Question: The longest radio-telescope separation available in 2016 is that between telescopes on Earths surface and a 10-m diameter radio telescope in the Russian RadioAstron satellite,
The longest radio-telescope separation available in 2016 is that between telescopes on Earth’s surface and a 10-m diameter radio telescope in the Russian RadioAstron satellite, which was launched into a highly elliptical orbit around Earth in summer 2011, with perigee ∼10,000 km (1.6 Earth radii) and apogee ∼350,000 km (55 Earth radii).
(a) Radio astronomers conventionally describe the specific intensity Iω(α, ω) of a source in terms of its brightness temperature. This is the temperature Tb(ω) that a blackbody would have to emit, in the Rayleigh-Jeans (low-frequency) end of its spectrum, to produce the same specific intensity as the source. Show that for a single (linear or circular) polarization, if the solid angle subtended by a source is △Ω and the spectral energy flux measured from the source is Fω ≡ ∫IωdΩ = Iω△Ω, then the brightness temperature is
where kB is Boltzmann’s constant.
(b) The brightest quasars emit radio spectral fluxes of about Fω = 10−25Wm−2 Hz−1, independent of frequency. The smaller such a quasar is, the larger will be its brightness temperature. Thus, one can characterize the smallest sources that a radio-telescope system can resolve by the highest brightness temperatures it can measure. Show that the maximum brightness temperature measurable by the Earth-to-orbit Radio Astron interferometer is independent of the frequency at which the observation is made, and estimate its numerical value.
= (2)cI kBW (2)3c2Fw kgw (9.29)
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