To Measure the Sky Hints and solutions to the end-of-chapter exercises Chapter 1 2. λ = hc E = 1239.85 eV nm E a) λ = 1240 13.6 = 91.2 nm (Lyman limit) b) 3. f λ has units of brightness/wavelength = W m 3 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 4. dF = f λ d λ = f υ d υ for any amount of flux, so f λ f υ = d υ d λ = − υ 2 c = − c λ 2 5. The observed flux from either star alone is proportional to its surface area times its surface brightness, so, if we treat both stars as blackbodies, the brightness maximum will be: F max = σ R 2 d 2 T 1 4 + a 2 T 2 4 ⎡ ⎣ ⎤ ⎦ When star 1 completely occults star 2, the relative brightness of the system is F 1 F max = T 1 4 T 1 4 + a 2 T 2 4 ⎡ ⎣ ⎤ ⎦ = 1 1 + a 2 ( T 2 4 T 1 4 ) = 1 1 + b When star 2 completely overlaps star 1, then the relative brightness is: F 2 F max = T 1 4 + a 2 T 2 4 − a 2 T 1 4 T 1 4 + a 2 T 2 4 ⎡ ⎣ ⎤ ⎦ = 1 + a 2 T 2 4 T 1 4 − a 2 1 + a 2 ( T 2 4 T 1 4 ) = 1 − a 2 1 + b 6. The energy of 1 photon at frequency 10 6 Hz is h ν =6.63 x 10 -34 x 10 6 =6.63 x 10 -28 J Flux is f υ d υ 1,000,000 1,000,001 ∫ ≅ 10 − 26 Wm -2 Hz -1 ⋅ 1Hz = 10 − 26 Wm -2 = 10 − 26 Js − 1 m -2 N (photons): 10 − 26 Js -1 m -2 6.63 × 10 -28 J/photon = 15 photons m -2 s -1 7. A magnitude of zero means: − 2.5 log(2.65 × 10 − 8 ) + K= 0 ⇒ K = − 18.94 . 8. Assume that the band-pass magnitude is directly proportional to the monochromatic flux at the center of the band, and use the magnitude difference formula to compute the monochromatic flux at the center of the B band in star X: f x = f 1 ⋅ 10 − 0.4(m x − m 1 ) = 375 ⋅ 10 − 0.4(17.79) = 2.87 x 10 − 5 Jy (b) The total energy collected is E = (monochromatic flux) x (bandwidth) x (area) x (exposure time) Or, converting units: E = ([2.87 x 10 -5 Jy] x [10 -26 J s -1 m -2 Hz -1 /1 Jy) x (2.5 x 10 14 Hz) x 100 s x 5 m 2 = 3.59 x 10 -14 J The corresponding number of photons is N = E λ hc = (3.59 × 10 − 14 )(4.4 × 10 − 7 ) (6.63 × 10 − 34 )(3 × 10 8 ) = 7942 9. The difference between the magnitude of two stars together and the magnitude of one star alone is: m both − m one = − 2.5 log ( F + F F ) = − 2.5 log 2 = − 0.753 , so m both = m one − 0.753 = 7.59 10. (a) Apply the reasoning of the previous problem: The magnitude difference between the total nebula and a one arcsec square is: Δ m = − 2.5 log144 = − 5.40 m 144 = m 1 − 5.4 = 17.77 − 5.40 = 12.37 (b) If distance to the nebula doubles, the surface brightness stays the same, the angular area decreases by a factor of 4, and the total apparent magnitude increases by 2.5 log(4) = 1.51 and becomes 13.88. 11. As in the previous problem, if distance to the nebula doubles, the intrinsic surface brightness stays the same, the angular area decreases by a factor of 10 4 , and the total apparent magnitude increases by 2.5 log(10 4 ) = 10 and becomes 22.37. However, in this case one cannot observe the intrinsic surface area or surface brightness because of the limited telescopic resolution. Note that the intrinsic diameter of the nebula would be D intrinsic = 4 π A intrinsic ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 2 = 4 π 144 10 4 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 2 = 0.141 arcsec This is considerably smaller that the actual observed image diameter of 1.2 arcsec, so the observed image will be spread over an area of A obs = π 4 (1.2) 2 = 1.13 arcsec 2 The observed surface brightness will therefore be reduced by a factor of A intrinsic A obs = 0.0144 1.13 = 1.27 × 10 − 2 over the intrinsic value. Or in magnitudes, the surface brightness is S obs = S intrinsic − 2.5 log(1.27 × 10 − 2 ) = 17.77 + 4.74 = 22.51 mag arcsec -2 12. (a) The distance modulus formula , m – M = 5 log (r) -5, gives: 13.25 – (-19.6) + 5 = 37.8 = 5log (r) r = 3.72 × 10 7 pc = 37.2 Mpc (b) To correct for the absorption, repeat the computation above but with the apparent magnitude brightened by 1.5 magnitudes: 13.25 – (-19.6) – 1.5 + 5 = 36.3 = 5log (r) r = 1.82 × 10 7 pc = 18.2 Mpc 13. Begin with the definition of flux: F = L 4 π r 2 , then let F=flux observed from an arbitrary distance , r m = apparent magnitude observed from a distance r F 10 = Flux observed from a distance of 10 pc M = apparent magnitude observed from a distance 10 pc (the absolute magnitude) Applying the formula for magnitude difference: Δ m = m − M = − 2.5 log F F 10 = − 2.5 log 10 2 r 2 = 5 log r − 5 14. Note that the definition of logarithms means F = e ln F and therefore log 10 F = ln F ⋅ log 10 e So, how much does the magnitude change if the flux changes by amount Δ F? Take the derivative of the magnitude equation: m = − 2.5 log 10 F + K , that is: dm = − 2.5 d (ln F ⋅ log 10 e ) = − 2.5 dF F log 10 e = 1.086 dF F , so Δ m ! Δ F F 15. It is important that the aperture used for each star be identical in size; i.e. use the same total number of pixels for each star image. One of many reasonable measurements might go like this: 34 16 26 33 37 22 25 25 29 19 28 25 22 20 44 34 22 26 14 30 30 20 19 17 31 70 98 66 37 25 35 36 39 39 23 20 34 99 229 107 38 28 46 102 159 93 37 22 33 67 103 67 36 32 69 240 393 248 69 30 22 33 34 29 36 24 65 241 363 244 68 24 28 22 17 16 32 24 46 85 157 84 42 22 18 25 27 26 17 18 30 29 35 24 30 27 32 23 16 29 25 24 30 28 20 35 22 23 28 28 28 24 26 26 17 19 30 35 30 26 Background: the 18 pixels 3 X 6 box in the lower left corner have a mean value of 24.11, a median of 25, and a mode 25. Either 24.11 or 25 is a reasonable choice for the background. Assume B=24.11. Brighter star: This star is very symmetric around a point midway between the pixels with values 393 and 363. Add the values of the 16 shaded pixels: Sum of star and background = 2680, so the total number of counts for the brighter star alone is 2680 – (16 x 24.11) = 2294.2 Fainter star: Also very symmetric, but around the center of the pixel with value 229. Take the 21 pixels indicated, but give only 13 pixels full weight (dark shading) and give the 8 outer pixels a weight of 3/8. Thus the effective number of pixels in both star apertures are the same. For the fainter star alone the total counts are: 1056 + ((3/8) x 253) - (16 x 24.11) = 765.1 Magnitude : If the magnitude of the brighter star is 9.000, the magnitude of the fainter is m f = − 2.5 log(765.1 / 2294.2) + 9.0 = 10.192 Chapter 2 1. (a) The tenth datum is unknown, so the population mean can’t be computed. The nine known values are not a random sample. Since they are the nine smallest values, the sample mean is below the population mean by an unknown amount. (b) The population median can be computed because the rank of the missing value is known. The median (the expected survival time) is μ 1 2 = (2.6 + 2.9)/2 = 2.75. . 2. (a) x = x 1 2 = 5. This is a sample, so s 2 = 1 10 2 25 + 16 + 9 + 4 + 1 ( ) { } = 11, and s = 3.31. (b) x = x 1 2 = 5. s 2 = 2 24 16 + 2 ⋅ 9 + 3 ⋅ 4 + 4 ⋅ 1 { } = 4.167, and s = 2.04. 3. (a) is a uniform distribution, with P ( i ) = 1 11 , for integers 0 ≤ i ≤ 10. (else P ( i ) = 0) (b) is a “triangular” distribution with P ( i ) = 1 25 5 − x − 5 ⎡ ⎣ ⎤ ⎦ , for integers 0 ≤ i ≤ 10. (else P ( i ) = 0) 4. (a) In the absence of further information, this appears to be a Poisson process, with a mean rate of 1.7 events per year per square km or r = 1.7 x 10 -4 events per yr per 100 m 2 roof. (b) The probability of zero events per year on the roof is given by the Poisson distribution: P = P P (0, r ) = 1 1 exp( − 1.7 × 10 − 4 ) = .99983 (c) The probability of more than one penetration in 40 years will be: P = 1 − P P (0, 40 r ) − P (1, 40 r ) = 1 − exp( − 6.8 × 10 − 3 ) − 6.8 × 10 − 3 1 exp( − 6.8 × 10 − 3 ) = 1 − 0.99322(1 + .0068) = 2.6 × 10 − 5 5. Because the integral of this function between the limits of zero and infinity diverges, it cannot be interpreted as a valid probability function. (The integral of a normalized continuous probability function