FROM NEWTON TO KEPLER. One simple derivation of Kepler s laws from Newton s ones.

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1 italian journal of pure and applied mathematics n ( ) 393 FROM NEWTON TO KEPLER. One simple derivation of Kepler s laws from Newton s ones. František Mošna Department of Mathematics Technical Faculty Czech University of Life Sciences, Prague Czech Republic mosna@tf.czu.cz Abstract. There is plenty of ways how to deduce Kepler s laws of planetary motion from Newton s ones (law of universal gravitation and law of motion). We offer one of them which is very simple and direct. It uses only mathematical tools and is suitable for teaching purposes.. Kepler s and Newton s laws Johannes Kepler (57 630) deduced his laws of planetary motion due to the exact measurement of planets performed by astronomer Tycho Brahe (546 60). The first and the second law were derived in Prague and published in work Astronomia Nova in 609. The third one was formulated in Linz and published Harmoniae Mundi in 69. Sixty five years later (in 684), Isaac Newton (643 77) deduced his law of universal gravitation from Kepler s laws. Plenty of literature deals with the relation between Kepler s and Newton s laws. The most of them use physical concepts, e.g. [] and []. The geometrical way of derivation was discussed in Richard Feynmann lecture presented on 3th March 964 at Caltech (its manuscript had been lost and it was found again in 99 at the office of author s colleague Leighton) and it is very interesting [3]. In this text, we would like to present one way how to derive Kepler s laws from the Newton s law of universal gravitation and motion. The aim and advantage of our deduction of especially the first one is its simplicity, directness, straightness and using only mathematical tools without introducing other physical variables. It can be presented maybe even in secondary schools.

2 394 f. mošna. Formulation of the laws Let us formulate this laws at first. Kepler s laws of planetary motion: (K) The orbit of every planet is an elipse with the Sun at one of the two foci. (K) A line joining a planet and the Sun sweeps out equal areas during equal intervals of time, i.e. area velocity is constant. (K3) Quotient T 3 a is constant, where T is orbital period of any planet and a is the major semi-axis of its orbit. Newton s laws of universal gravitation and the second law of motion: (NG) Any point mass M attracts every single other point mass m situated in the distance r from M by a force F = κ mm r, where κ = Nm /kg (acting in direction of radius-vector of both points). (N) The acceleration a of a body is parallel and directly proportional to the net force F acting on the body, F = am, where m is mass of a body. 3. Derivation of Kepler s laws At first, we will describe the situation in Figure. In the beginning, the point mass m (planet) in the distance from the point mass M (the Sun) has the velocity v 0 in the direction perpendicular to the radius-vector of both points (M and m). Figure : Situation and start condition Let the point mass M be situated to origin of axes and let point mass m be described by the position vector r(t) = (x(t), y(t), z(t)). According to (NG),

3 from newton to kepler 395 the force by which the point mass m is attracted to the point mass M can be expressed by the formula By (N) it holds F = κ mm r 3 m r r = k, where k = κm. () r 3 F = m r. () From these two relations () and () we obtain equation r + k r = 0. (3) r Trajectory is a plane curve At first, we derive that the trajectory of point mass m is situated in some plane containing the point mass M. This fact will be proved if we show that the velocity r of the point m is situated in the same plane as radius-vector r of that point, i.e., when r r is a constant vector. Let us calculate the derivation of this cross product using relation (3) d dt ( r r) = r r + r r }{{} =0 = k ( r r) = 0. r 3 That is why the cross product r r is really constant. So, it is proved that the motion of a point in so called central-force field is a plane curve. 3.. Trajectory is an ellipse (. Kepler law) We can assume for further calculation that the motion is situated in the plane xy and we shall describe this motion by position vector r(t) = (x(t), y(t)) and r(0) = (, 0) and r(0) = (0, v 0 ) in consistency with initial conditions. We transform the position vector to the polar coordinates r and ϕ We can differentiate twice r = (x, y) = (r cos ϕ, r sin ϕ). r = (ẋ, ẏ) = ṙ(cos ϕ, sin ϕ) + r ϕ( sin ϕ, cos ϕ), r = (ẍ, ÿ) = ( r r ϕ )(cos ϕ, sin ϕ) + (ṙ ϕ + r ϕ)( sin ϕ, cos ϕ) and rewrite the initial conditions ṙ(0) = 0 and ϕ(0) = v 0. This derivatives can be put into (3) and we obtain ( r r ϕ + k )(cos ϕ, r sin ϕ) + (ṙ ϕ + r ϕ)( sin ϕ, cos ϕ) = 0. r

