Astr 2320 Tues. Jan. 24, 2017 Today s Topics Review of Celestial Mechanics (Ch. 3) Copernicus (empirical observations) Kepler (mathematical concepts) Galileo (application to Jupiter s moons) Newton (Gravity and Physics)
Copernicus: 1473 1543 Proposed heliocentric model Circular orbits and uniform motion Less accurate for predicting positions but more physically realistic Simple explanation for retrograde motion De Revolutionibus Orbium Coelestium published in 1543 Computed the scale of the Solar System relative to Earth s orbit (i.e., in AU) From: Horizons, by Seeds
Copernicus continued Copernicus also claimed: Planets were all round worlds, like the Earth Earth was just another planet Copernicus model was just an alternative model. It was simpler and elegant but there was no physical evidence. Proof came 100 years after his death.
Johannes Kepler: 1571 1630 The nature of planetary orbits Problem: Copericus heliocentric model just wouldn t fit the precise data from Brahe. Realized Mars orbit must be elliptical, not circular Everything now fit: 1) The orbits of all planets are elliptical with the Sun at one focus. 2) The planet moves faster near the Sun (equal areas in equal times). 3) P 2 ~ a 3
Kepler s Laws, #1 and #2 1609 Published two laws showing: #1 Planets orbit the sun in ellipses, with the Sun at one focus #2 Motion is faster when they are near the Sun, in such a way that a line from the planet to the sun sweeps out equal areas in equal times From our text: Horizons, by Seeds
Properties of Ellipses Ellipse defined by two constants semi-major axis a 1/2 length of major axis eccentricity e 0=circle, 1 = line Two ellipses with the same a but different e e=0 e=0.98 Same focus, at the sun
Kepler s Laws, #3 1619 Publishes third law, showing that there is a relationship orbital period and semi-major axis: Exact relationship is P 2 a 3. Outer planets orbit more slowly than inner ones Example: Earth P = 365 days, a = 1.00 AU. Mars p = 687 days, a = 1.524 AU 687 days 365 days 2 1.524 AU = 1.000 AU Orbital Period of some asteroid with a = 9 AU? 3 ( ) ( ) 3 P P asteroid Earth 2 3/ 2 a = a asteroid Earth a Asteroid PAsteroid = PEarth = 1year = aearth 1.88 2 = 1. 524 3.54 = 3. 54 3 ( 9) 3/ 2 3 = (3) 27 years
Kepler s Laws Continued Form (P 2 =Ka 3 ) of the law results from gravity so it is valid for any orbit: Units used determine K K = 1 if a is in AU and P is in years (Solar System units) For other units K must be computed. Moon and artificial satellites around the Earth Satellites around other planets Stars orbiting each other Stars orbiting the Galaxy
Galileo Galilei 1564-1642 Galileo s earlier work Invented Physics Developed concept of inertia and force 1590 Masses fall at same rate heavier do not fall faster (unless affected by air resistance) 1604 Observes a supernova (Kepler s), no parallax must be beyond the Moon Telescopes: 1609 Hears of invention of telescope, which at that point just use eyeglass lenses Works out details of better lenses and lens placement, builds improved ones himself
Galileo: First telescopic observations Sidereus Nuncius (The Starry Messenger) published in 1610 reporting: Moon isn t perfect (violating Aristotelian principles for heavens) Shows mountains and valleys» Uses shadows to estimate heights Milky Way made up of myriad faint stars Doesn t directly violate Aristotelian principles, but suggests that a few simple phenomena can explain many features of the heavens Discovers 4 moons (Galilean Satellites) orbiting Jupiter Violates idea that all motion is centered on the Earth Shows that orbiting objects can follow a moving body 4 moons will also be seen to follow Kepler s 3 rd law P 2 a 3 (but with a different proportionality constant)
Galileo s additional observations Detects sunspots and the rotation of the Sun. Further evidence of the imperfect heavens Detects the phases of Venus Phases show that Venus must orbit the Sun. Full Venus when it is on far side of Sun. Crescent Venus when it is on near side of Sun. From our text: Horizons, by Seeds
Galileo s critical observations Jupiter s moons show orbits which are not earthcentered and follow Kepler s laws (Gravity). Venus phases show it must circle the Sun Several objects (Moon, Sun) show imperfections which are not supposed to be present in the heavens Galileo s observations clearly support Copernican model, but so far his printed work has mostly been reporting what he sees, rather than directly arguing for Copernican model.