equals unity.) This distribution with γ > 0 becomes arbitrarily large as a approaches zero. A student might argue that the mean will be driven this limit, and so in a sense this represents a “typical” member of the population. But the function also permits indefinitely large values a, so in another sense a mean of zero does not represent the population at all. Imposing upper and lower limits on the permitted values of the random variable will allow a computation of a valid mean and standard deviation. (If one investigates the mathematical definitions of the mean and standard deviations of continuous distributions even slightly, it is easy to show that imposing only a lower limit on a —and no upper limit — leads to a well-defined mean if γ > 2, and a well-defined variance if γ > 3.) 6. Follow the reasoning in the example on page 48. Transform the variable q into the standard normal variable z : z = ( q − 0.8) / 0.6, z low = − 1.50, z high = − 0.833 Now compute the probability that a single trial will result in the discovery of an earth-like planet. Use the tabulation of the standard normal distribution P SN (z) = G(z) and its integral given in Appendix C: Prob( earthlike ) = G ( z ) dz − 1.5 − 0.833 ∫ = P (1.5) − P (.833) = .933 − .796 = .137 So in 500 trials, on should expect 68.5 earthlike planets discovered. 7. Counting photons is a Poisson process. The fractional uncertainty in counting N events (equation 2.15) is (a) 0.05 = 1 N , so N =400. (b) N =40,000. 9. The standard deviation of the sample of four measurements is 13.6 km/s, which implies an uncertainty in the mean of 13.6 / 4 = 6.8 km/s. If the astronomer measures N additional stars, to reach an uncertainty of 2.0 km/s, then: 2.0 = 13.6 N + 4 and therefore N=43. 8. Assume she spends equal amounts of time measuring the target and the background. The counts are N * =N b =N meas /2. The uncertainty in the star brightness can be computed from the variance: σ ∗ 2 = σ meas 2 + σ b 2 = 3 N ∗ So the relative uncertainty is σ ∗ N ∗ = 3 N ∗ = 6 N meas and (a) for 5% uncertainty N meas = 2400, (b) N meas = 240,000. 9. The uncertainty in the mean is related to the scatter in the population and the number of samples, N, by : σ mean = σ N We estimate the scatter in the population from the standard deviation (N-1 weighting) of the four measurements as 13.6 km/s. The above equation then implies a sample size of N = 46 would yield an uncertainty in the mean of 2 km/s. 10. Because we know that the fluctuations in the value of r in 10 second exposures are normally distributed with scatter of .05 mV, we can use the Central Limit Theorem to conclude that uncertainty of r in 100-s exposures will be reduced by a factor of 1 10 . We will assume that this detector accumulates voltage in a linear fashion as exposure time increases, so the values for each of the Ns in equation 2.39 on the longer exposures will each increase by a factor of 10. Making these substitutions in equation 2.39: σ * 2 = 1660 + 850 + .05 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 1 10 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2.7556 + .7225 ( ) × 10 6 ⎡ ⎣ ⎤ ⎦ = 2510 + 3.5 = 2513 N * σ * = 810 50.1 = 16.2 11. (a) A straight-forward average of the five trials and a treatment of the five results as random variables gives: x = 21.2, σ mean = 37.2 5 = 16.6 (b) An alternative approach is to use that fact that these are (presumably) counts of photons, so an arrival 106 photons in five seconds suggests (Poisson distribution) an uncertainty in the average of 106 5 = 2.06 The divergence of these two methods suggests that something may be amiss, as does the fact that the mean and median values are quite different. The fact that the result of trial 2 differs from the mean by more than almost 2 σ in method (a) is a little suspect. However, if we really believe that we are counting photons (as in method b) then the second trial is 29 σ larger than the mean, and it is highly unlikely that we are correct in including it as a valid measurement of the same process as the other trials. Often a single deviant event like trial number 2 will alert us to a systematic error. 