106 : Fall Application of calculus to planetary motion

Transcription

1 106 : Fall 2004 Application of calculus to planetary motion 1. One of the greatest accomplishments of classical times is that of Isaac Newton who was able to obtain the entire behaviour of planetary bodies from his universal law of gravitation. In particular he derived the laws of Kepler using his method of Fluxions which is essentially differential calculus. However he framed his proofs in the language of euclidean geometry which was very difficult to follow and very often obscured the essential ideas. Subsequent to Newton s work mathematicians used differential calculus and gave a more transparent treatment of the derivation of Kepler s laws. We shall begin by proving two things. I. In a central field of force, motion is always in a plane. II. In a central field of force, equal areas are swept by the radius vector in equal intervals of time. Notice that neither of these results assumes the inverse square law. By a central field of force we mean that there is a point O with the property that at any point P the force exerted depends only on the magnitude OP of the vector OP. The force is attractive or repulsive according as the direction of the force is towards O or away from O. Gravitational fields are examples of force fields which are central and attractive. Electromagnetic forces are attractive or repulsive depending on whether the charges are like or unlike. Furthermore, the gravitaional attractive force is always proportional to 1 OP 2, namely, it is inversely proportional to the square of the distance from the center of attraction. This is often called the inverse square law. In deriving the two facts mentioned above we do not need to assume the inverse square law. We shall discuss later what special consequences arise when we assume the inverse square law. 2. Motion is always in a plane. We need the concept of product of vectors in R 3. Given vectors a = (a 1, a 2, a 3 ), b = (b 1, b 2, b 3 ) 1

2 their product a b := c is defined by c 1 = a 2 b 3 a 3 b 2, c 2 = a 3 b 1 a 1 b 3, c 3 = a 1 b 2 a 2 b 1. Notice the cyclic symmetry of the definition which makes it easy to remember it. There is another product operation on vectors but this time the result is a scalar and so it is called the scalar product as opposed to the one just defined, which is the vector product. The scalar product a b of two vectors is defined by The condition a = (a 1, a 2, a 3 ), b = (b 1, b 2, b 3 ) a b = a 1 b 1 + a 2 b 2 + a 3 b 3. a b = 0 is the condition that the two vectors are orthogonal to each other. More generally, if θ is the angle between the two vectors, then cos θ = a b a. b. Here the number a = a a = a a2 2 + a2 3 is the distance of the point from the origin. The vector product and scalar product have the following easily verified properties. vector product. (1) a a = 0. (2) a b = b a. (3) (a (b c)) + (b (c a)) + (c (a b)) = 0. 2

3 (4) a (αb + βc) = α(a b) + β(a c). scalar product. (1) a b = b a. (2) a (b + c) = a b + a c. Finally, it is easy to see that the vector a b is orthogonal to both a and b. If the vectors depend on a parameter t (time), then the rules for differentiating them are easy to derive. We denote differentiation by primes so that a = da dt. Then (a b) = a b + a b (a b) = a b + a b. Notice the similarity of these laws to the rules for differentiating a product of functions. This is not an accident; the vector and scalar products are built interms of ordinary products and so the rules for differentiating them are the same as for ordinary products. To write the equations of motion of a particle under Newton s laws the cartesian coordinate frame is awkward especially when the force is centrally directed. Nevertheless we shall start with the cartesian coordinate frame of reference. We assume that there is a function V (r) on R 3, depending only on r, such that the force vector F is given by: F(x) = V (r) x r. It is clear from this that if V (r) > 0, this force is always attractive, of magnitude V (r), and directed towards the origin. If V (r) < 0 it is a rfepulsive force, directed away from the origin, and of magnitude V (r). Gravitational forces are attractive and so we shall assme that V (r) > 0 hereafter. Newton s law that force=mass acceleration then leads to the following equations of motion: mx 1 = V (r) x 1 r, Let U be a function such that mx 2 = V (r) x 2 r, V = du dr. 3 mx 3 = V (r) x 3 r.

4 Then U is called the potential. It is easy to calculate that F(x) = U(x). Theorem 1. The motion lies always in a fixed plane through thge origin. Proof. Let u be the position of the particle and u its velocity. To prove that u lies in a fixed plane through the origin it is enough to show that there is a fixed vector w 0 such that u and u are always orthogonal to w 0. If we define the vector w by w = u u, then u and u are always orthogonal to w. Hence if we prove that w is independent of time, then u will always be in the plane orthogonal to w 0 := w(0). But w = (u u ) = u u + u u = U (r) mr u u = 0. Hence w is independent of t, as we wanted to prove. 3. Kepler s second law. We shall now prove Kepler s second law. In view of the above theorem we may assume that the motion is in a plane and we take this plane to be the xy-plane. We can state Kepler s law also as follows: the rate of change of the area swept by the radius vector is constant. Now let P be the posiiton of the particle at time t and r = OP ; let θ be the angle that OP makes with the x-axis. Then at time t + dt the position is r + dr, and the area between OP and OP where P is the posiiton of the particle at time t + dt is, up to terms of the second order, 1 2 r2 dθ. Hence Kepler s second law is the statement that r 2 θ = constant. 4

