Physics 115/242 The Kepler Problem

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1 Physics 115/242 The Kepler Problem Peter Young (Dated: April 21, 23) I. INTRODUCTION We consider motion of a planet around the sun, the Kepler problem, see e.g. Garcia, Numerical Methods for Physics, Sec by Let the planet have mass m and velocity v, and the sun have mass M. The total energy is given E = 1 2 mv2 m, (1) r where r is the distance from the sun to the planet. The energy is conserved, as is the angular momentum: L = m r v. (2) The motion is in a plane so the only non-zero component is The force on the planet is given by L z L = m(xv y yv x ). (3) The negative sign indicates that this is an attractive (inwards) force. F( r) = m r 2 ˆr. (4) II. CIRCULAR MOTION For the special case of circular motion the sum of the gravitational force in Eq. (4) and the centripetal force mv 2 circ /r circ is zero, i.e. so mv 2 circ = m, (5) r circ v circ = r 2 circ r circ. (6)

2 2 It is easy to check that the potential energy is minus twice the kinetic energy so the total energy is E circ = 1 m. (7) 2 r circ The angular momentum, L circ, and period of the circular orbit, T circ, are given by L circ = mv circ r = m r circ (8) T circ = 2π r circ v circ = 2π r 3/2 circ. (9) III. THE GENERAL CASE: ELLIPTICAL MOTION In general the motion is not circular but an ellipse with the sun at one focus, see the figure below. y 2a 2b r θ x q If a and b are the semi-major and semi-minor axes (b a) then the eccentricity, ǫ, is defined to be ǫ = 1 b2 a2. (10) We now quote without proof that the formula for the ellipse is r(θ) = a(1 ǫ2 ) 1 ǫcosθ. (11) Hence, the perihelion (distance at closest approach, θ = π) is given by q = (1 ǫ)a, (12) while the aphelion (distance when furthest away, θ = 0) is given by = (1+ǫ)a. (13)

3 3 We also quote without proof that the eccentricity, angular momentum and energy are related by ǫ = 1+ 2EL2 G 2 M 2 m3. (14) From Eqs. (7) and (8) we see that Eq. (14) correctly gives ǫ = 0 for a circular orbit. One can also show that the period is related to the semi-major axis a by [ T = 2π a3/2 = 2π ( ) ] 3/2, (15) 1+ǫ which is known as Kepler s third law. Furthermore, the energy can also be expressed in terms of a by E = 1 m. (16) 2 a Comparing the general expressions for the energy and period in Eqs. (16) and (15) with those for a circular orbit in Eqs. (7) and (9), we see that they are the same except that r circ is replaced by a in the general case. IV. RELATION TO SPEED AT APHELION In the numerical problems in the homework, we start the planet at the aphelion with speed v i, in a direction perpendicular to that of the star, see the figure below. y v i x q Hence we have L = mv i, E = 1 2 mv2 i m. (17a) (17b)

4 4 We would like to express the eccentricity in terms of v i. To do this, we define v circ to be the speed of a circular orbit of radius, i.e. from Eq. (6), so E in Eq. (17b) can be written as v circ =, (18) E = 1 2 mv2 i mv 2 circ. (19) From Eqs. (17a), (18) and (19), the expression for the eccentricity in Eq. (14) can be written as ǫ 2 = 1+ 2EL2 G 2 M 2 m 3, = 1+ ( v 2 i 2v 2 circ) v 2 i = v 4 circ Hence the eccentricity can be written in the following simple form ( ) 2 1 v2 i vcirc 2. (20) ǫ = 1 v2 i vcirc 2, (21) where we have assumed v i v circ (so that the planet starts at the aphelion rather than the perihelion). Similarly, from Eqs. (11) and (13) we can write and, from Eqs. (9) and (15), r(θ) = 1 ǫ 1 ǫcosθ, (22) T T circ = where T circ is the period of a circular orbit of radius, i.e. from Eq. (9) 1, (23) (1+ǫ) 3/2 T circ = 2π 3/2. (24) Remember that is the aphelion, see the figure.

5 5 V. NUMERICS that In the numerical calculations we set m = M = G = = 1. From the above results it follows v circ = 1, T circ = 2π, ǫ = 1 v 2 i, T = 2π (2 v 2 i )3/2 = r = 1 ǫ 1 ǫcosθ. 2π (1+ǫ) 3/2, (25a) (25b) (25c) (25d) (25e) Orbits of different eccentricity are produced by different values of v i in the region 0 < v i 1. For v i = 1 the orbit is circular (ǫ = 0). If we were to set set v i = 0 (ǫ = 1) the planet would fall into the sun and crash. For ǫ 1, the orbit is close to circular and the speed is nearly constant. However, for ǫ 1, the ellipse becomes more and more squashed, and the magnitude and direction of the velocity change rapidly when the planet approaches the sun. This requires a smaller time step to preserve accuracy. In fact, for ǫ 1, it would be better to use adaptive step-size control so the time-step would be automatically reduced when the planet gets close to the sun and increased again when the planets moves away. (However, adaptive step-size control will not be needed for the homework.)

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