PH507 Astrophysics Dr Dirk Froebrich 1

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1 PH507 Astrophysics Dr Dirk Froebrich 1 MASS MEASUREMENT PLANET REVIEW The NEW Solar System 003UB313 Eris A Planet is: A celestial body that is (a) in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit In 003, The International Astronomical Union (IAU) Working Group on Extrasolar Planets made a position statement on the definition of a planet. - Objects with true masses below the limiting mass for thermonuclear fusion of deuterium (currently calculated to be 13 times the mass of Jupiter for objects with the same isotopic abundance as the Sun) that orbit stars or stellar remnants are "planets" (no matter how they formed). The minimum mass and size required for an extrasolar object to be considered a planet should be the same as that used in our Solar System. - Substellar objects with true masses above the limiting mass for thermonuclear fusion of deuterium are "brown dwarfs", no matter how they formed nor where they are located. - Free-floating objects in young star clusters with masses below the limiting mass for thermonuclear fusion of deuterium are not "planets", but are "sub-brown dwarfs" (or whatever name is most appropriate).

2 PH507 Astrophysics Dr Dirk Froebrich In the picture above we see the positions of the asteroid belt (green) and other near-earth objects The material in the plane of the Solar System is known as the Kuiper Belt ( AU) or Trans-Neptunian Objects. Surrounding this is a much larger region known as the Oort Cloud ( ,000 AU), that contains material that occasionally falls in, under the influence of gravity, towards the Sun as comets.

3 PH507 Astrophysics Dr Dirk Froebrich 3 The Sun Over 1.4 million kilometers (869,919 miles) wide. Contains percent of the mass of the entire solar system: well over a million Earths could fit inside its bulk. The total energy radiated by the Sun averages 383billion trillion kilowatts, the equivalent of the energy generated by 100 billion tons of TNT exploding each and every second. i.e * 10 6 W Planetary configurations Ecliptic is the intersection of the celestial sphere with the ecliptic plane, which is the geometric plane containing the mean orbit of the Earth around the Sun. Some of the definitions below make the assumption of coplanar circular orbits. True planetary orbits are ellipses with low eccentricity and inclinations are small so the concepts are applicable in real cases. Copernicus correctly stated that the farther a planet lies from the Sun, the slower it moves around the Sun. When the Earth and another planet pass each other on the same side of the Sun, the apparent retrograde loop occurs from the relative motions of the other planet and the Earth.

4 PH507 Astrophysics Dr Dirk Froebrich 4 As we view the planet from the moving Earth, our line of sight reverses its angular motion twice, and the three-dimensional aspect of the loop comes about because the orbits of the two planets are not coplanar. This passing situation is the same for inferior or superior planets. A Retrograde loop occurs when a superior planet moves through opposition, and occurs as the earth's motion about its orbit causes it to overtake the slower moving superior planet. Thus, close to opposition, the planet's motion relative to fixed background stars, follows a small loop. Lecture 5: Measuring Mass Mass can be measured in two ways. 1 We could count up the atoms, or count up the molecules and grains of dust and infer the number of atoms. This method can be used if the object is optically thin and we have good tracer: a radiation or scattering mechanism in which the number of photons is related to the number of particles. Otherwise, measuring the mass of an object relies upon its gravitational influence on nearby bodies or on itself (self-gravity). Newton s second law: F=m*a. The first law relates the acceleration to a change in speed or direction (law of inertia). Third law: action = reaction. Kepler s empirical laws for orbital motion describe the nature of the acceleration from which masses can be derived.

5 PH507 Astrophysics Dr Dirk Froebrich 5 Kepler's Laws First Law: The orbit of each planet is an ellipse with the Sun at one focus p b r F Q C S a f q S = Sun, F = other focus, p = planet. r = HELIOCENTRIC DISTANCE. f = TRUE ANOMALY a = SEMI-MAJOR AXIS = mean heliocentric distance, which defines the size of the orbit. b = SEMI-MINOR AXIS. e = ECCENTRICITY, defines shape of orbit. Ellipse, by definition: SP + PF = a (1) e = CS / a () Therefore b = a * (1-e) (3) When CS=0, e = 0, b = a, the orbit is a circle. When CS=, e = 1, the orbit is a parabola. q = PERIHELION DISTANCE = a - CS = a ae q = a(1-e) (4) Q = APHELION DISTANCE = a + CS = a + ae Q = a(1+e) (5) Second Law: For any planet, the radius vector sweeps out equal areas in equal time intervals