Galileo and the Dialog Written as a debate between 3 people Salviati Copernican advocate (really Galileo) Sagredo Intelligent but uninformed Simplicio Aristotelian philosopher not very bright Hoped to avoid earlier ruling by not directly advocating Copernican model Actually made things worse by convincing accusers they were Simplicio 1633 Inquisition condemns him for violating 1616 order Something like modern contempt of court ruling Proceeding not a re-argument of Copernican vs. Aristotelian debate But forced to recant, admitting errors Sentenced to life imprisonment actually house arrest Dies in 1642 Pope John Paul II finally makes some amends 350 years later.
Newton: 1642-1727 Principia published in 1687 3 Law of motion 1. A body continues at rest or in uniform motion in a straight line unless acted upon by some force. 2. A body s change of motion is proportional to the force acting on it and is in the direction of the force. 3. When one body exerts a force ( F = on maa ) second body, the second body exerts an equal and opposite force back on the first body. Universal gravitation There is an attractive force between all bodies, proportional to their mass, and inversely proportional to the square of F their distance. Mm = G G 2 r = 6.67 10 11 m 3 /(s 2 kg)
Explanation for Kepler s Laws Momentum keeps the planets moving you do not need some force to do this. Gravity provides the force which makes orbits curve Gravity of Sun curves orbits of Planets Gravity of Earth curves orbit of moon (and also makes objects on earth fall downward) Conservation of Angular Momentum explains why motion is faster when closer to the sun. The inverse square law of gravity explains P 2 a 3 and the details of why the orbits are ellipses.
Circular Orbits: Limiting case of an ellipse. Centripetal acceleration (v 2 /r) caused by Gravity 2 mv r Period found by Mm = G 2 r v = GM r Period = distance velocity 2π r = v 2π r GM r 2π GM Kepler s 3 rd Law just comes from this = = r 3/ 2 P 2 π = GM 2 4 r 3 From our text: Horizons, by Seeds Given P and a (and G) we can find the mass of a planet or star
Geometric Properties of the Ellipse FF ' = 2ae (definition of e) Consider triangle BcF: b 2 + a 2 e 2 = r 2 = a 2 (r + r ' = 2a) so: b 2 = a 2! a 2 e 2 = a 2 (1! e 2 ) b = a(1- e 2 ) 1/2 (relationship between b & a) Furthermore: R min = a! ae = a(1! e) R max = a + ae = a(1+ e) (distances at perihelion, aphelion) Applying law of cosines to F!PF gives: r '2 = r 2 + (2ae) 2 + 2r(2ae)cos! But since r ' = 2a - r we have: 4a 2! 4ar + r 2 = r 2 + 4a 2 e 2 + 4raecos! a - r = ae 2 + recos! a - ae 2 = r + recos! a(1- e 2 ) = r(1+ ecos!) so: r = a(1- e 2 ) / (1+ ecos!) (equ. for ellipse in polar coordinates)
What About the Velocity? Kepler's 2nd law: 1/ 2 r 2 dq / dt = constant (must hold for entire period) 1/ 2r 2 dq / dt =!ab / P (area/period) Since b = a(1- e 2 ) 1/2 : r 2 dq / dt = (2!a / P)[a(1- e 2 ) 1/2 ] Or: d" /dt = (2! /P)(a/r) 2 (1-e 2 ) 1/2 Recall s = rq so ds / dt = r dq / dt = V" V " = r dq / dt = r(2! / P)(a 2 / r 2 )(1- e 2 ) 1/2 = (2! / P)[a 2 (1- e 2 ) 1/2 ] /[a(1- e 2 ) / (1+ ecos")] So finally: V " = (2!a / P)(1+ ecos") / (1- e 2 ) 1/2 Since 1- e 2 = (1+ e)(1- e) so we consider 2 cases: Perihelion velocity (" = 0 o ): V peri = (2!a / P)(1+ e) / (1- e 2 ) 1/2 Aphelion velocity (" = 180 o ): V aph = (2!a / P)(1- e)(1- e 2 ) 1/2
Newtonian Derivation of Kepler s Laws #1: The general form of a planetary orbit is an ellipse/conic section Extensive derivation requiring calculus (see Mechanics) #2: A planet in orbit about the Sun sweeps out equal areas in equal amounts of time Recall that the area of a sector is given by: Area = θr 2 /2 (θ in radians) 3 4 2 θ 1 r s
Newtonian Derivation of Kepler s Laws Consider the motion of a planet between points 1 & 2 and between points 3 & 4. The orbital path length is given by s 1 and the angular difference is given by q and a given time interval:!t = t 2 " t 1 = t 4 " t 3 : The conservation of angular momentum requires: mv 1 r 1 = mv 2 r 2 = mv 3 r 3 = mv 4 r 4 so: v 1 r 1 = v 3 r 3 and multiplying by!t gives:!tv 1 r 1 =!tv 3 r 3 but since distance = velocity x time we have: s 12 r 1 = s 34 r 3 but s 12 = r 1! 12 and s 34 = r 3! 34 so:! 12 r 12 =! 34 r 32 dividing by 2 gives: (! 12 r 12 ) / 2 = (! 34 r 32 ) / 2 (area of sectors) 2-nd law results from conservation of angular momentum.
