Question

A convenient parameter for specifying the sensitivity of a radio telescope is its sensitivity in units of K/Jy; that is, the number of Kelvins of antenna temperature TA produced by an unpolarized point source whose flux density is 1 Jy.

(a) (3 points) What is the effective collecting area Ae of a radio telescope whose sensitivity is 1 K/Jy?

Answer 1

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Answer on Question 55055, Physics / Astronomy | Astrophysics

Question:

A convenient parameter for specifying the sensitivity of a radio telescope is its sensitivity in units of K/Jy; that is, the number of Kelvins of antenna temperature TA produced by an unpolarized point source whose flux density is 1 Jy.

(a) (3 points) What is the effective collecting area Ae of a radio telescope whose sensitivity is 1 K/Jy?

Solution:

By definition, $Ae = W_{\nu} / S_{\text{(matched)}} = 2W_{\nu} / S$. Since we can equate incoming energy with a noise temperature, $W\nu = kT_A$. Substituting $k = 1.38 \times 10^{-16} \, \text{ergK}^{-1}$, $T_A = 1K$, and $S = 1 \, \text{Jy}$ (i.e. $10 - 23 \, \text{erg} \, s^{-1} \, \text{cm}^{-2} \, \text{Hz}^{-1}$), we get $Ae = 2kTA / S = 2.76 \times 10^{7} \, \text{cm}^{2}$ (equivalent to a circle of radius about $29.6 \, \text{m}$).