Tycho Brahe ( )

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1 Tycho Brahe ( ) Foremost astronomer after the death of Copernicus. King Frederick II of Denmark set him up at Uraniborg, an observatory on the island of Hveen. With new instruments (quadrant), Brahe could measure positions of stars and planets to 4 minutes of arc (1/8th the diameter of the full moon). With new instruments (quadrant, similar to present-day sextant), Brahe could measure positions of stars and planets to better than 4 minutes of arc (1/8th the diameter of the full moon). Made the most accurate observations of planet positions ever recorded.

2 Using the Quadrant (similar to a Sextant) to measure a star s altitude. An observer sights a star along the Quadrant while a plumb line measures the angle.

3 Tycho Brahe ( ) Also Credited with observing the supernova of 1572, demonstrating that the stars do change. He coined the term Nova for new star in a book. Term still used today. Brahe failed to find evidence for the Earth s motion and presumed the Copernican model was false. Brahe endorsed the Geocentric cosmology.

4 Johannes Kepler ( ) Brahe moved to Prague in Kepler joined him to develop cosmological model consistent with Brahe s observations. Kepler initially obtained excellent agreement with Brahe s data with the planets moving in spheres and equants, similar to Ptolemy s model, except for 2 data points which were off by 8 minutes of arc (twice the accuracy of Brahe s measurements). Kepler kept struggling and finally rejected Ptolemaic model and eventually realized that that Brahe s data were only consistent with a model where planets (1) orbit the Sun and (2) their orbits are ellipses.

5 Observed Planet and uncertainty on the measurement

6 Observed Planet and uncertainty on the measurement Position Predicted by Model

7 Johannes Kepler ( ) Kepler s Laws: published in 1609 in Astronomica Nova (The New Astronomy). 1st Law: A planet orbits the Sun in an ellipse, with the Sun at one focus of the ellipse. 2nd Law: A line connecting a planet to the Sun sweeps out equal areas in equal time intervals. 3rd Law: The square of the orbital period, P, of a planet is equal to the cube of the average distance, a, of the planet from the Sun.

8 Ellipses defines an ellipse Aphelion Sun Perihelion

9 Sir Isaac Newton ( ) Developed Universal theory of Gravity, and set his three Laws of Motion, which are framework for classical mechanics, including basis for modern engineering. -1st Law: An object at rest, remains at rest unless acted on by an outside force. An object with uniform motion, remains in motion unless acted on by an oustide force. - 2nd Law: An applied force, F, on an object equals the rate of change of its momentum, p, with time. - 3rd Law: For every action there is an equal and opposite reaction.

10 Start with Kepler s 3rd law: For circular orbit: Newton s Theory of Gravity Insert this into Kepler s 3rd law: Multiply both sides by factors: Yields: Force!

11 Newton s Theory of Gravity By Newton s 3rd law, this is the force on mass M from mass m, so the force on mass m should be: Equating these two formula gives: where Defining and we get the usual formula:

12 Derive Kepler s Laws from Newton s Laws Define Center of Mass Frame for 2 objects for i=1..n objects Differentiate with time, Differentiate with time, again: by Newton s 3rd Law

13 Derive Kepler s Laws from Newton s Laws Choose Frame with R=0 =0 Define Reduced Mass of the System : μ= m1 m2 / (m1+m2)

14 Derive Kepler s Laws from Newton s Laws Now write out total Energy of the System Insert: And : r = r 2 - r 1

15 Derive Kepler s Laws from Newton s Laws Total Energy Becomes: Grouping Terms : And: Which Gives: Total Energy of the System is the sum of the Kinetic Energy of the reduced mass and the potential energy of the total mass/reduced mass system.

16 Introduce Angular Momentum: Look at rate-of-change of L as a function of t: = 0, Conservation of Angular Momentum

17 Derive Kepler s Laws from Newton s Laws Similarly for the Angular Momentum: Insert: Leads to: The Two-Body problem may be treated as a One-body problem with μ moving about a fixed mass M at a distance r.

