1. Solar System Science Lecture 1: Properties of the solar system Lecture 1: Properties of the Solar System oTopics in this lecture: oPlanetary orbits oMass distribution oAngular momentum distribution

2. Solar System Science Lecture 1: Properties of the solar system Properties of the Solar System 1.Planets orbit roughly in the ecliptic plane. 2.Planetary orbits are slightly elliptical, and very nearly circular. 3.Planets and Sun revolve and orbit in a west-to-east direction. The planets obliquity (tilt of rotation axes to their orbits) are small. Uranus and Venus are exceptions. 4.The planets differ in composition. Their composition varies roughly with distance from the Sun: dense, metal-rich planets are in the inner part and giant, hydrogen- rich planets are in the outer part. 5.Meteorites differ in chemical and geologic properties from the planets and the Moon. 6.The rotation rates of the planets and asteroids are similar (5 to 15 hours). 7.Planet distances from the Sun obey Bode's law. 8.Planet-satellite systems resemble the solar system. 9.The Oort Cloud and Kuiper Belt of comets. 10.Planets contain ~99% of the solar system's AM but Sun contains >99% of solar system's mass.

3. Solar System Science Lecture 1: Properties of the solar system Orbits of the planets oPlanets moves around the Sun in an orbit effected by the Sun’s mass, and to a less extent, by other bodies in the Solar System. oLaws governing planetary motion was formulated by Johannes Kepler and based on Tycho Brahe’s observations. oKepler’s Laws: 1.Planets have elliptical orbits with the Sun at one focus. 1.As a planet orbits, a line connecting the planet to the Sun sweeps out equal areas in equal times. 1.The square of the orbital period is proportional to the cube of the semimajor axis of the orbit.

4. Solar System Science Lecture 1: Properties of the solar system Kepler’s 1 st Law: Law of Orbits oPlanets have elliptical orbits with the Sun at one focus. oEquation of ellipse: r + r ’ = 2 a oa is semimajor axis, b is semiminor axis of ellipse, F ’ and F are focal point. oDistance of focus from ellipse centre is a e, where e is the eccentricity: oe = 0 => circle o0 ellipse oe = 1 => parabola oe > 1 => hyperbola

5. Solar System Science Lecture 1: Properties of the solar system Kepler’s 1 st Law (cont.) oImplies that a planet’s distance from the Sun varies during its orbit. oClosest point to Sun: perihelion. oFarthest point from Sun: aphelion. oAverage of perihelion and aphelion is called the semimajor axis.

6. Solar System Science Lecture 1: Properties of the solar system Elliptical orbits oConsider a point at either end of the semiminor axis where r = r’. oUsing the Pythagorean Theorem, r 2 = b 2 + (ae) 2 oSetting r = a, we may write: a 2 = b 2 + a 2 e 2 => b 2 = a 2 ( 1 - e 2 ) r’r’ oRelates semiminor axis to eccentricity and the semimajor axis.

7. Solar System Science Lecture 1: Properties of the solar system Elliptical orbital path oFrom figure below, r’ 2 =r 2 sin 2  + (2ae + rcos  ) 2 Eqn 1 oBut as r + r’ = 2a or r’ = 2a - r, we may write r’ 2 = 4a 2 - 4ra + r 2 Eqn 2 oEquating the RHS of Eqns 1 and 2: 4a 2 - 4ra + r 2 = r 2 sin 2  + (2ae + rcos  ) 2 = r 2 (sin 2  +cos 2  ) + 4a 2 e 2 + 4aer cos  oAs sin 2  +cos 2  = 1 =>4a 2 - 4ra = 4a 2 e 2 + 4aer cos  oRearranging gives, oFor 0<e<1, this is the equation of an ellipse in polar coordinates. Eqn 3

8. Solar System Science Lecture 1: Properties of the solar system Perihelion and aphelion distances oIf  = 0 o, cos  = 1 and oThe planet is at perihelion, the closest point to the Sun. oIf  = 180 o, cos  = -1 and oThe planet at the aphelion, the most distant point from the Sun. oExample: The semimajor axis of Mars is 1.5237 AU and the eccentricity is 0.0934. What is the distance of Mars at perihelion? r = a(1-e) = 1.5237(1-0.0934) = 1.3814 AU oWhat is the distance of Mars at aphelion?   = 0 o  = 180 o

9. Solar System Science Lecture 1: Properties of the solar system Kepler’s 2 nd Law: Law of areas oAs a planet orbits, a line connecting the planet to the Sun sweeps out equal areas in equal times. => Planet movies faster at perihelion.