12. As in the Chapter 1 problem, it is important that the aperture used for each star be identical in size; i.e. use the same total number of pixels for each star image, and centering of the apertures will require the use of fractional pixels. Repeating the measurements made in the solutions for Chapter 1: 34 16 26 33 37 22 25 25 29 19 28 25 22 20 44 34 22 26 14 30 30 20 19 17 31 70 98 66 37 25 35 36 39 39 23 20 34 99 229 107 38 28 46 102 159 93 37 22 33 67 103 67 36 32 69 240 393 248 69 30 22 33 34 29 36 24 65 241 363 244 68 24 28 22 17 16 32 24 46 85 157 84 42 22 18 25 27 26 17 18 30 29 35 24 30 27 32 23 16 29 25 24 30 28 20 35 22 23 28 28 28 24 26 26 17 19 30 35 30 26 Background: the 18 pixels 3 X 6 box in the lower left corner has a total count of 440 and a mean value of 24.44. Uncertainty in the total count is 440 = 20.98 and uncertainty in the average background is 440 / 18 = 1.165. Note that the standard deviation of the sample of 18 pixels is s =4.49, which implies an uncertainty in their mean value of 1.06. This is consistent with the uncertainty computed under the assumption of Poisson statistics. Brighter star: This star is very symmetric around a point midway between the pixels with values 393 and 363. Add the values of the 16 shaded pixels: Sum of star and background = 2680, so the total number of counts for the brighter star alone is F STD =2680 – (16 x 24.44) = 2288.9 with an uncertainty given by: σ std 2 = 2680 + 16 18 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 440 = 3027.7 so σ std = 55.0. Fainter star: Also very symmetric, but around the center of the pixel with value 229. Take the 21 pixels indicated, but give 13 pixels full weight (dark shading) and give the 8 outer pixels a weight of 3/8. Thus the effective number of pixels in both star apertures are the same. For the fainter star alone the total counts are: F f = 1056 + ((3/8) x 253) - (16 x 24.44) = 759.8 and the uncertainty is: σ 2 f = 1056 + 3 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 253 + 16 18 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 440 = 1439.25 so σ f = 37.9 Magnitude : If the magnitude of the brighter star is 9.000, the magnitude of the follows from the ratio R = F f / F STD = 759.8/2288.9 = 0.332: m f = − 2.5 log(759.8 / 2288.9) + 9.0 = 10.197 To get the uncertainty in the magnitude of the fainter star, first note that for the uncertainty in the ratio, R (for a product or ratio, the relative variances add), we have σ 2 F f F STD F f F STD ( ) 2 = σ 2 F f F STD 0.332 ( ) 2 = 37.9 759.8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 + 55 2288.9 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = (0.055) 2 Then either use the result from Problem 1.13, or note, since m f = − 2.5 log( F f / F STD ) + 9.0 = − 2.5 log( R ) + 9.0 σ m 2 = ( − 2.5) 2 σ R log e R ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = (1.086) 2 σ R R ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = (0.060) 2 This assumes that the cataloged standard magnitude of the brighter star is perfectly known. (Not always the case!). So m f = 10.197 ± 0.060. Chapter 3 N α δ δ φ ∆ B A 1 2 N δ + ∆δ θ α ∆ B A r δ B A r ∆δ C Problem 1 Problem 4 1. Draw the great circle connecting the two points, and the hour circle of each point. The result is a spherical triangle whose vertices are the celestial pole and the two points: triangle NAB in the diagram. The sides of this triangle are, respectively, the angle of intertest φ , and the complements of the declinations of the objects, (90°- δ 1 ) and (90° - δ 2 ). The angle opposite side φ is Δα . Apply the law of cosines and simplify: cos ( ⇤ ) = cos (90 ⇥ 1 ) cos (90 ⇥ 2 ) + sin (90 ⇥ 1 ) sin (90 ⇥ 2 ) cos ( ) cos ( ⇤ ) = sin ( ⇥ 1 ) sin ( ⇥ 2 ) + cos ( ⇥ 1 ) cos ( ⇥ 2 ) cos ( ) 2. Apply the result of the previous problem, letting: 1 = latitude of NYC 2 = latitude of Mexico, or LA = longitude difference, NYC and other, so for example: cos φ NY-LA = sin 41 ° sin 34 ° + cos 41 ° cos 34 ° cos (118° – 74°) = 0.818 The NY-LA distance (35.2°) is longer than the NY-Mexico City distance (30.6°). 