Radio Telescope Resolution - Centaurus A

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SUMMARY

The discussion focuses on calculating the physical size of Centaurus A, which is 15 million light years away, using the resolution capability of radio telescopes. Specifically, it establishes that a structure measuring 1 arcsecond across corresponds to a physical size of approximately 75 light years. The calculation involves understanding the relationship between angular resolution and physical distance, where 1 arcsecond is defined as 1/3600 of a degree.

PREREQUISITES
  • Understanding of angular measurements, specifically arcseconds and degrees.
  • Basic knowledge of geometry, particularly circumference calculations.
  • Familiarity with astronomical distances, particularly light years.
  • Concept of radio telescope resolution and its implications in astronomy.
NEXT STEPS
  • Research the formula for calculating the circumference of a circle based on radius and angular measurements.
  • Explore the principles of radio telescope resolution and its impact on astronomical observations.
  • Learn about the methods used to measure astronomical distances, including parallax and standard candles.
  • Investigate the characteristics and significance of Centaurus A in the context of extragalactic astronomy.
USEFUL FOR

Astronomers, astrophysics students, and anyone interested in understanding the measurement of astronomical objects and the capabilities of radio telescopes.

Timiop2008
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Problem:
Centaurus A is 15 million light years away and a radio telescope can resolve structures of 1s of arc across. Show that the Structure is 75 light years across.
Please Help!
 
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1 sec of arc is 1 / 60 * 1/60 of a degree.
Imagine a circle 15M light years in radius with you at the centre, what length of the circumference would 1arcsec measure?
 

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