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Home , Kepler orbit

PLANETARY

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1. History was an exact science since antiquity. Hipparchos of Nicea (160-127 b.c.), had accurate astronomical tables of planetary . Among his feats are that 1 he corrected the common estimates length of the of the year of 365 4 days to an estimate that is accurate to few minutes. He also he measured the distance to the moon to be about 60 earth radii which shows that 1600 before Columbus, scholars not only knew that earth was round, but also had good estimates of its radius. That earth was round was in fact known to Aristo, 350 bc, based on the shadow it casts on the moon at eclipses. (1571-1630), building on the helliocentric hypothesis, took the astronom- ical data about planetary orbits and summarized them in three simple laws. His first law is that planetary orbits are with the sun at the of the . His second law is that the planetary orbit sweeps equal areas at equal times, and the third law says that T 2, with T the period of the orbit, is proportional to R3 with R the semi-major axis. The third law is a relation between the orbits of different planets while the first two are relations for each individual planetary orbit. It is interesting to note that Kepler did not know the distance to the sun, nor to any other planet, in absolute units, (say in terms of the earth radius). Kepler measured distances in the astronomical unit, which uses the earth sun distance as a yardstick. In some ways the first law is the most remarkable, and was the greatest puzzle. Kepler found it using accurate data of Tycho Braha about the orbit of . Mars had always been the most fascinating planets to astronomers since its motion in the sky, when viewed from earth, changes direction (retrograde). To appreciate the accuracy and trust Kepler put into Braha observations, look at the figure of the planetary orbit of Mars, which I have drawn to scale. To the naked eye you can’t tell that this is not a circle. This is because the orbit of Mars is an ellipse with very small eccentricity. In fact, the ellipticity of most planets are quite small Venus Earth Mars Saturn Neptune Pluto 0.2056 0.0068 0.0167 0.093 0.0483 0.056 0.0097 0.2482 Newton (1642-1727) derived Kepler first law from his equation of motion F = ma and the assumption that the gravitation between the planet and the sun is GMm F(r) = − r r3

Date: December 4, 2006. I thank M. Berry and A. Mann. 1 2 YA

Figure 1. The ellipse of Mars, with eccentricity ² ≈ 0.1. To the naked eye the orbit looks like a circle. However, the sun, which is at one of the foci, is clearly not at the center. with r the vector from the sun to the planet, M and m their and G a universal constant. I do not know what brought Newton to think of the 1/r2 force law and I shall discuss some speculations below. The aim of this section is to derive the result the planetary orbits are ellipses.

2. Central The force between the sun and a planet is an example of a central force. More generally, a central force is of the form F(r) = f(r)r with r = |r|.

3. Notation Vectors are boldface: x. Hats are unit vectors: xˆ. Dots are time derivatives:x ˙

4. Reduced The first step is to reduce the problem of the motion of the sun and the motion of the planet to the relative distance. For two bodies interacting with forces that depend on relative distance and satisfy Newton third law F(x2 − x1) = −F(x1 − x2) Newton’s equation of motion is

m1x¨1 = F(x1 − x2), m2x¨2 = F(x2 − x1) = −F(x1 − x2) PLANETARY ORBITS 3

By adding the two equations (and multiplying by factor) we find that the center of mass m x + m x X = 1 1 2 2 M moves with constant speed. For the relative coordinate one finds µ¨r = F(r) where 1 1 1 r = x2 − x1, = + µ m1 m2 µ is known as the reduced mass. It is always smaller than the mass of the lighter object. Now the two body problem is reduced to an equation of motion for an effective one body.

5. Conservation of angular momentum Since r˙ k p and p˙ k F it follows that for central force p˙ k r. This implies conservation of angular momentum L = r × p L˙ = r˙ × p + r × p˙ = 0 + 0 and implies that the motion takes place in the plane perpendicular to L. This in turn implies Kepler second law.

6. Cylindrical coordinates The unit vectors θˆ and ˆr depend on the position of the particle. Since ˙ ˆr˙ = θ˙θ,ˆ θˆ = −θ˙ˆr The position, velocity and are r = rˆr, r˙ =r ˙ˆr + rθ˙θ,ˆ ¨r = (¨r − rθ˙2)ˆr + (2r ˙θ˙ + rθ¨)θˆ The component of the equation of motion in the θˆ direction reproduces conservation of angular momentum L = µr2θ˙zˆ Since r2θ˙ is the rate at which area is swept by the planetary orbit, conservation of angualr momentum implies Kepler’s second law.

