NGC 5102
NGC 5102 imaged by the Hubble Space Telescope
Observation data (J2000 epoch)
ConstellationCentaurus
Right ascension13h 21m 57.6s[1]
Declination−36° 37′ 49″[1]
Redshift468 ± 2 km/s[1]
Distance12.1 ± 0.7 Mly (3.70 ± 0.23 Mpc)[2][3][4][5]
Apparent magnitude (V)10.4[1]
Characteristics
TypeSA0−[1]
Apparent size (V)8.7′ × 2.8′[1]
Other designations
PGC 46674[1]

NGC 5102 is a lenticular galaxy in the Centaurus A/M83 Group of galaxies. It was discovered by John Herschel in 1835.

Distance measurements

At least two techniques have been used to measure the distance to NGC 5102. The surface brightness fluctuations distance measurement technique estimates distances to spiral galaxies based on the graininess of the appearance of their bulges. The distance measured to NGC 5102 using this technique is 13.0 ± 0.8 Mly (4.0 ± 0.2 Mpc).[2] However, NGC 5102 is close enough that the tip of the red giant branch (TRGB) method may be used to estimate its distance. The estimated distance to NGC 5102 using this technique is 11.1 ± 1.3 Mly (3.40 ± 0.39 Mpc).[3] Averaged together, these distance measurements give a distance estimate of 12.1 ± 0.7 Mly (3.70 ± 0.23 Mpc).[5]

References

  1. 1 2 3 4 5 6 7 "NASA/IPAC Extragalactic Database". Results for NGC 5102. Retrieved 2006-12-15.
  2. 1 2 J. L. Tonry; A. Dressler; J. P. Blakeslee; E. A. Ajhar; et al. (2001). "The SBF Survey of Galaxy Distances. IV. SBF Magnitudes, Colors, and Distances". Astrophysical Journal. 546 (2): 681–693. arXiv:astro-ph/0011223. Bibcode:2001ApJ...546..681T. doi:10.1086/318301. S2CID 17628238.
  3. 1 2 I. D. Karachentsev; V. E. Karachentseva; W. K. Hutchmeier; D. I. Makarov (2004). "A Catalog of Neighboring Galaxies". Astronomical Journal. 127 (4): 2031–2068. Bibcode:2004AJ....127.2031K. doi:10.1086/382905.
  4. ↑ Karachentsev, I. D.; Kashibadze, O. G. (2006). "Masses of the local group and of the M81 group estimated from distortions in the local velocity field". Astrophysics. 49 (1): 3–18. Bibcode:2006Ap.....49....3K. doi:10.1007/s10511-006-0002-6. S2CID 120973010.
  5. 1 2 average(4.0 ± 0.2, 3.40 ± 0.39) = ((870 + 760) / 2) ± ((602 + 802)0.5 / 2) = 820 ± 50


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