6 PH507 Astrophysics Dr Dirk Froebrich 6 Time interval t for planet to travel from p to p1 is the same as time taken for planet to get from p to p3. Let the time interval t be very small. Then the arc from p to p1 can be regarded as a straight line and the area swept out is the area of the triangle S p p1. If f 1 is the angle to p 1, and f is the angle to p: p 1 p p 3 r 1 r f p S Area = 1/ * r * r 1 * sin (f1-f). Since t is very small, r~ r 1 and sin (f1-f) ~ (f1-f) = f Area = 1/ * r * f The rate this area is swept out is constant according to Kepler's second law, so r * df/dt = h (6) where h, a constant, is twice the rate of description of area by the radius vector. It is the orbital angular momentum (per unit mass). The orbital angular momentum is conserved. The total area of the ellipse is π*a*b, which is swept out in the orbital period P, so using Eq. (6) π *a*b/p = h. The average angular rate of motion is n = π /P, so n * a * (1-e ) 1/ = h (7)

7 PH507 Astrophysics Dr Dirk Froebrich 7 Kepler s Third Law Kepler's third law took another ten years to develop after the first two. This law relates the period a planet takes to travel around the sun to its average distance from the Sun. This is sometimes called the semi major axis of an elliptical orbit. P = k * a 3 where P is the period and a is the average distance from the Sun. Or, if P is in years and a is in AU: P = a 3 Kepler s Third Law follows from the central inverse square nature of the law of gravitation. First look at Newton's law of gravity - stated mathematically as F = Gm m r 1 Newton actually found that the focus of the elliptical orbits for two bodies of masses m1 and m is at the centre of mass. The centripetal forces of a circular orbit are r 1 F 1 v m 1 X m Centre of Mass v 1 F F 1 = and F = where v = π r P m 1 v 1 m v r 1 = r 4 π m 1 r 1 P r = 4 π m r P

8 PH507 Astrophysics Dr Dirk Froebrich 8 and since they are orbiting each other (Newton s nd law) r 1 = m r m 1 Let's call the separation a = r1 + r. Then a = r 1 + m 1 r 1 m = r 1 m 1 m + 1 a n d m u l t i p l y i n g b o t h s i d e s b y m, a m = m 1 r 1 + m r 1 o r, s o l v i n g f o r r 1, r 1 = a m ( m 1 + m ) Now, since we know that the mutual gravitational force is G m 1 m F g r a v = F 1 = F = a then substituting for r1, a 3 = G (m 1 + m ) P Solving for P: P = π 4π a 3 G (M 1 + M ) Third Law is therefore: The cubes of the semi-major axes of the planetary orbits are proportional to the squares of the planets' periods of revolution. Example Europa, one of the Jovian moons, orbits at a distance of 671,000 km from the centre of Jupiter, and has an orbital period of 3.55 days. Assuming that the mass of Jupiter is very much greater than that of Europa, use Kepler's third law to estimate the mass of Jupiter. Using Kepler's third law: m jupiter a + meuropa = 4 π 3 GP The semi-major axis, a=6.71x10 5 km = 6.71x10 8 m, and the period, P = 3.55 x 3600 x 4 = 3.07 x 10 5 seconds

9 PH507 Astrophysics Dr Dirk Froebrich π ( 6.71*10 m) Thus: m jupiter + meuropa = *10 m kg s ( 3.07 *10 s) and since m jupiter >> m europa, then m jupiter ~ 1.9 x 10 7 kg. Summary of Kepler s Laws = 1.9 *10 7 kg Summary: Measuring the mass of a planet Kepler s third law gives G(M+m) = 4 π a 3 / P (To remember, do a dimensional analysis: GM/R and R/P are both accelerations). Since M >> m for all planets, it isn't possible to make precise enough determinations of P and a to determine the masses m of the planets. However, if satellites of planets are observed, then Kepler's law can be used. Let mp = mass of planet ms = mass of satellite Ps = orbital period of satellite as = semi-major axis of satellite's orbit about the planet. Then: G(mp+ms) = 4π as 3 /Ps

