Lecture 21 Gravitational and Central Forces 21.1 Newton s Law of Universal Gravitation According to Newton s Law of Universal Graviation, the force on a particle i of mass m i exerted by a particle j of mass m j is, F ij = Gm ( ) im j rij (21.1) rij 2 r ij This is a central force because it only acts on the line connecting particle i and j (i.e. no angular or orientational dependence). is the universal constant of gravitation. G = 6.67259 10 11 Nm 2 kg 2 Implicit in the gravitational force is Newton s third law. Centripetal acceleration of the moon toward the Earth and the gravitational acceleration of an apple falling on the surface of the Earth have a common origin this leap allowed for the development of the law of universal gravitation. Why does the Earth attract as if all of its mass were concentrated at a single point? Consider a thin shell of uniform mass M and radius R and divide the shell up into rings or width R θ: The circumference of the ring is, 2π(R sin θ Therefore, if the mass density (mass per unit area) is, then the mass of the ring is, ρ M A, M ρ A = ρ(2πr sin θ)(r θ) = ρ2πr 2 sin θ θ
Q θ R θ O r u φ F Q P What is the gravitational force exerted on a point mass m at P by a single element of the ring at Q (acting in the PQ direction)? We resolve the force into components, one along PO of magnitude F Q cos φ, the other perpendicular to PO, magnitude F Q sin φ. But from symmetry, the vector sum of all of the perpendicular components exerted at P by the whole ring vanishes. Hence, the force F exerted by the entire ring is in the PO direction with magnitude, F = Gm M cos φ = Gm2πρR2 sin θ cos φ φ Taking the limit that θ θ and adding up all the rings, we obtain the total force exerted at P by the whole shell: F = Gm2πρR 2 π o sin θ cos φdθ To perform this integral, we see from the figure above that, = r 2 + R 2 2rR cos θ, from the law of cosines. But since r and R are constant, udu = rr sin θdθ Similarly, cos φ = u2 + r 2 R 2 2ru
Therefore, r+r F = Gm2πρR 2 + r 2 R 2 du r R 2Rr 2 = GmM r+r ) (1 + r2 R 2 du 4Rr 2 r R = GmM r 2 where we have substituted in the mass of the shell,m = 4πρR 2. In vector form, this is just, F = GMm ˆr r 2 This is an incredible result! It says that a uniform spherical shell of matter attracts a particle as if the whole mass of the shell were concentrated at its center. Therefore, this will be true for every shell of a solid uniform sphere! 21.2 Kepler s Laws 1. Each planet moves in an ellipse with the sun as a focus. 2. The radius vector sweeps out equal areas in equal times. 3. The square of the period of revolution about the sun is proportional to the cube of the major axis of the orbit.. Newton showed that Kepler s three laws are a consequence of the law of gravity. Before we prove these laws, let us remind ourselves of some properties of central fields. 21.2.1 Conservative Fields F = f(r)ˆr Take the curl of this force in spherical coordinates: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ dv = r 2 sin θdθdφdr to give us, F = 1 r 2 sin θ ˆr rˆθ r sin θ ˆφ r θ φ F r F θ r F φ r sin θ
Now, since F θ = F φ = 0, we are left with, F = 1 f(r) ˆφ 1 f(r) ˆφ = 0 r sin θ r r θ Therefore, since the curl of F vanishes, we say that F is conservative. This implies that there exists a potential V (r) such that, In integral form, this implies V (r) = V (r) F = V = r r r o r F d r = f(r)dr r o where r o is some reference value where the potential is take to be zero (typically at infinity). 21.2.2 Central Fields If the force acts on a line connecting the particles, then it is obvious that, r F = 0 But recall our torque equation, dl dt = r F = N Thus, in a central field, dl dt = 0 L = constant and the angular momentum is conserved (as we have discussed before). We now proceed to Kepler s Laws which, for convenience, we will not prove in order 21.2.3 Proof of Kepler s Second Law The area da is swept out by the radius vector r in a time dt as a planet orbits the sun. This area is equal to half of the area of a parallelogram formed by r d r (by definition). Therefore, we arrive at Kepler s 2nd law: da = 1 2 r d r = 1 L r vdt = 2 2m dt da dt = A = L 2m = constant (21.2) which is a consequence of the fact that we are dealing with central forces and the angular momentum is conserved.
Sun r d r = v d t da 21.2.4 Proof of Kepler s 1st law To derive Kepler s 1st law, we start with Newton s second law expressed in polar coordinates: m r = f(r)ˆr (21.3) Recall from the beginning of the course that we derived the radial component of the acceleration to be: a r = r r = r r θ 2 m( r r θ 2 ) = f(r) and the θ component of the acceleration was given as: m(2ṙ θ + r θ) = 0. From the last expression above, we see that: hence, a θ = r θ = 2ṙ θ + r θ d dt (r2 θ) = 0, r 2 θ = constant = l = L m where l is the angular momentum per unit mass. This just reaffirms the fact that the angular momentum is constant when a particle moves in a central force. Lets now play a trick, let r = 1/u, therefore ṙ = u = θ du dθ = ldu dθ
since θ = l. Let us now differentiate again to find, r = l d du dt dθ = l θ d2 u dθ = 2 l2 d2 u dθ 2 Therefore, the radial equation of motion, as given above, becomes: d 2 u dθ 2 + u = 1 ml 2 f(u 1 ) (21.4) This is the differential equation (or orbit equation) of the orbit of a particle moving under a central force. The solution of this equation will give u as a function of θ and we will then be able to show that the orbits obtained give us ellipses.