AST111 PROBLEM SET 2 SOLUTIONS Home work problems 1. Angles on the sky and asteroid motion An asteroid is observed at two different times. The asteroid is located at RA=02h23m35.65s, DEC=+25d18m42.3s (Epoch J2000). on 2003-Aug-25, UT=02h00m00s (universal time). At UT=03h00m00s the asteroid is located at RA=02h23m36.95s, DEC=+25d19m10.6s. (a) How many radians is equivalent to an arc second in declination (DEC) at the position of the asteroid?. π radians 1 180.0 60 1 6 radians = 4.848 10 60 (b) How many radians is equivalent to a second of time in right ascension (RA) for the above asteroid? Your answer should depend on the cos or sin of the declination. Declination in degrees is δ = 25.0 + 18.0/60.0 + 42.3/60.0 2 = 25.31175 In radians (multiply by π/180) this is δ = 0.44177337694 radians. Taking the cosine of this cos δ = 0.903995. On the celestial equator there are 15 per second in RA giving 15 0.903995 = 13.55992 per second at this declination. 1
2 AST111 PROBLEM SET 2 SOLUTIONS Recall that 1 is 4.848 10 6 radians (this number is π/180/60/60). We multiply our value for 1 second in time giving 6.574036353 10 5 radians per second of time. (c) What is the angular distance in arcseconds between the two observations? DEC = 60.0 + 10.6 42.3 = 28.3 RA = 1.3s We convert the RA into arcseconds by multiplying by 13.55992 giving 17.628. The total angular distance in arcseconds is 17.6282 + 28.3 2 = 33.34 (d) If the object is 1.5AU away, how fast is it moving in km/s (in the plane perpendicular to our line of sight)? The angular distance corresponds to a distance of this d = rθ = 1.5AU 33.34 4.848 10 6 radians = 0.00024244AU 0.00024244AU 1.5 108 km = 36367km AU The time difference between observations was 1 hour = 60 60s = 3600 s, giving a speed of v = 36367km/3600s = 10 km/s 2. On the center of mass (a) How far away from the center of the Sun is the center of mass of the Sun- Jupiter system? The mass of Jupiter is about 1/1000th of the Sun and its semi-major axis is 5.2 AU. r Sun = r J M J /M = 5.2 10 3 AU This is equivalent to 7.8 10 10 cm or 7.8 10 8 m or 7.8 10 5 km. (b) If a planetary system is oriented with orbital plane perpendicular to the line of sight, no Doppler shift can be detected from the star s reflex motion. However no matter what the orientation of the planet-star system, the proper motion of the star on the sky about the center of mass is detectable (at sufficiently good angular resolution). Draw diagrams (of the orbit) illustrating why this is true.
AST111 PROBLEM SET 2 SOLUTIONS 3 A closed orbit cannot lie on a line, it must be an ellipse and so extends in two dimensions. No matter which way you tilt it it does not project to a point as seen from by a distant observer. So the trajectory on the sky is always extended. This contrasts with radial velocity measurements. As the velocity vector lies in the orbital plane if the observe sees the system edge on the line of sight velocity component is zero. So planets can exist in systems that cannot be found in Doppler shift survey but a very good astrometric survey would find all planets (if it could see full orbits). 3. Circular and escape velocities A baseball pitcher can throw a fastball at a speed of 150km/hour. What is the largest size (in radius) spherical rocky asteroid of density ρ = 3g cm 3 from which he can throw the ball fast enough that it (a) escapes from the asteroid into heliocentric orbit? (b) goes into a stable circular orbit about the asteroid? The speed v = 150 km hour hour 3600s = 0.0417 km s 1 = 4.177 10 3 cm s 1 The mass M = 4π 3 ρr3. The escape velocity v escape = 2GM/R. Using the escape velocity we solve for radius 3vescape 2 R escape = (1) 8πGρ Using the circular velocity v c = GM/R we solve again 3vc R circular = 2 4πGρ (2) a) In cgs using the escape velocity 3 (4.2 103 ) R escape = 2 4π6.67 10 8 3.0 = 3.2 106 cm = 32 km b) The circular velocity is 2 smaller than the escape velocity. And we can see a difference of 2 in the formulas of equation 1 with 2. However in each case
4 AST111 PROBLEM SET 2 SOLUTIONS we are using the same velocity. Using the same velocity equation 2 gives a larger radius than equation 1. We multiply the radius R escape by 2 giving a radius of R circular = 46 km. Workshop problems 1. Keplerian orbits (a) Pluto has a semi-major axis of 39.48 AU and an eccentricity of 0.2488. What are the radii of its perihelion and aphelion (in AU)? a(1 + e) = 49.3026 AU is aphelion. a(1 e) = 29.6573 AU is perihelion. (b) Is its angular momentum greater or lesser than that of an object with the same mass and semi-major axis in a circular orbit? Angular momentum per unit mass L = GMa(1 e 2 ) If we lower eccentricity to zero the angular momentum increases. The circular orbit has the most possible angular momentum at a given energy or semimajor axis. (c) Find a formula for the velocity at perihelion and that at aphelion for an orbit with semi-major axis a and eccentricity e. E = GM 2a = v2 2 GM r Insert r = a(1 + e) and solve for v 2. vap 2 = GM ( ) 2 a 1 + e 1 and v ap = GM (1 e) a (1 + e)
Using r = a(1 e) we find AST111 PROBLEM SET 2 SOLUTIONS 5 v peri = GM (1 + e) a (1 e) 2. Orbital resonance Io, Europa, and Ganymede are at semi-major axes of 421.77, 671.08 and 1070.4 10 3 km, respectively, with respect to the center of Jupiter. (a) Show that Europa and Io are in an orbital resonance, and identify which one. In other words find two integers i, j such that ip Europa jp Io where P Europa and P Io are the orbital periods of Europa and Io. Orbital period P a 3/2 so we take the ratio ( ) 3/2 P Eu 671.08 = = 2.007 P Io 421.77 They are near a 2:1 resonance. (b) Show that Europa and Ganymede are also locked in an orbital resonance and identify which one. ( ) 3/2 P Gan 1070.4 = = 2.014 P Eu 671.08 Also a 2:1 resonance. (c) Suppose we have an angle λ Io such that λ Io = λ dt Io = n Io where n Io is the mean motion of Io. And similarly angles λ Europa a and λ Ganymede. Assuming that each pair of bodies is exactly in resonance, show that the Laplace angle φ Laplace = λ Io 3λ Europa + 2λ Ganymede does not vary. In other words show the φ Laplace = 0. Take the time derivative of the angle φ = n Io 3n Europa + 2n Gany The 2:1 resonance between Io and Europa gives n Io = 2n Europa and the 2:1 resonance between Europa and Ganymede gives n Europa = 2n Gany. Getting rid of n Io and n Gany we find φ = 2n Europa 3n Europa + n Europa = 0