10 PH507 Astrophysics Dr Dirk Froebrich 10 If the mass of the satellite is small compared with the mass of the planet then mp = 4π as 3 /(G Ps ) All the major planets have satellites except Mercury and Venus. Their masses were determined from orbital perturbations on other bodies and later, more accurately from changes in the orbits of spacecraft. So: we can determine the masses of massive objects if we can detect and follow the motion of very low mass satellites. That doesn t lead very far. How can we determine the masses of distant stars and exoplanets? BASIC STELLAR PROPERTIES - BINARY STARS For solar type stars, single:double:triple:quadruple system ratios are 45:46 : 8 : 1. Binary nature of stars deduced in a number of ways: 1. VISUAL BINARIES: - Resolvable, generally nearby stars (parallax likely to be available) - Relative orbital motion detectable over a number of years - Not possible (yet) for exoplanets!. ASTROMETRIC BINARY: only one component detected 3. SPECTROSCOPIC BINARIES: - Unresolved - Periodic oscillations of spectral lines (due to Doppler shift) - In some cases only one spectrum seen 4. ECLIPSING BINARY: - Unresolved - Stars are orbiting in plane close to line of sight giving eclipses observable as a change in the combined brightness with time (light curves). Some stars may be a combination of these.

11 PH507 Astrophysics Dr Dirk Froebrich Visual Binaries Angular separation 0.5 arcsec (close to Sun, long orbital periods - years remember: at 1 parsec, 1 arcsec corresponds to 1 AU) EXAMPLE: Sirius: Also known as Alpha Canis Majoris, Sirius is the fifth closest system to the Sun, at 8.6 light-years. Sirius is composed of a main-sequence star and a white dwarf stellar remnant. They form a close binary, Alpha Canis Majoris A and B, that is separated "on average" by only about 0 times the distance from the Earth to the Sun astronomical units (AUs) of an orbital semi-major axis -- which is about the same as the distance between Uranus and our Sun. The companion star is so dim that it cannot be perceived with the naked eye. After analyzing the motions of Sirius from 1833 to 1844, Friedrich Wilhelm Bessel ( ) concluded that it had an unseen companion

12 PH507 Astrophysics Dr Dirk Froebrich 1 Observations: Relative positions: ρ = angular separation θ = position angle Absolute positions: Harder to measure orbits of more massive star A and less massive star B about centre of mass C which has proper motion µ. Secondary Declination N Motion of centre of mass = proper motion µ ρ θ B E Right Ascension Primary C A NB: parallax and aberration must be accounted for! RELATIVE ORBITS: - TRUE orbit: q = peri-astron distance (arcsec or km) Q = apo-astron distance (arcsec or km) a = semi-major axis (arcsec or km) a = (q + Q)/ - APPARENT orbits are projected on the celestial sphere Inclination i to plane of sky defines relation between true orbit and apparent orbit. If i 0 then the centre of mass (e.g. primary) is not at the focus of the elliptical orbit. Measurement of the displacement of the primary gives inclination and true semi-major axis in arcseconds (a").

13 PH507 Astrophysics Dr Dirk Froebrich 13 i i True orbit Incline by 45 Apparent orbit If the parallax p in arcsec is observable then a can be derived from a". Earth radius of Earth's orbit B a Sun a" r = distance of binary star p A For i=0, a = a"/p AU In general a correction for i 0 is required. Now let s go back to Kepler s Law From Kepler's Law, the Period P is given by P 4 π a 3 = G (ma + m B ) For the Earth-Sun system P=1year, a=1a.u., ma+mb~msun so 4π/G = 1 P a 3 = (ma + m B ) provided P is in years, a in AU, ma, mb in solar masses. The total system mass is determined: a" 3 1 m A + m B = ( ) p P

14 PH507 Astrophysics Dr Dirk Froebrich 14 ABSOLUTE ORBITS: B B c r A d * e r B f A q Q A Semi-major axes aa = (c+e)/ ab = (d+f)/ Maximum separation = Q = c + f Minimum separation = q = d + e So aa + ab = (c+d+e+f)/ = (q + Q)/ = a a = aa + ab (1) (and clearly r = ra + rb) From the definition of centre of mass, ma*ra=mb*rb (ma*aa=mb*ab) ma/mb = ab/aa = rb/ra So from Kepler s Third Law, which gives the sum of the masses, and the equation above, we get the ratio of masses, ==> ma, mb. Therefore, with both, we can solve for the individual masses of the two stars.