10. Solar System Science Lecture 1: Properties of the solar system Kepler’s 2 nd Law: Law of areas oAngular momentum of planet: L = r  p = m (r  v). oDuring  t, radius vector sweeps through  = v t  t / r, where v t is the component of v perpendicular to r. oDuring this time, the radius vector has swept out the triangle, of area A = rv t  t / 2. oAs  t  0, dA/dt = rv t / 2 = 1⁄2 r 2 (d  /dt). oNow, the magnitude of L is given by L = m v t r= m r 2 d  /dt. => dA/dt = L / 2m = const oi.e. the rate of sweeping out area is a constant. F vrvr  AA v vtvt r

11. Solar System Science Lecture 1: Properties of the solar system Kepler’s 3 rd Law: Law of Periods oThe square of the orbital period is proportional to the cube of the semimajor axis of the orbit: P 2  a 3 oP is the period measured in years and a is the semimajor axis in AU. oConsider m 1 and m 2 orbiting at r 1 and r 2. Both complete one orbit in period P. Forces due to centripetal accelerations are: F 1 = m 1 v 1 2 / r 1 = 4  2 m 1 r 1 / P 2 F 2 = m 1 v 2 2 / r 2 = 4  2 m 2 r 2 / P 2 using v = 2  r / P. m2m2 m1m1 r1r1 r2r2 a v1v1 v2v2 COM

12. Solar System Science Lecture 1: Properties of the solar system Kepler’s 3 rd Law: Law of Periods oAs F 1 = F 2 => r 1 / r 2 = m 2 / m 1 (more massive body orbits closer to centre of mass). oSeparation of the bodies is a = r 1 + r 2, and r 1 = m 2 a / (m 1 + m 2 ) oCombining with F 1 and F = F 1 = F 2 = Gm 1 m 2 /a 2 : P 2 = 4  2 a 3 / G(m 1 + m 2 ) oAs M sun (= m 1 ) >> m planet (=m 2 ), const = 4  2 / GM Sun. m2m2 m1m1 r1r1 r2r2 a v1v1 v2v2 COM

13. Solar System Science Lecture 1: Properties of the solar system Bode’s Law oEmpirical prediction of planet distances from Sun. oBegin with: 0, 3, 6, 12, 24, 48, 96, 192, 384 o Now add 4: 4, 7, 10, 16, 28, 52, 100, 196, 388 oThen divide by 10: 0.4, 0.7, 1.0, 1.6, 2.8, 5.2, 10.0, 19.6, 38.8 oSequence is close to mean distances of planets from the Sun. oBode’s Law or Titus-Bode’s Law: r n = 0.4 + 0.3  2 n PlanetDistance (AU) Bode’s Law (AU) Mercury0.4 Venus0.7 Earth1.0 Mars1.51.6 Ceres2.8 Jupiter5.2 Saturn9.610.0 Uranus19.219.6

14. Solar System Science Lecture 1: Properties of the solar system Bode’s Law (cont.) Bode’s Law (cont.) n 0123456 Planet MercuryVenusEarthMarsCeresJupiterSaturnUranus s-m axis0.40.71.01.52.85.29.619.2 rnrn 0.40.71.01.62.85.210.019.6 r n = 0.4 + 0.3  2 n

15. Solar System Science Lecture 1: Properties of the solar system Bode’s Law (cont.) oLaw lead Bode to predict existence of another planet between Mars and Jupiter - asteroids belt later found. oUranus fitted law when discovered. oNeptune was discovered in 1846 at the position predicted by Adams, to explain the deviation of Uranus from its predicted orbit. oPluto’s orbit when discovered in 1930 did not fit the relation. Planet Distance (AU) r n (AU) Uranus19.219.6 Neptune30.0738.8 Pluto39.577.2

16. Solar System Science Lecture 1: Properties of the solar system The Solar System to scale

17. Solar System Science Lecture 1: Properties of the solar system Mass distribution oThe density of a planet is measured in g cm -3 (cgs units). oConvenient because the density of water is 1 g cm -3. oTo determine volume, need: 1. Distance (from parallax) 2. Angular extent of the planet. oTo determine the mass (from Kepler’s 3 rd Law) we need: 1. Distance (from parallax), 2. Angular size of the planet’s orbit, 3. Orbital period.