18 Johannes Kepler ( ) Kepler s Laws: published in 1609 in Astronomica Nova (The New Astronomy). 1st Law: A planet orbits the Sun in an ellipse, with the Sun at one focus of the ellipse. 2nd Law: A line connecting a planet to the Sun sweeps out equal areas in equal time intervals. 3rd Law: The square of the orbital period, P, of a planet is equal to the cube of the average distance, a, of the planet from the Sun. P 2 = Constant x a 3

19 Ellipses

20 Ellipses

21 Eccentricity

22 Eccentricity

23 Eccentricity Eccentricity=1 Parabolic orbit (goes through perihelion only once)

24 Kepler s First Law Each planet moves on an elliptical orbit with the Center of Mass (the Sun) at one focus: Perihelion distance = (1-eccentricity) x (semimajor axis)

25 Kepler s First Law Each planet moves on an elliptical orbit with the Center of Mass (the Sun) at one focus: Perilously close Perihelion distance = (1-eccentricity) x (semimajor axis)

26 Planets Follow Eliptical Orbits The Inner Solar System The Outer Solar System

27 Kepler s First Law Not just planets! Comets have elliptical orbits too.

28 Kepler s First Law Not just planets! Comets have elliptical orbits too. Halley s Comet (76 yr orbit)

29 Kepler s First Law Not just planets! Comets have elliptical orbits too. Halley s Comet in 2024 Halley s Comet (76 yr orbit)

30 Kepler s Second Law Equal Areas = Equal Times Conservation of Angular Momentum Near perihelion, in any particular amount of time (such as 30 days) a planet sweeps out an area that is short by wide. Near aphelion, in the same amount of time (such as 30 days) a planet sweeps out an area that is long by narrow. perihelion The areas swept out during any 30-day period are all equal

31 Kepler s Second Law Equal Areas = Equal Times Conservation of Angular Momentum The areas swept out during any 30-day period are all equal

32 Velocity Inversely Proportional to Orbital Radius

33 Kepler s Third Law Square of orbital periods (P) is proportional to cube of semi-major axis (a): P 2 a 3 Orbital Period squared (years 2 ) Mercury Earth Mars Venus Jupiter Saturn Semimajor axis Average distance cubed (AU 3 ) from Sun.

34 Kepler s Third Law

35 Orbital Period vs Orbital Radius

36 Derive Kepler s Laws from Newton s Laws The Two-Body problem may be treated as a One-body problem with μ moving about a fixed mass M at a distance r.

37 Derive Kepler s Laws from Newton s Laws Take cross product: A x (B x C) = (A C) B - (A B) C

38 Derive Kepler s Laws from Newton s Laws D is a constant ^ v x L and r lie in the same plane, so must D D is directed toward Perihelion, D determines e (eccentricity) of ellipse. Maximum reached with ^r and D point in same direction.

39 Derive Kepler s Laws from Newton s Laws A (B x C) = (A x B) C Kepler s 1st Law!

40 Derive Kepler s Laws from Newton s Laws Integrate from Focus to r: da = ½ r 2 dθ da dt = ½ r2 dθ dt Let : v = vr + vθ = dr ^ ^ dt r + r dθ dt θ da dt = ½ r vθ r and vθ are perpendicular, therefore : r vθ = r x v = L / μ = L / μ da L = Kepler s 2nd Law! dt 2μ = Constant

41 Derive Kepler s Laws from Newton s Laws Integrate 2nd Law with time: A = ½ L/μ dt A = ½ (L/μ) P, where P = period b P 2 = A = π ab 4π 2 a 2 b 2 μ 2 L 2 a L 2 / μ 2 1st Law: r = L = μ [ GMa (1-e 2 ) ] ½ GM (1 + e cosθ) Kepler s 3rd Law P 2 4π 2 a 3 = G (m1 + m2)

42 Calculate Orbital Speed at Perihelion and Aphelion Use Kepler s 1st Law At Perihelion Aphelion Sun Perihelion At Aphelion Total Energy:

43 Total Energy in a Bound Orbit Etot = -GMμ 2a = -Gm 1m2 2a = <U> / 2 That is the total energy is 1/2 of the average potential energy. This is the Virial Theorem. More in depth derivation by Rudolf Clausius (reproduced in book). It has many applications in astrophysics, including providing evidence for Dark Matter in Clusters of Galaxies! Rudolf Julius Emanuel Clausius (January 2, 1822 August 24, 1888), was a German physicist. He is one of the central founders of the science of thermodynamics.

