Ministry of Higher Education & Scientific Research University of Salahaddin-Erbil

1. Ministry of Higher Education& Scientific Research University of Salahaddin-Erbil College of Science Department of Physics 2nd Year Physics Subject: Analytical Mechanics Chapter 7: Kepler’s laws of planetary motion Asst. Prof. Dr. Tahseen G. Abdullah

2. Name Distance from Sun(AU) Revolution Period(Years) Diameter(km) Mass(1023 kg) Density(g/cm3) Mercury 0.39 0.24 4878 3.3 5.4 Venus 0.72 0.62 12102 48.7 5.3 Earth 1.00 1.00 12756 59.8 5.5 Mars 1.52 1.88 6787 6.4 3.9 Jupiter 5.20 11.86 142984 18991 1.3 Saturn 9.54 29.46 120536 5686 0.7 Uranus 19.18 84.07 51118 866 1.2 Neptune 30.06 164.82 49660 1030 1.6 Pluto 39.44 248.60 2200 0.01 2.1 Principle Characteristics of the Planets

3. Kepler’s first law Planets follow elliptical orbits, with the Sun at one focus of the ellipse.

4. Orbit Examples • planet’s orbit the Sun in ellipses, with the Sun at one focus. • the eccentricity of the ellipse, e, tells you how elongated it is. • e=0 is a circle, e<1 for all ellipses e=0.02 e=0.4 e=0.7 eccentricity of the planets • Pluto has the highest eccentricity of any planet • ePluto = 0.25 • Halley’s comet has an orbit with high eccentricity • eHalley’s comet = 0.97

5. Notes About Ellipses Planet Orbits • The Sun is at one focus • Nothing is located at the other focus • Aphelion (afelion) 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 (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 (perigee)

6. Kepler’s second law The line joining the Sun and a planet sweeps out equal areas in equal time intervals. As a result, planets move fastest when they are near the Sun (perihelion) and slowest when they are far from the Sun (aphelion).

7. Same Areas Q. If the planet sweeps out equal areas in equal times, does it travel faster or slower when far from the Sun?

8. Law of Areas A line that connects a planet to the sun sweeps out equal areas in the plane of the planet’s orbit in equal time intervals; that is the rate dA/dt at which it sweeps out area A is constant Area A of the wedge is the area of the triangle i.e. A=(r2)/2; dA/dt = r2(d/dt)/2= r2/2 But angular momentum L=mr2 Then dA/dt= r2/2=L/2m In central field L is constant because in central field: Or in Fig.(b): dA=|r x dr|/2=….=L/2m

9. Example: Kepler’s Second Law and Angular Momentum Conservation. Consider a planet of mass Mp moving around the Sun in an elliptical orbit. Since the gravitational force acting on the planet is always toward radial direction, it is a conservative central force field. Therefore the torque acting on the planet by this force is always 0. Since torque is the time rate change of angular momentum L, the angular momentum is constant.

10. Because the gravitational force exerted on a planet by the Sun results in no torque, the angular momentum L of the planet is constant. Since the area swept by the motion of the planet is: This is Keper’s second law which states that the radius vector from the Sun to a planet sweeps out equal areas in equal time intervals.

11. Kepler’s Third Law • The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet. • T is the period of the planet • a is the length of the semi-major axis • For orbit around the Sun, K = KS = 2.97x10-19 s2/m3 • K is independent of the mass of the planet and K=1 in A.U. • 1 A.U. = 1.5 x 108 km G is the gravitational constant = 6.673 x 10-11 m3 kg-1 sec-2

12. T2αa3 Kepler’s Third Law

13. Some Important Points of Kepler’s Third Law • Approximately, T2earth/a3earth = T2planet/a3planet • Sometimes we use Earth-years and Earth-distance to the Sun (1 A.U.) as units. • The constant of proportionality depends on the mass of the Sun--and that’s how we know the mass of the Sun. • We can apply this to moons (or any satellite) orbiting a planet, and then the constant of proportionality depends on the mass of the planet. • Newton’s law of gravitation, and Newton’s second law (net force = mass x acceleration) can be used to derive Kepler’s three laws of planetary motion.