5 This is the same as saying that d dt (r2 θ ) = 0. The equations of motion are mx = V (r) x r, my = V (r) y r. Theorem 2. Kepler s second law is valid. Proof. It is clearly convenient to work in polar coordinates. Now, from x = r cos θ, y = r sin θ we get, on differentiation with respect to t, x = r cos θ rθ sin θ, y = r sin θ + rθ cos θ. We can solve for r and θ and obtain r = x cos θ + y sin θ, θ = 1 r ( x sin θ + y cos θ). Then r 2 θ = r( x sin θ + y cos θ). Hence, writing s = sin θ, c = cos θ for brevity, (r 2 θ ) = (r( sx + cy ) = r ( sx + cy ) + r( sx + cy ) = (cx + sy )( sx + cy ) + r( sx + cy ) + r( cx sy )θ = (cx + sy )( sx + cy ) + r( sx + cy ) + ( cx sy )( sx + cy ) ( ) V (r) V (r) = r s x + c y r r = xy V (r) xy V (r) r r = 0. This is what we wanted to prove. 5

6 4. Inverse square law and its consequences. We now assume that the force field is F(x) = km x r 2 r where k > 0 is a constant. This corresponds to the potential Notice that the magnitude of the force is U(r) = km r. and is directed towards the center, namely the origin. It is thus attractive. km r 2 We have already computed r, θ. But the equations of motion are of the second order in time and so we must calculate r, θ. Actually it is enough to calculate r. The calculation is a little tedious but elementary. No new principles are involved. The equations of motion are We have x = k r 2 x r, y = k r 2 y r. r = cx + sy, rθ = sx + cy where we continue to write c, s for cos θ, sin θ respectively. Then Thus we get r = sθ (cr sθ r) kc 2 r 2 + cθ (sr + cθ r) ks 2 r 2 = rθ 2 k r 2. r rθ 2 = k r 2. The important step is now to go over to the variable z = 1 r and also use θ as the independent variable rather than t. In view of Kepler s second law we assume that r 2 θ = h 6

7 wher h is a constant. Then Hence r = 1 z 2 z = r 2 θ dz dθ = hdz dθ. r = h d2 z dθ 2 θ = h 2 z 2 d2 z dθ 2. Therefore the equation for r becomes an equation for z viewed as a function of θ. We have r rθ 2 = h 2 z 2 d2 z dθ 2 rh2 z 4 = kz 2 leading to d 2 z dθ 2 + z = k h 2. This is a classical equation with solutions z = A sin θ + B cos θ + k h 2. Here A and B are constants depending on the initial conditions. So far we have not really chosen the line θ = 0. Now the distance r to the orbit is a minimum at some point and we choose this point to have coordinate θ = 0. But this point is then a maximum for z and so we have dz θ = 0, d 2 z < 0 when θ = 0. dθ2 From the formula for z these conditions imply that Hence the orbit has the equation A = 0, B > 0. l r = 1 + e cos θ, l = h2 k, e = Bh2 k. It is well known that this is an equation of a conic with eccentricity e and one focus at the origin. 7

8 We shall now compute the energy of the orbit. Recall that the energy is the sum of the kinetic energy and the potential energy. The potential U is the potential energy and so, writing E for the energy, Using the formulae for x, y we get E = 1 2 m(x 2 + y 2 ) km r. E = 1 2 m(r 2 + r 2 θ 2 ) km r = 1 2 mr mh2 r 2 km r where we write h = r 2 θ which is a constant. Now the energy is constant on the orbit (see the homework problem) and so we can calculate this at any point of time. Let us calculate this at the instant when θ = 0. At that point r is a minimum and so r vanishes; moreover the relation between r and θ gives Hence A simple calculation gives Thus r = h2 /k 1 + e cos θ = h2 /k 1 + e. E = 1 2 mh2 r 2 km r, r = h2 /k 1 + e. E = mk2 h 2 (e2 1). ( 2h 2 e = 1 + E mk 2 Now the orbit is an ellipse, parabola, or a hyperbola according as e < 1, = 0, > 1. Thus the orbit is an ellipse, parabola, or a hyperbola, according as ). E < 0, = 0, or > 0. For the planets we have periodic orbits and so the planetary orbits must be ellipses. This is Kepler s first law. By a closer analysis one can also establish Kepler s third law. The hyperbolic orbits are, for example, those of comets. Halley s comet is a rare example of a comet with a periodic orbit. We have treated the sun as a point mass fixed at the origin. This is not strictly true, but the sun s motion is negligible when we are treating the motion of the earth 8

9 around the sun because the sun is so much more massive than the earth. Also it can be shown that the gravitational field of the sun is the same as if its entire mass is concentrated at its center. This was proved by Newton and it allowed him to treat the motion as that of point particles. It can aslo be shown that the first and second law of Kepler already imply that the force must obey the inverse square law. This was also Newton s discovery and the starting point of his great investigations. 9