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Astronomy 5500 Exercises

Problem . What is the distance to the star Spica ( α Virginis), which has a measured parallax according to Hipparcos of π abs = 12.44 ±0.86 mas? Solution . The distance to Spica is given by the parallax equation, i.e. The uncertainty is: The distance to Spica is 80.39 ±5.56 parsecs.

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Astronomy 5500 Exercises

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  1. Problem. What is the distance to the star Spica (α Virginis), which has a measured parallax according to Hipparcos of πabs = 12.44 ±0.86 mas? Solution. The distance to Spica is given by the parallax equation, i.e. The uncertainty is: The distance to Spica is 80.39 ±5.56 parsecs. Astronomy 5500Exercises

  2. Problem. How distant is Spica (α Virginis), a B1 III-IV star with apparent visual magnitude V = 0.91, given that B1 III-IV stars typically have an absolute magnitude of MV = –4.1 ±0.3 (Turner ZAMS)? Solution. The distance modulus for Spica is given by: Thus: The uncertainty is: The spectroscopic parallax distance to Spica is 100.5 ±13.9 parsecs. Note the small disagreement with the star’s Hipparcos parallax distance of 80.39 ±5.56 parsecs.

  3. Problem. Find the mass of the Galaxy given the local circular velocity of 251 km/s at the Sun’s location roughly 8.5 kpc from the Galactic centre. Solution. Use Kepler’s 3rd Law in Newtonian fashion, i.e. (MG + M)(in M) = a(in A.U.)3/P(in yrs)2. For an orbital speed of 251 km/s and orbital radius of 8.5 kpc the orbital period is: The semi-major axis is: So the mass of the Galaxy is:

  4. So ~1011M is derived for the mass of the Galaxy internal to the Sun. If the orbital velocity curve is flat to ~16 kpc from the Galactic centre, then one can redo the calculations to find that ~21011M is derived for the mass of the Galaxy internal to ~16 kpc from the centre. Where did the extra ~1011M come from, or is it proper to apply Kepler’s 3rd Law in situations like this? Recall that it applies to the case of a two-body situation only, not to a multi-body situation. Mass/Light Ratios, M/L: The mass-luminosity relation varies roughly as L ~ M4 (M3 for cool stars), so the mass-to-light ratio should vary as M/L ~ 1/M3 or1/M2. The typical star near the Sun is a cool M-dwarf with a mass of only 0.25 M or less, implying a typical mass-to-light ratio for our Galaxy of ~16. Since most stars are probably less massive than that, the actual mass-to-light ratio for the Galaxy could be in excess of ~25 or so.

  5. Problem. What is the mass-to-light ratio for the Milky Way, given the properties that have been derived for it, i.e. estimated mass of 2  1011M and LSR velocity of 251 km/s, and Mbol() = 4.79, from MV() = 4.82 and (B−V) = 0.63? Solution. Most supergiant Sb galaxies have MV −20.5. If the Milky Way is identical, then its luminosity in solar units is: So for the Milky Way, L/L = 10(25.32/2.5) = 1010.128 = 1.3  1010, and M/L = 2  1011M/M/1.3  1010L/L 15, larger than values typically adopted in research papers.

  6. Problem. Use the rotation curve for stars within 1" of the centre of M32 shown in Carroll & Ostlie to estimate the mass lying within that region. Compare your answer with the value obtained using the velocity dispersion data in Example 25.2.1 of the textbook and with the estimated range cited in that example. Solution. We did an example using orbital velocity to estimate galaxy mass. Here, vr = 50 km/s and r = 3.7 pc.

  7. Use Kepler’s 3rd Law in Newtonian fashion, i.e. (MG + Mstar) = a3/P2, for a in A.U. and P in years. For an orbital speed of 50 km/s and orbital radius of 3.7 pc the orbital period is: The semi-major axis is: So the mass of the Galaxy is: as compared to an estimate of ~107M obtained from the velocity dispersion and values ranging from 1.5−5106M from the rotation curve, similar to the present value.

  8. Problem. The star Delta Tauri is a member of the Hyades moving cluster. It has a proper motion of 0".115/year, a radial velocity of 38.6 km/s, and lies 29º.1 from the cluster convergent point. a. What is the star’s parallax? b. What is the star’s distance in parsecs? c. Another Hyades cluster member lies only 20º.0 from the convergent point. What are its proper motion and radial velocity? Solution. The relationship applying to all stars in a moving cluster is: With the values given, the distance is found from:

  9. The star’s parallax is: And its space velocity is: The other star must share the same space velocity as Delta Tauri, so its proper motion and radial velocity are:

  10. Problem. The open cluster Bica 6, at l = 167º, has a radial velocity of 57 km/s, or 48 km/s relative to the LSR. Its distance from main-sequence fitting is 1.6 kpc. Is the cluster’s motion in the Galaxy consistent with Galactic rotation? Solution. See predictions at right. The cluster should have a LSR velocity of −9 km/s if it coincided with Galactic rotation. The cluster’s actual LSR velocity of 48 km/s is therefore completely inconsistent with Galactic rotation. Can you think of a reason why?

  11. Problem. Interstellar neutral hydrogen gas at l = 45º has a radial velocity of +30 km/s relative to the LSR. What are its distance from Earth and the Galactic centre if R0 = 8.5 kpc and θ0 = 251 km/s? Solution. Recall: For a flat rotation curve, θ = θ0 = 251 km/s, so:

  12. For plane triangles the cosine law is: Here: Angle A = 45º Side b = 8.5 kpc = R0 Side a = 7.27 kpc = R Side c = distance d to cloud So: Yielding: Which is a quadratic equation with solution:

  13. Problem. The distance modulus of the Large Magellanic Cloud (LMC) is 18.43, and its estimated mass is 2  1010M. It orbits the Galaxy in the Magellanic Stream. What is its time scale for dynamical capture by the Milky Way for C = 23? Solution. The distance to the LMC, its “orbital radius” r, is determined from: so: The orbital speed of the LMC can be inferred from Kepler’s 3rd Law: for m1 and m2 in M, a in AU, P in yrs. where:

  14. Since 1 pc = 206265 AU, MMW 2  1011M, then: So: and:

  15. The time scale for dynamical capture can now be evaluated: Which implies dynamical capture of the LMC in only half of its orbital period. But that assumes that Milky Way matter extends as far as the LMC so that there is friction occurring constantly, whereas there is some evidence that the LMC has orbited the Milky Way several times.

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