Newtonian Derivation of Kepler s Laws Law #3: The square of the orbital period is proportional to the cube of the semi-major axis of it's orbit. Consider a circular orbit for simplicity. Equate the centripital and gravitational forces (F c = F g ): M p V p 2 ) / r = (GM s M p ) / r 2 dividing by M p and 1/ r : V p 2 = GM p / r V p = 2!r / P but the circular velocity is: where P is the orbital period so: (2! ) 2 r 2 / P 2 = GM p / r and solving for p we have: P 2 = (4! 2 / GM s )r 3 but the circle is a special case of an ellipse so: P 2 = k a 3 or P 2 = a 3 / M (P is in years, a in AU and M is in solar masses)
Newton s Test of Universal Gravitation Recall the form of Newton!s Gravitational Law: F g = GMm / r 2 so a g = F g / m = GM / r 2 a (apple) = 9.807 m/s 2 (at R E ) Since R E = 6378 km and d m = 3.844 x 10 5 km: R E /d m = 60.27 so the acceleration at dm should be: a m = a g /(60.27) 2 = a g /3632 But what is it? a m = V m 2 / d m V m = (2!d m )/P = 1.023 x 10 3 m/s So a m = 2.723 x 10-3 m/s 2 a g /3632 = 2.698 x 10-3 m/s 2 (within 1%!)
Two-Body Problem Center of Mass: location where Fg = 0, and lies along the line connecting the two masses. Each mass must have the same orbital period and so: P 1 = 2!r 1 /v 1 = P 2 = 2!r 2 /v 2 so r 1 /v 1 = r 2 /v 2 and r 1 /r 2 = v 1 /v 2 Newton's 3rd law means F 1 = F 2 so: m 1 v 1 2 /r 1 = m 2 v 2 2 /r 2 substituting for V gives: (m 1 4! 2 r 1 2 )/r 1 P 2 = (m 2 4! 2 r 2 2 )/r 2 P 2 or: m 1 r 1 = m 2 r 2 thus: r 1 /r 2 = m 2 /m 1 = v 1 /v 2 Now we define a relative orbit where the more massive object, i.e., the Sun, lies near the center of mass. Let a = r 1 + r 2 and v = v 1 + v 2 Since r 1 = r 2 + m 2 /m 1 and r 1 + r 2 = m 2 r 2 /m 1 + r 2 so a = r 2 (1 + m 2 /m 1 ) The displacement is small for planets. Note: r 2 = a/(m 1 /m 1 +m 2 /m 1 ) and r 2 = m 1 a/(m 1 + m 2 ) combining gives: r 1 =m 2 a/(m 1 +m 2 ) (note the symmetry) Recall that F g = F c F 1 = m 1 v 1 2 /r 1 = Gm 1 m 2 /(r 1 +r 2 ) 2 (gravity = centripital force) substituting for v 1 (circular orbit) 4! 2 m 1 r 1 /P 2 = 4! 2 m 1 m 2 a/p 2 (m 1 +m 2 ) = Gm 1 m 2 /a 2 Thus: P 2 = [4! 2 /G(m 1 +m 2 )]a 3 (Newtonian form of Kepler!s 3-rd Law) Note: Masses can be derived given the period and semi-major axis of the orbits.
Chapter 3 Homework Chapter 3: #1, 2, 5, 6 Due Tues. Feb. 7