitude and direction. Examples are the displacement from the origin of some coordinate system, velocity, force etc. We will write vectors in bold print and scalars (and the magnitudes of vectors) in regular print. The direction of a vector can be represented by dividing it by its magnitude, i.e. a/a. The magnitude of a vector can be represented as a = a Vectors can be combined in various ways. Addition (and subtraction) is straightforward: b a+b a Box 1 (continued) Vectors can also be multiplied in two ways: The dot-product is a scalar given by a. b ab cos where is the angle between the vectors. The dot-product is the magnitude of one vector times the component of the other vector that is parallel to the first (and vice versa). It is easy to see that a.b = b.a. The cross-product is a vector that has direction perpendicular to both a and b and has magnitude a b ab sin The direction of a b is set by the right-hand rule such that a, b and a b are like the thumb, first and middle fingers of your right hand. This means that (a b) (b a) . Clearly parallel vectors (where = 0) have zero cross-product (as they must since the direction perpendicular to both is then not defined!). Vector differentiation occurs in the normal way. If we have a time dependent vector a(t). Then da/dt. is given by da a(t t ) a(t ) dt t a(t+dt)-a(t) a(t) a(t+dt) da/dt is thus itself a vector. Finally, differentiation of cross-products proceeds in the usual way by the chain rule d a b a db da b dt dt dt Consider the simple case where the gravitational field is dominated by a single point-like object of mass M (i.e. in the case of the Solar System, the Sun). For an orbiting object of mass m, whose instantaneous position, relative to the central object is given by r and whose instantaneous velocity is v = dr/dt, Newtonian gravity describes the force and resulting acceleration of the orbiting object as: (1.4) m d 2r dv GMm m 3 r 2 dt dt r If we take the cross product of this equation (1.4) with r, and cancel out m, we get: (1.5) r d 2r GM 3 rr 0 2 dt r since r r 0 by definition If we now look at the quantity r v (i.e. r dr dt ) we get (1.6) d dr d 2r dr dr r r 2 0 from (1.5) and a a 0 dt dt dt dt dt Thus the quantity r v is constant for motion in a gravitational field. The quantity r v is the specific angular momentum (i.e. angular momentum per unit mass), and the angular momentum, J, is simply m r v . (1.7) J mr v constant Remembering that the direction of r v is perpendicular to r and v, the constancy of r v means that the orbit lies in a plane. Equation (1.6) now explains Kepler's Second Law since the rate at which area swept out by the vector r is just (1.8) dA 1 1 rvsin r v dt 2 2 If we now take the dot-product of equation (1.4) with v ( = dr/dt) , we get: (1.9) m v. dv GMm 3 v. r dt r With a few moments thought you should be able to convince yourself that (1.10) d 1 2 dv d GMm GMm dr mv m v. and 3 r. dt 2 dt dt r dt r Therefore, we can rewrite (1.9) as (1.11) d 1 2 GMm mv 0 dt 2 r You may recognize that the quantity in brackets is the total energy, E, of the orbiting object (i.e. kinetic energy plus negative potential energy). GMm 1 E mv 2 r 2 Equation (1.11) therefore simply proves the conservation of energy. Because the orbit is in a plane, from (1.6), we can conveniently adopt (r,) coordinates and drop the vector notation. We have defined the angular momentum, J, which we have seen is constant, to be (1.12) J m vr J mr 2 d dt We can decompose v into a component parallel and perpendicular to r using Pythagoras. J2 dr d dr v2 r m2r 2 dt dt dt 2 2 2 Thus, we can rewrite our energy equation as 2 (1.13) E 1 dr 1 J2 GMm m constant 2 2 dt 2 mr r Rearranging (1.13) we get: 1/ 2 (1.14) dr 2 E 2GM J2 2 2 dt m r m r Note in passing that the left hand side of (1.14) is just the radial velocity of the particle towards or away from the central object. If E is positive then this is still finite as the distance tends to infinity and the particle can completely escape from the influence of the central object. On the other hand, if E is negative, then the radial velocity must become zero at some finite distance from the central object and in this case the particle never gets further away, and is bound to the central object. The energy E is thus often referred to as the binding energy of the particle. Dividing (1.14) by J ( = mr2d/dt) we get: 1/ 2 (1.15) 1 dr dt 2 E 2GM 1 2 2 2 2 2 mr dt d mJ J r m r If we now make the following substitutions: 1 1 du 2 dr mr mr GMm wu dw du J2 u 2 2 E GMm w constant mJ 2 J 2 2 0 then (1.15) becomes (1.16) dw w02 w2 d 1/ 2 Equation (1.16) has solution w w0 cos i.e. substituting back into the above, (1.17) 1 1 1 cos r r0 with 1/ 2 (1.18) J2 2 EJ 2 r0 and 1 G 2 M 2 m3 GMm2 The quantity is known as the eccentricity of the orbit. A circle is an ellipse with = 0. For E < 0, i.e. for an object which has negative net energy (i.e. which is bound to the central object), equation (1.17) is in fact the equation of an ellipse with the origin at one focus. Thus is Kepler's First Law explained. You may be more familiar with the equation of an ellipse in Cartesian coordinates: (1.19) x 2 y2 1 a2 b2 In this case, the origin of the Cartesian system (x,y) is offset from the origin of the polar coordinate system (r,) by (a,0). The semi-major and semi-minor axes of the ellipse are given by: a (1.19) b r0 1 2 r0 1 2 1/ 2 The E = 0 case corresponds to a parabola and the unbound E > 0 case to a hyperbola. The circular = 0 case has E = -G2M2m3/2J2 We can easily now recover Kepler's Third Law by noting that the area of an ellipse is (1.20) A ab Now, the time taken to go around the orbit is, from (1.8) (1.21) T dt 2m 2mA m dA 2ab J J J Now, from (1.18), we know r0 = J2/GMm2, and from (1.19) we have that b2 = ar0, it follows that: (1.22) 42 3 T a GM 2 This is exactly Kepler's Third Law with a = A. Note that the mass of the planet does not enter in to equation (1.22) so it applies for any planetary system in which the gravitational field is dominated by a single massive object. E: Gravitational fields from symmetrically distributed mass systems. Newton's analysis in Section D beautifully explained Kepler's Three Laws of planetary motion in terms of a simple 1/r2 force field emanating from the Sun. Newton next wanted to demonstrate that this was the same force as familiar gravity on Earth. Galileo's observations that the motions of objects on Earth was independent of their masses, could be explained by Newton's new force since the size of the force was proportional to the mass of the body it was acting on, so the acceleration would be independent of mass. However, for several years he was stymied by the fact that the Earth is obviously not a point mass and he did not know how to calculate the gravitational field of an extended distribution of mass such as the Earth. After several years he was able to prove the following two theorems which will here be stated without proof: For any spherically symmetric shell of matter, the gravitational field arising from the mass on the shell is (a) zero within the shell and (b) the same, outside the shell, as if all the mass of the shell was a point mass at the centre of the shell. Thus, the gravitational field at the surface of the Earth is the same as if the Earth's mass was concentrated into a point at the centre of the Earth, so that one can in fact use the simple equation for the point mass (1.4). Furthermore, the gravitational field at some radius r within an extended spherically symmetric object (such as an idealized galaxy) is given by (1.4) with M the mass enclosed within the radius r. The effect of mass further out in the galaxy is zero.