44 High Tide in Bay of Bundy Low Tide

45

46 What Causes Ocean Tides?

47 Saturn as seen by Spacecraft Cassini, October 2004

48 Tidal Forces Force of Moon on m at center of Earth Force of Moon on m on surface of Earth Fc,x = -G Mm / r 2 Fc,y = 0 Fp,x = (G Mm/s 2 )cosϕ ΔF = Fp - Fc = G Mm s 2 = (r - Rcosθ) 2 + (Rsinθ) 2 = r 2 M= mass of Moon; m= test mass Fp,y = -(G Mm/s 2 )sinϕ cosϕ 1 G Mm i - s 2 r 2 ^ s 2 ( - ) ( ) sinϕ ^j ( 1-2Rcosθ ) R << r r

49 ΔF = Fp - Fc = G Mm Tidal Forces s 2 = (r - Rcosθ) 2 + (Rsinθ) 2 = r 2 cosϕ 1 G Mm i - s 2 r 2 ^ s 2 ( - ) ( ) 1-2Rcosθ r ΔF G Mm/r 2 [ cosϕ ( 2Rcosθ 1 + ) -1 ] ^ i r - G Mm/r 2 [ 1 + 2Rcosθ ] sinϕ j^ r ( sinϕ ^j ) R << r For the Earth-Moon, cosϕ 1, r sin ϕ = Rsinθ, sin ϕ = (R/r) sinθ ΔF G MmR/r 3 [ 2cosθ ^i - sinθ ^j ]

50 Tidal Forces ΔF G MmR/r 3 [ 2cosθ ^i - sinθ ^j ] Force of the Moon on the Earth Differential (relative) force on the Earth, relative to the center

51 Tidal Forces ΔF G MmR/r 3 [ 2cosθ ^i - sinθ ^j ] When will the tidal force equal the force of gravity? Consider the Earth-Moon system: Mm, Rm = mass of moon, radius of moon ME, RE = mass of Earth, radius of Earth G Mmm/Rm 2 = ( 2GMEm/r 3 ) Rm Assume constant density: Mm = (4/3)πRm 3 ρm r 3 = (2ME/Mm ) Rm 3 = (2RE 3 ρe) / (Rm 3 ρm) Rm 3

52 Tidal Forces G Mmm/Rm 2 = ( 2GMEm/r 3 ) Rm Assume constant density: Mm = (4/3)πRm 3 ρm r 3 = (2ME/Mm ) Rm 3 = (2RE 3 ρe) / (Rm 3 ρm) Rm 3 r = ( 2ρE ρm ) ⅓ RE or r = fr ( ρe ρm ) ⅓ RE, fr = 2 ⅓ = Roche, using a more sophisticated assumption for the density of the Moon calculated fr = 2.456

53

54

55

56

57 Comet Shoemaker-Levy 9 after passing through Jupiter s Roche Limit

58 Saturn as seen by Spacecraft Cassini, October 2004

59 Planetary Moons form when a moon crosses the Roche Limit Saturn Janus Epimetheus Rings Mimas Enceladas Tethys Other Moons Rings Dione Rhea Rings Roche Limit Distance from Saturn Center in RS

60

61

62 How do you know the Earth Rotates?

63 How do you know the Earth Rotates?

64 How do you know the Earth Rotates? In 1851, Léon Foucault proved the Earth s rotation directly. A pendulum swinging on the Earth feels the rotation due to the Coriolis Force. FC = -2 m Ω x v Versions of the Foucault Pendulum now swings in the Panthéon of Paris and the Houston Museum of Natural Science v=cpxdadtjx2s v=49jwbrxcpjc

65 How do you know the Earth Rotates?

66 How do you know the Earth Rotates? Coriolis Force affects rotation of weather patterns (cyclones rotate counter clockwise in southern hemisphere; hurricanes rotate clockwise in northern hemisphere).

67 How do you know the Earth Rotates? Hurricane Gustav Aug 31, 2008

68 How do you know the Earth Rotates? Hurricane Gustav Aug 31, 2008 Coriolis Force affects rotation of weather patterns (cyclones rotate clockwise in southern hemisphere; hurricanes rotate counter-clockwise in northern hemisphere).

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