4 396 f. mošna From here, we have the equations r r ϕ + k = 0 and ṙ ϕ + r ϕ = 0. (4) r We multiply both sides of the second part of (4) by r and get 0 = rṙ ϕ + r ϕ = d dt (r ϕ), and so, r ϕ is a constant. The initial conditions give us We put this result into the first part of (4) ϕ = v 0 r. (5) r v 0 r 3 + k r = 0 (6) and we deal similarly with this equation (6). We multiply both sides of it by ṙ and we obtain 0 = ṙ r r0v 0 ṙ r + k ṙ 3 r = d ( ṙ + dt v0 r k ). r This derivative is equal to zero and it again proves that ṙ + r0v 0 r k r is a constant. With respect to the initial conditions we get We can express the derivative ṙ from (7) ṙ + r 0v 0 r k r = v 0 k. (7) ṙ = v 0 k v 0 r + k r. (8) Now, we would like to express ϕ as a function of r, so we are looking for the function g such that ϕ = g(r) (see Figure ). The chain rule for derivatives gives us dϕ dt (t) = dg dr (r(t)) dr dt dϕ (t) or ϕ(t) = (r(t)) ṙ(t). dr We put (5) and (8) into this relation and we obtain v 0 = dϕ r dr v0 k v0 + k r r. (9)

5 from newton to kepler 397 Figure : Functions We can rewrite the equation (9) into form where and the discriminant We shall solve (0) by integration dϕ dr = v 0 r Ar + Br + C, (0) A = v 0 k, B = k, C = r 0v 0 < 0 D = 4 ( k v 0) > 0. ϕ = v 0 Br + C arcsin C r D + c = arcsin kr v0 r (k v0) + c. The constant c = π can be got from initial condition, hence We continue the calculation and ϕ = arcsin r k v0 r ( r ) π k 0v0. cos ϕ = sin(ϕ + π ) = r k v0 r ( r ) k 0v0 r = This is an equation of conic section k v 0 ( r ). () k 0v0 cos ϕ r = p + ɛ cos ϕ, where p = k v 0 0 and ɛ = k v 0.

6 398 f. mošna Parameter ɛ = 0 corresponds to the circle (and to the formula about the first κm cosmic speed v = ), parameter ɛ = corresponds to the parabola (and κm to the relations concerning the second cosmic speed v = ). The elliptic trajectory remains for ɛ <, i.e., v 0 < v. So we got the. Kepler law from here Area velocity is constant (. Kepler s law) In order to derive the. Kepler law, we must express the area, which is swept out by radius-vector of point mass m between time t and t in polar coordinates, t = t t. We use again the expression of r by ϕ, i.e., r = h(ϕ), function h is inverse to the function g (see Figure ). Figure 3: Area of radius-vector Let us calculate the area of Ω. We use step by step the translation to the polar coordinates (see Figure 3), the Fubini theorem and the substitution ϕ = ϕ(t) Ω = dxdy = r drdϕ Ω D ( ϕ ) h(ϕ) ϕ = r dr dϕ = ϕ 0 ϕ h (ϕ) dϕ t t = t h (ϕ(t)) ϕ(t) dt = t r (t) ϕ(t) dt. Now, we use relation (5) and get Ω = t t v 0 dt = v 0 (t t ) = v 0 t, () which gives us the. Kepler law about the constant area speed of radius-vector of motion.

7 from newton to kepler Relation between period and semi-major axes (3. Kepler s law) The 3. Kepler s law is a simple consequence of obtained results. We shall use () for t = 0 and t = T, where T is a period of motion of point mass m, and we can compare it with formula of ellipse area with semi-axes a and b πab = v 0 T. (3) The relations a = p and b = a ɛ ɛ (4) hold for parameters p, ɛ used for expression of ellipse in polar coordinates and semi-axes a, b used in cartesian coordinates. From (4) we have b = ap and (3) gives us πa ap = v 0 T. Because p = k v 0r 0, we have a πa k v 0 = v 0 T and from here we obtain which gives the 3. Kepler s law Mathematical apparatus T a 3 = 4π k, Finally, we summarize the mathematical apparatus used in our deductions:. properties of vectors, the dot (scalar, inner) product and cross (vector) product;. equation of ellipse x a + y = and formula of ellipse area P = abπ, where b a and b are semi-axes; 3. polar coordinates x = r cos ϕ, y = r sin ϕ; 4. equation of conic section in polar coordinates r( + ɛ cos ϕ) = p, where p is parameter and ɛ is angular eccentricity, conic section is a hyperbola for ɛ >, parabola for ɛ =, ellipse for ɛ <, (circle for ɛ = 0); relations b = a ( ɛ ), p = a( ɛ ), where a, b are semi-axes of ellipse; 5. chain rule for derivative, derivative of cross product; 6. integral dx x Ax + Bx + C = Bx + C arcsin C x D + c, where C < 0, D = B 4AC > 0;

8 400 f. mošna 7. double integral, its property, Fubini theorem and substitution theorem; 8. formula of ellipse area P = abπ. References [] Arnold, V.I., Mathematical Methods of Classical Mechanics, Springer, New York, 989. [] Sommerfeld, A., Mechanics, Academic Press, New York, 95. [3] Goodstein, D.L., Goodstein, J.R., Feynman s lost lecture, Norton, New York, 996, 999. Accepted:

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