3. The distance the Earth travels in a sidereal year (the orbit circumference) is 2 π au. But the amplitude of the radial velocity variation is the orbital velocity of the Earth. If a is the length of the au in kilometers, then its estimated value is a = 1 2 π (29.167 ± .057 km/s)(31557940 s)=(1.488 ± .003) × 10 8 km 4. Again, from the spherical triangle with points A, B and the north celestial pole (N). Then apply the law of sines, noting that r, Δδ , and Δ α are all small: sin Δ α sin r = Δ α r = sin θ sin(90 − δ − Δ δ ) = sin θ sin(90 − δ ) thus: Δ α = r sin θ cos δ For triangle ABC, all sides are very small, so plane trig results apply, i.e.: Δ δ = r cos θ (You can do this problem less directly by first applying the law of cosines on △ NCB, and making use of the double-angle formulae and the series expansion of cos for small angles) 5. This is identical to problem 4 with θ = 90° and r = 1000 arcsecs. Thus: Δ α = 1000arcsec cos δ ⋅ 1 sec time 15 arcsec = 66.7 s cos δ e.g. at δ = 85 ° : Δ α = 66.7 s cos(85) = 764 s = 12 m : 44 s 6. Mean ST = Apparent ST – EOT. During December, EOT decreases from about +11 minutes on the 1 st to -2 minutes on the 30th . On the apparent clock, earliest sunset is at the solstice. However, the declination of the sun changes very little close to the solstice, and consequently the apparent time of sunset changes very little near the solstice. Because of more positive values in EOT before the solstice, the mean solar clock will read earlier at sunset on the days before the solstice than on the solstice itself. Zone time will differ from local mean solar time with longitude within a time zone, so the zone time of sunset will likewise differ with location by some constant offset, but the date of earliest sunset will be the same at any longitude. (This date does depend on latitude) 7. At the solstice the sun is at RA = 18 hrs and a midnight the sun has HA = - 12 hrs, so at that time RA = 6 hrs is on the meridian. So at midnight the sidereal time is 6 hrs. At sunset at latitude = + 40, the celestial sphere indicates the sun is at about HA +4:40 when on the horizon. Therefore the local solar time at sunset is 4:40 PM). 8. d= 1 4.17 × 10 -3 = 240pc , θ = a d ⇒ a = θ d = 11 arcsec ⋅ 240 pc = 2640 au 9. Rate of increase of radius is v r = 1 2 Δ λ λ c = 0.08 656 ⋅ 3 × 10 5 = 36.6km/s . (b) The time needed to reach this present radius at this speed is the age of the remnant: T = a v r = 2640 au 37km/s ⋅ 1.49 × 10 8 km au = 1.06 × 10 10 s 3.15 × 10 7 s/year = 338 years (c) Assume the proper motion of the edge of the remnant is due to a tangential velocity equal to the measured radial velocity. Using the relationship for proper motion and distance, the new estimate for the distance is d μ = υ T 4.74 Δ θ 60 = 257 pc 10. (a) from the parallax equation, the distance is: d = a p 206265 = (1.535 × 10 8 )(206265) / (0.32) = 9.984 × 10 13 km The relative uncertainty in d is σ d d = σ p p ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 + σ a a ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ 1 2 = (.04 / .32) 2 + .05) 2 ( ) { } 1 2 = 0.135 or 1.34x10 13 km. (b) The modern estimate is (10.748 ±0.019 )x 10 13 km. 11. Refraction does not change the apparent angular diameter of the sun in the horizontal direction. In the vertical direction, the apparent angular diameter is reduced by the difference between the refractive effect on the top of the solar image and the refraction at bottom of the solar image If R(z) is the refraction