15 PH507 Astrophysics Dr Dirk Froebrich 15 3 Spectroscopic Binaries To measure the periodic line shifts requires Orbital period relatively short (hours - months) and i 0. Doppler shift of spectral lines by component of orbital velocity in line of sight (nominal position is radial velocity of system): Time wavelength Time wavelength Stars observable 1 Star observable See:

16 PH507 Astrophysics Dr Dirk Froebrich 16 Data plotted as RADIAL VELOCITY CURVE: Shape of radial velocity curves depends on orbital eccentricity and orientation If the orbit is tilted to the line of sight (i<90 ), the shape is unchanged but velocities are reduced by a factor sin(i). Take a circular orbit with i=90 a = ra + rb v = va + vb Orbital velocities: va = π ra / P vb = π rb / P v = π r / P Since ma*ra = mb*rb Ratio of masses: ma/mb = rb/ra = vb/va ()

17 PH507 Astrophysics Dr Dirk Froebrich 17 r B r = v A v r A v B In general, measured velocities are vb sin i and va sin i, so sin i terms cancel: so we measure the mass ratio accurately. From Kepler's law ma + mb = a3/p (in solar units). Observed quantities: va * sin(i) => ra * sin(i) vb * sin(i) => rb * sin(i) a * sin(i) So can only deduce (ma + mb) = (a sin(i))3/p For a spectroscopic binary, only lower limits to each mass can be derived, unless the inclination i is known independently. DETAILED DERIVATION when only one radial velocity is known (eliminate VB using mass ratio) Assume a planet and star, both of considerable mass, are in circular orbits around their centre of mass. Given the period P, the star s orbital speed v* and mass M*, the mass of the planet, Mp is given by Mp 3 / (M* + Mp) = v 3 * P / ( π * G) Note that there are 9 unknowns: P, a*, ap, M*, Mp, v*, vp, a, M However: a = a* + ap M = M*+ Mp Centre of mass; M * a * = Mp * ap Kepler s law relates: P, a, M P = π a * /v * P = π ap/vp.so that is 6 equations. Note: we usually only know vr* = v* sin(i) and we assume the planet mass is small.

18 PH507 Astrophysics Dr Dirk Froebrich 18 4 Eclipsing Binaries Since stars eclipse i ~90 For a circular orbit: 1, 1' FIRST CONTACT, ' SECOND CONTACT 3, 3' THIRD CONTACT 4, 4' FOURTH CONTACT 4' 3' ' 1' v Observer in plane Variation in brightness with time is LIGHT CURVE. Timing of events gives information on sizes of stars and orbital elements. Shape of events gives information on properties of stars and relative temperatures. If smaller star is hotter, then:

19 PH507 Astrophysics Dr Dirk Froebrich 19 Case 1 Smaller star is hotter Case Larger star is hotter F or magnitude Secondary minimum Primary minimum Case 1 t' 1 t' t' 3 t' t 1 t t 3 t 4 4 Case t 1 t t 3 t 4 t' 1 t' t' 3 t' 4 time If orbits are circular: minima are symmetrical ie t-t1 = t4-t3 = t'-t1' = t4'-t3'; minima are half a period apart; eclipses are of same duration. Asymetrical and/or unevenly spaced minima indicate eccentricity and orientation of orbit. For a circular orbit: Distance = velocity x time RS = v(t t1) and RS + R = v(t4 - t1) => RL = v(t4 - t) (5) RS/RL = (t -t1)/ (t4 - t) (6)

20 PH507 Astrophysics Dr Dirk Froebrich 0 t1 t t3 t4 R L R S Light curves are also affected by: Non-total eclises No flat minimum Limb darkening (non-uniform brightness) Ellipsoidal stars (due to proximity) "rounds off" eclipses "rounds off" maxima Reflection effect (if one star is very bright) 5 Eclipsing-Spectroscopic Binaries For eclipsing binaries i 70 (sin3i > 0.9) If stars are spectroscopic binaries then radial velocities are known. So: masses are derived, radii are derived, ratio of temperatures is derived Examine spectra and light curve to determine which radius corresponds with which mass and temperature: Since Luminosity L = 4π R σ T4, (Stefan-Boltzmann equation) the ratio of luminosities is derived from L A = R A T A 4 L B R B T B How else to determine mass? 1. Mass-Luminosity relation derived from binaries.. Lensing/micro-lensing. 3. Virial expression