Figure 2. Orbit in general central forces do not close. The Kepler case, where they do after one turn, is special. 4 YA

7. The radial motion Using the conservation of angular momentum in the ˆr component of the equation of motion one gets

L2 (1) µr¨ − = f(r) µr3 Define the potential, as usual, by

f(r) = −U 0(r)

Conservation of energy for this one dimensional problem takes the form

µ L2 r˙2 + + U(r) = E 2 2µr2

8. Orbits Time derivatives of r and the derivative of r with respect to the angle are related by L r˙ = r0θ˙ = r0 µr2 Putting this in the equation of motions, 2, gives µ ¶ L L 0 L2 µ r0 − = f(r) µr2 µr2 µr3

The substitution 1 u = r gives µ −u00 − u = r2f(r) L2 In the special case k f(r) = − r2 a miracle happens. The right hand side is independent of r and equation for the orbit is formally like the equation of motion of a harmonic oscillator in a (fixed) gravitational field kµ u00 + u = L2 whose general solution is kµ³ ´ u(θ) = 1 − ² cos(θ − θ ) L2 0 with PLANETARY ORBITS 5

9. why 1/r2 law? Musings 9.1. Why did Newton pick 1/r2 as the basic force law? It appears that the significance of 1/r2 was “in the air”. The Russian mathematician, V. Arnold says that Newton actually owed this to Hooke but begrudged to acknowledge this debt. It is natural to expect that the force diminishes with distance, but why at this particular rate? One can motivate 1/r2 by something like dimensional analysis if one accepts Kepler third law. The argument runs as follows: In circular orbits r˙ = 0. A force k law − rn gives by the radial eq. 1 L2 k − = − µr3 rn which can be written in terms of the period (µωr2)2 k k − = ⇒ ω2rn+1 = µr3 rn µ Since k = GMm with m the mass of the planet, M the sun and G the universal , the right hand side m ³ m ´ GM = GM 1 + ≈ GM µ M a universal constat for all planets. The heaviest planet, Jupiter, is 1000 times lighter than the sun, so the right hand side is almost the same for all planets. Kepler third law implies n = 2. Ady Mann tells me that, according to hid high school teacher Newton figured 1/r2 from comparison of g on earth, which he new, and the earth gravitational pull on the moon. Newton did not know that mass of the earth, nor the value of G, but he did know that for 1/r2 law GM g = R2 If one assumes that the moon orbit due to the earth pull is approximately circular (a rather poor assumption, since the Sun’s pull can not be neglected) one gets ω2r3 = GM = gR2 hence, using Hipparrchos estimate of the distance to the moon as 60 earth radii (and the known value of the radius of the earth, and g), the month is computable from the fall of the apple as µ ¶ g R 3 ω2 = R r which is about right (with about 15% error).

10. Closed and open orbits There are two cases depending on whether |²| < 1 or not. Recall that r ≥ 0. If |²| < 1 then θ is allowed to takes any value from 0 to 2π and the orbit encircles the origin. If the inequality fails θ is confined to values so that cos θ is small enough. The orbit does not encircle the origin and is confined to a sector. 6 YA

10.0.1. Remark. The case ² = 1 is a special limiting case which turns out, corre- sponds to . We can always choose coordinates so that θ0 = 0 and 0 < ² then closed orbit can be written as kµ¡ ¢ (2) u(θ) = 1 − ² cos(θ) L2 0 ≤ ² < 1 describes ellipses, while ² > 1 . This proves Kepler first law which states that planetary orbits are ellipses with the sun at one of the foci.

11. Ellipses and hyperbolas Geometrically, ellipse (hyperbolas) is the locus of points whose sum (difference) of distances from the two foci is constant 2R. If the two foci are d apart then in polar coordinates the corresponding equations are p 2R = r(θ) ± (r(θ) sin θ)2 + (r(θ) cos θ − d)2

Rearranging and squaring this equation to get rid of the root gives an equation for either and ellipse or a µ ¶ R 4R2 d 1 = 1 − cos θ = (1 − ² cos θ) r(θ) 4R2 − d2 2R 1 − ²2

d For an el1lipse R > 0 and ² = 2R < 1, is the eccentricity of the ellipse. For an hyperbola R can have any sign and ² > 1.

11.0.2. Example: Write the equation for ellipses/hyperbolas in cartesian coordi- nates. Solution:

¡ ¢2 R(1 − ²2) = r(θ) − ²x ⇒ x2 + y2 = R(1 − ²2) + ²x

Rearranging and deviding by (1 − ²2) gives

y2 (x − ²R)2 + = R2(1 + ²2) 1 − ²2 The canonical form of an ellipse in cartesian coordinates is

x2 y2 + = 1 a2 b2 Hence a2 = R2(1 + ²2), b2 = R2(1 − ²4) It follows that b2 = 1 − ²2 a2 PLANETARY ORBITS 7

12. Ellipticity of Planets Musings 12.1. If you look at the figure it looks remarkable that Kepler identified the orbit of Mars as an ellipse rather than a circle: The elliptical orbit of Mars is almost indistinguishable from a circle: eccentricity of 0.1 means that the ellipse differs from a circle by ²2 which is a 1% effect. Kepler could tell the orbit is an ellipse because the difference between the nearest and farthest distances from the sun is first order in ², which is a 10% effect. You can see in the figure that the focus is not at the center. Kepler took Barah data seriously, and presumably insisted on having the sun at a special point, and not eccentric to that orbit.