at zenith angle z, then the reduction in the angular size of the minor axis when the center of the sun is at zenith angle is Δ b=R ( z – 0.267°) – R (z + 0.267°) 12. (a)There is no month 6 frame because the object is on the ecliptic, so it will be behind the sun at that time. (b) The conservative approach is to set up a reference frame based on the three “background” stars. Since the scales and orientations of the four images are identical, we only need to set the zero point of each frame to be the MEAN POSITION (center of “mass”) of the three background stars on each frame. Then measure x,y coordinates for each object relative to this center on each frame. The combined results are illustrated graphically in the included figure. Each of the three background stars has roughly the same position in each frame, and the moving object does not. From the deviations of individual measurements, it looks like these three stars (open circles) probably have very small but detectable relative systematic motions in latitude (y-direction), but because these are small and not systematic we will ignore them for purposes of the problem. The figure shows the proper motion over one year (dashed line) of the one rapidly moving object. The proper motion measures, in arcsec/yr: μ x =1.75, μ y =3.2, The displacement of the moving object from the dotted line at month 3 measures the parallax (length of line p 3 ) as does, independently, the length of p 9 . The results, again in arc sec, are about: p 3 =1.12, p 9 =0.8, Averaging these, the parallax p = 0.96 arcsec, implies a distance of d =1.04 pc, and μ = 3.65 arcsec/yr. Uncertainties, estimated from the scatter in four measurements of the three background stars, as well as from the disagreement of the two parallax estimates, are about 0.15 arcsec. (c) The distance of one parsec means that this object, although very nearby, is not a member of the solar system. An object bound to the sun in a circular orbit would at a distance of 1 pc, by Kepler’s third law, have a period and proper motion of about P = d 3 2 ≈ (2 × 10 5 ) 3 2 = 9 × 10 7 yr μ = 360 ⋅ 60 ⋅ 60 9 × 10 7 = 0.013 arcsec/yr Therefore, unless some very peculiar effects have mimicked the heliocentric parallax, the object is moving TOO FAST to be an asteroid. (d) The tangential velocity is about 17.6 km/s. 14. (no problem 13).If the uncertainty in the wavelength of an absorption line of wavelength λ is δλ , then the uncertainty in the radial velocity determined from this measurement is just: δ v = c δλ λ = c R Here R is the spectrograph resolution and the extreme right-hand side assumes that δλ is completely determined by spectrograph resolution. If N lines are measured, then the uncertainty in the mean radial velocity should be (assuming constant R): δλ mean = 1 N c R = 0.67 km s − 1 0 3 9 12 p p 3 9 15. Statistical parallax cannot be used to find the distance of this cluster, since there is no reason to suppose that the magnitude of the tangential velocity of the cluster is roughly equal to its radial velocity. If one assumes that the two are in fact equal, then the tangential velocity equation suggests a distance of d = 51 4.74 × 0.145 = 74 pc the propagation of errors gives σ 2 = d 2 1 N σ v 2 v 2 + σ μ 2 μ 2 ⎧ ⎨ ⎩ ⎪ ⎫ ⎬ ⎭ ⎪ = 8.6 pc This, however, is a meaningless result, since the distance is completely unknown. Chapter 4 2. Beta Cen has a parallax of 6.21 mas, as measured by HIPPARCOS. 3. (a) Only seven stars are currently listed: four confirmed and three suspected variables in the GCVS. The Cepheids are CE Cas A, CE Cas B and CF Cas . (b) CF Cas has V=10.80, (c) P=4.87522 days, Julian date epoch of max light = 2437022.191, (e) The most recent treatment is An D.; Terndrup D.M.; Pinsonneault M.H.: Astrophys. J., 671, 1640-1668 (2007). 5. Between 2015 and 2020, Eris will be within two degrees of RA 1:43, DEC -3°, Makemake within three degrees of 12:52 +25°.