14. Can be predicted from the inverse square law Start by assuming a circular orbit The gravitational force supplies a centripetal force Ks is a constant Example: Derivation of Kepler’s Third Law from Newton’s law of gravitation, and Newton’s second law • This can be extended to an elliptical orbit • Replace r with a • Remember a is the semimajor axis • Ks is independent of the mass of the planet, and so is valid for any planet • If an object is orbiting another object, the value of K will depend on the object being orbited • For example, for the Moon around the Earth, KSun is replaced with KEarth

15. Calculate the mass of the Sun using the fact that the period of the Earth’s orbit around the Sun is 3.16x107s, and its distance from the Sun is 1.496x1011m. Example Using Kepler’s third law. The mass of the Sun, Ms, is

16. Orbital Paths • Extending Kepler’s Law, Newton found that ellipses were not the only orbital paths. • possible orbital paths • circle (bound) – e=0 • ellipse (bound) - e<1 • parabola (unbound) – e=1 • hyperbola (unbound) – e>1

17. Orbit of a particle in a central force field Angular momentum per unit mass (Keper’s second law)

18. To find the equation of the orbit from Eq.(1), we shall use the variable u defined by: Using the above variables Eq.(1) becomes:

19. Thus, the differentia equation of the orbit of a particle moving under a central field is:

20. Energy Equation of the orbit In a central force field the force is a function of r only and it is conservative. Thus, using the energy equation of the orbit of a particle moving under a central field is:

21. Ex.: A particle in a central field moves in the spiral orbit r=cΘ2. • Determine the form of the force function using differential equation of the orbit. • Determine the form of the force function using energy equation of the orbit. • Determine how the angle Θvaries with the time. (a)

22. (b)

23. (c) • Ex.: A particle in a central field moves in a logarithmic spiral orbit given by, r=k eα Θ where k and αare constants. • Determine the form of the force function using differential equation of the orbit. • Determine the form of the force function using energy equation of the orbit. • Determine how the angle Θvaries with the time.

24. Ex. (H.W.) (a) Orbits of a particle in an inverse square field. (b) Orbital energies of a particle in the inverse square field.

25. End of the Lecture Let Learning Continue Thank You

26. Kepler's Third Law - Examples • Suppose you found a new planet in the Solar System with a semi major axis of 3.8 A.U. • A planet with a semi major axis of 3.8 A.U. would have an orbital period of 7.41 years years

27. Kepler's Third Law - Examples • Suppose you want to know the semi major axis of a comet with a period of 25 years • A planet with an orbital period of 25 years would have a semi major axis of 8.55 A.U. A.U.

28. Nine Planets Kepler’s laws of planetary motion

33. If it sweeps out equal areas in equal times, does it travel faster or slower when it is far from the Sun?

34. If it sweeps out equal areas in equal times, does it travel faster or slower when far from the Sun?

35. Same Areas Answer: The planet travel slower when far from the Sun

36. Quick quiz • A planet has two moons with identical mass. Moon 1 is in a circular orbit of radius r. Moon 2 is in a circular orbit of radius 2r. The magnitude of the gravitational force exerted by the planet on Moon 2 is • four times as large • twice as large • the same • half as large • one-fourth as large as the gravitational force exerted by the planet on Moon 1.

37. Quick quiz • Suppose an asteroid has a semimajor axis of 4 A.U. How long does it take the asteroid to go around the sun? • (a) 2 years • (b) 4 years • (c) 6 years • (d) 8 years

38. Quick quiz

39. Example 8.1 – Log-spiral Find the force law for a central-force field that allows a particle to move in a logarithmic spiral orbit given by, where k and a are constants. Solution: Calculate: Now use To find: Force is Attractive and Inverse cube!

40. Example 8.2 – r(t), q(t) Solution: Start with: Determine the functions r(t) and q(t) for the problem in Ex 8.1. Rearrange, integrate: Answer:

41. Similarly for r(t), remember And write Answer (2): Where l and C are determined by the initial conditions

42. Example 8.3 – Total Energy Solution: Need U… What is the total energy of the orbit of the previous two examples?

43. Given the reference

44. Ch 13 Checkpoint 5 Satellite 1 is in a certain circular orbit around a planet, while satellite 2 is in a large circular orbit. Which satellite has (a) the longer period and (b) the greater speed T2=(42/GM)R3 Since R1<R2 then T1<T2 Longer period forsatellite 2; (b) K=mv2/2=GmM/2R v2=GM/R Greater v for smaller R i.e R1 the greater speed for satellite 1

45. Ch 13-8 Satellites Orbits and Energy • For an orbiting satellite, speed fixes its kinetic energy K and its distance from earth fixes its potential energy U. Then mechanical energy E (E = K+U) of the Earth-satellite system remains constant. • K=mv2/2 but mv2/R=GmM/R2 Then K=mv2/2=GmM/2R U=-GmM/R = -2K ; we have K=- U/2 and E=K+U= K+(-2K)=-K(circular orbit) • For an elliptical orbit R=a Then E=-K=-GmM/2a For same value of a, E is constant

46. iClicker question Earth is closer to the Sun in January than in July. Therefore, in accord with Kepler’s 2nd Law … (A) Earth moves at a higher orbital speed in July than in January; (B) Earth moves at a higher orbital speed in January than in July; (C) Since Earth’s orbit is circular with an eccentricity of zero, the original statement is false.

47. Question • A circular orbit has an eccentricity of _____. • A) exactly 0 • B) between 0 and 0.5 • C) exactly 1 • D) infinity