18. Solar System Science Lecture 1: Properties of the solar system Mass Distribution (cont.) oThe volume is determined from: V = 4/3  R 3 where 2R = 2  d  / 360. oMass determined from Newton’s form of Kepler’s 3 rd Law: P 2 = 4  2 a 3 / G (m + M ) => m = ( 4  2 a 3 / G P 2 ) - M o  = m / V g cm -3 oCompare to:Cork:0.2 Wood:0.5 Water:1.0 Basalt:3.3 Lead:11.0 Gold:19.0  2R d

19. Solar System Science Lecture 1: Properties of the solar system Properties of the planets oFrom consideration of size and density, can divide the planets into two categories: 1.Terrestrial Planets oSmall size, high density and in the inner solar system. oMercury, Venus, Earth, and Mars. 2.Jovian Planets oLarge size, low density and in outer solar system: oJupiter, Saturn, Uranus, Neptune. oPluto oPluto is in category of own. It has small size and low density.

20. Solar System Science Lecture 1: Properties of the solar system Compression vs. composition: The inner planets oFrom their densities, inner planets likely to be composed of rock and some metal in cores. oMight expect that planets less massive than the Earth would have lesser densities, because they are less compressed at the center by gravity. oAmongst the terrestrial planets, this is true for both Mars and the Moon, which are both smaller and less dense than the Earth. o Venus is roughly the same size and density as Earth. o But, Mercury is both less massive, and more dense than the Earth. => Has Mercury a different composition than the Earth?

21. Solar System Science Lecture 1: Properties of the solar system oWhat about the densities of the outer planets? oMight expect the outer planets, which are very massive, to be much more compressed than the inner planets, and so more dense. oIn fact, these heavier bodies are less dense than the inner, terrestrial planets. oThe only composition which we can use to construct such massive bodies with such low densities is a mixture of hydrogen and helium, the two lightest elements. oThe composition of the outer planets is hence more similar to the Sun and stars than to the inner planets. Compression vs. composition: The outer planets

22. Solar System Science Lecture 1: Properties of the solar system Angular momentum distribution oThe Sun has a relatively slow rotational period of ~26-days. => Like most G-, K- and M-class stars. oThe orbital AM is:h orb = mvr = 2  mr 2 /P (r = distance) oThe spin AM of a inhomogeneous nonspherical rotating body is more difficult to evaluate. oThe mass of the body is 4  ave R 3 /3, where R is the radius and  is the mean density.

23. Solar System Science Lecture 1: Properties of the solar system Angular momentum of the Sun oAn average density is adopted for the Sun and it is assumed that the mass is mostly within 0.6R. This value is 0.72 for a perfect sphere, but the Sun is oblate. oMean density = 1.41 g cm -3. oMass = 2 x 10 33 g. oPeriod at equator = 26.5 days = 2289600 s. oRadius = 6.96 x 10 10 cm. oThe spin AM is therefore: oSetting R = 0.6R => oDetailed modelling gives ~1.7 x 10 47 g cm 2 s -1.

24. Solar System Science Lecture 1: Properties of the solar system Angular momentum of the planets oThe orbital AM is: oEarth om = 5.97 x 10 27 g or = 1 AU = 1.5 x 10 13 cm oP = 1 year oJupiter om = 1898 x 10 27 g or = 5.2 AU = 5.2 x 1.5 x 10 13 cm oP = 11.86 years oSaturn om = 586 x 10 27 g or =9.61 AU oP = 29.5 years oJupiter therefore carries ~50% of the total AM of the Solar System, while the Jovian planets together make up ~99.27% of the total!

25. Solar System Science Lecture 1: Properties of the solar system Angular momentum distribution o=>Sun only has 0.4% of the total AM in solar system. PlanetMass (x10 27 kg) Period (years) AM (gcm 2 s -1 ) Mercury0.330.248.6x10 45 Venus4.870.611.9x10 47 Earth5.9712.6x10 47 Mars0.641.883.4x10 47 Jupiter1898.811.861.9x10 50 Saturn568.419.57.8x10 49 Uranus86.9719.311.7x10 49 Neptune102.85302x10 49 Pluto0.012939.913.7x10 45 0.4% AM 99.2% AM

26. Solar System Science Lecture 1: Properties of the solar system Orbital angular momenta of the planets oNote the overwhelming importance of the Jovian planets. oThe symbol associated with each planet:

27. Solar System Science Lecture 1: Properties of the solar system Mass and AM distributions oAlthough Sun contains 99.9% of the mass of Solar System, the outer planets have 98% of system’s angular momentum. oThis is a serious problem: material accreting onto the Sun cannot have retained all its original AM. o There are two parts to the problem: 1.How does material lose AM and fall into the star in the first place? 2.How does the star lose AM and slow down? Solar-type stars all rotate at about the same speed at the Sun.