13. Eccentricity From the equation of the ellipse at θ = 0 1 kµ (3) = (1 − ²) r(0) L2 Conservation of energy can be written as µ L2 k (r0θ˙)2 + − = E 2 2µr2 r Since r0 = 0 at θ = 0, π we can substitute and find a relation that fixes the eccentricity of the ellipse in terms of the energy k2µ ¡ ¢ 1 − ²2 = −E 2L2 Hence E negative says that ² < 1 and the orbit is an ellipse. If E is positive ² > 1 and the orbit is a hyperbola. From the equation of motion, Eq. 2, we determine R L2 1 k R = = kµ 1 − ²2 −2E The energy alone fixes R. The (R,E) relation, being independent of eccentricity, is the same as one gets for a circular orbit. In the case of an with fixed angular momentum the energy is min- imized for circular orbit ² = 0. This can be used to argue why in old planetary systems, where energy is dissipated in a way that conserves total angular momen- tum, the orbits should be almost circular.

14. Periods and Kepler’s third law

Let r± be the nearest and farthest points of the planet from the sun. Then the period is Z Z r+ dr r+ rdr T = 2 = 2µ p 2 2 2 r− r˙ r− 2Eµr − 2µr U(r) − L k In the case U(r) = − r , and for E < 0, this reduces to r Z 2µ r+ rdr T = p −E r− (r − r−)(r+ − r) 8 YA

14.1. Integral. I have asked Mathematica to do the definite integral Z r+ rdr p = h(r−, r+) r− (r − r−)(r+ − r) The result is h(r−, r+) = π(r− + r+) 1 It is remarkable that the integral depends only on (r− + r+). To see that this. is the case without any explicit computation, let

2c = (r− + r+), 2d = r+ − r− then Z Z Z Z r+ rdr d (x + c) dx d x dx d dx p = √ = √ + c √ 2 2 2 2 2 2 r− (r − r−)(r+ − r) −d d − x −d d − x −d d − x The first integral vanishes since the integrand is anti-symmetric. The second in- tegral, by scaling, is just a number (independent of d) since by the substitution x = dt Z Z d dx 1 dt √ = √ 2 2 2 −d d − x −1 1 − t 14.2. Third law for ellipses. Putting together gives r r r 2µ r + r 2µ µ T = π − + = π R = 2π R3/2 −E 2 −E k The mass of the planet dropped from this relation, which is now universal. It states that the square of the period in proportional to the cube of the major axis, with the same constant of proportionality for all planets in our .

15. Laplace-Runge-Lenz All time independent problems have the energy as a constant of motion, and all central force problems have angular momentum conserved. The has yet another, special, constant of motion discovered by Laplace and popularized by Runge and Lenz: The vector connecting the two focal points of the ellipse. We shall denote this vector by e for the eccentricity. We want to find an expression for e. One should be able to express e as a linear combination of two linearly independent vectors in the plane. One is r and another is p. The latter is not quite the right object since an orbit and its time reverse both share the same eccentricity. So, the expression for e should involve only objects that are even under time reversal. r is, but p is not. We therefore replace p by L × p and the former by ˆr since the eccentricity is dimensionless. We therefore write (4) e = ˆr + αL × p and we can find α from the condition that e = 0 for a circular orbit. For a circular orbit (rθ˙)2 k (5) µ = r r2

1I owe this argument to M.V. Berry PLANETARY ORBITS 9

Substituting this in the equation e = 0 gives (6) 0 = 1 − α µ2(r2θ˙)(rθ˙) = 1 − αµk and α is a constant. This is, of course, not a proof that e is a constant of motion. But, it gives us a candidate. We can check that it is by differentiating and substituting the equation of motion: 1 ³ 1 ´ 1 ³ ´ (7) e˙ = ˆr˙+αL×p˙ = r2r˙−r (r˙·r)+ L×r = r2r˙−r (r˙·r)+(r×r˙)×r = 0 r3 µ r3 Since 2r˙ · r = (r˙2) = (r˙2) = 2rr˙ and we have used the bac minus cab rule: A × (B × C) = B(A · C) − C(A · (B) 15.0.1. Exercise. Write the LRL vector in cylindrical coordinates p × L = µ2 (r2θ˙) (r ˙ˆr + rθ˙θˆ) × zˆ = µ2 (r2θ˙)(−r˙θˆ + rθ˙ˆr) Hence A = µ2 (r2θ˙)(−r˙θˆ + rθ˙ˆr) − µ kˆr Since A is constant, we may evaluate it for any point on the orbit. Take the point which is furthest away from the focal point, where r˙ = 0. Using this fact and the equation 3 µ ¶ L2 A = µ(µ r3θ˙2 − k)ˆr = − µk ˆr = −kµ² ˆr r(0) So, up to factors, the vector is related to the eccentricity.

16. Kepler orbits from Runge-Lenz Using the fact that Runge-Lenz is a conserved it is a very easy exercise to get the Kepler orbit: (8) e r cos θ = e · r = r + αL × p · r = r − α`2 which is rearranged to the equation of the ellipse

`2 (9) = µk (1 − e cos θ) r

References 1. Abell, Morrison, Wolff Exploration of the Universe

Faculty of Physics, Technion, Haifa E-mail address: [email protected]