1. Sect. 13.3: Kepler’s Laws & Planetary Motion

2. German astronomer (1571 – 1630) Spent most of his career tediously analyzing huge amounts of observational data (most compiled by Tycho Brahe) on planetary motion (orbit periods, orbit radii, etc.) Used his analysis to develop “Laws” of planetary motion. “Laws” in the sense that they agree with observation, but not true theoretical laws, such as Newton’s Laws of Motion & Newton’s Universal Law of Gravitation. Johannes Kepler

3. Kepler’s “Laws” are consistent with & are obtainable from Newton’s Laws Kepler’s First Law –All planets move in elliptical orbits with the Sun at one focus Kepler’s Second Law –The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals Kepler’s Third Law –The square of the orbital period of any planet is proportional to the cube of the semimajor axis of the elliptical orbit Kepler’s “Laws”

4. The points F 1 & F 2 are each a focus of the ellipse –Located a distance c from the center –Sum of r 1 and r 2 is constant Longest distance through center is the major axis, 2a a is called the semimajor axis Shortest distance through center is the minor axis, 2b b is called the semiminor axis Math Review: Ellipses  Typical Ellipse The eccentricity is defined as e = (c/a) –For a circle, e = 0 –The range of values of the eccentricity for ellipses is 0 < e < 1 –The higher the value of e, the longer and thinner the ellipse

5. The Sun is at one focus –Nothing is located at the other focus Aphelion is the point farthest away from the Sun –The distance for aphelion is a + c For an orbit around the Earth, this point is called the apogee Perihelion is the point nearest the Sun –The distance for perihelion is a – c For an orbit around the Earth, this point is called the perigee Ellipses & Planet Orbits

6. All planets move in elliptical orbits with the Sun at one focus A circular orbit is a special case of an elliptical orbit –The eccentricity of a circle is e = 0. Kepler’s 1 st Law can be shown (& was by Newton) to be a direct result of the inverse square nature of the gravitational force. Comes out of N’s 2 nd Law + N’s Gravitation Law + Calculus Elliptic (and circular) orbits are allowed for bound objects –A bound object repeatedly orbits the center –An unbound object would pass by and not return These objects could have paths that are parabolas (e = 1) and hyperbolas (e > 1) Kepler’s 1 st Law

7. Fig. (a): Mercury’s orbit has the largest eccentricity of the planets. e Mercury = 0.21 Note: Pluto’s eccentricity is e Pluto = 0.25, but, as of 2006, it is officially no longer classified as a planet! Fig. (b): Halley’s Comet’s orbit has high eccentricity e Halley’s comet = 0.97 Remember that nothing physical is located at the second focus –The small dot Orbit Examples

8. Kepler’s 2 nd Law can be shown (& was by Newton) to be a direct result of the fact that N’s Gravitation Law gives Conservation of Angular Momentum for each planet. The Gravitational force produces no torque (it is  to the motion) so that Angular Momentum is Conserved: Kepler’s 2 nd Law The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals 

9. Geometrically, in a time dt, the radius vector r sweeps out the area dA = half the area of the parallelogram The displacement is dr = v dt Mathematically, this means That is: the radius vector from the Sun to any planet sweeps out equal areas in equal times Kepler’s 2 nd Law

10. If the orbit is circular & of radius r, this follows from Newton’s Universal Gravitation. This gravitational force supplies a centripetal force for user in Newton’s 2 nd Law    K s is a constant Kepler’s 3 rd Law The square of the orbital period T of any planet is proportional to the cube of the semimajor axis a of the elliptical orbit K s is a constant, which is the same for all planets.

11. Can be shown that this also applies to an elliptical orbit with replacement of r with a, where a is the semimajor axis. K s is independent of the planet mass, & is valid for any planet Note: If an object is orbiting another object, the value of K will depend on the mass of the object being orbited. For example, for the Moon’s orbit around the Earth, K Sun is replaced with K Earth, where K Earth is obtained by replacing M Sun by M Earth in the above equation. Kepler’s 3 rd Law

12. Table 13-2, p. 370 Solar System Data

13. “Weighing” the Sun! We’ve “weighed” the Earth, now lets “weigh” the Sun!! Assume: Earth & Sun are perfect uniform spheres. & Earth orbit is a perfect circle. Note: For Earth, Mass M E = 5.99  10 24 kg Orbit period is T = 1 yr  3  10 7 s Orbit radius r = 1.5  10 11 m So, orbit velocity is v = (2πr/T), v  3  10 4 m/s Gravitational Force between Earth & Sun: F g = G[(M S M E )/r 2 ] Circular orbit is circular  centripetal acceleration Newton’s 2 nd Law gives: ∑F = F g = M E a = M E a c = M E (v 2 )/r OR: G[(M S M E )/r 2 ] = M E (v 2 )/r. If the Sun mass is unknown, solve for it: M S = (v 2 r)/G  2  10 30 kg  3.3  10 5 M E