Airglow - How To Calculate The Effects of Airglow

How To Calculate The Effects of Airglow

We need first to convert apparent magnitudes into fluxes of photons; this clearly depends on the spectrum of the source, but we will ignore that initially. At visible wavelengths we need the parameter S₀(V), the power per square centimetre of aperture and per micrometre of wavelength produced by a zeroth-magnitude star, to convert apparent magnitudes into fluxes -- W cm−2 µm−1. If we take the example of a V=28 star observed through a normal V band filter ( µm bandpass, frequency Hz), the number of photons we receive per square cm of telescope aperture per second from the source is :

(where is Planck's constant; is the energy of a single photon of frequency ).

At V band, the emission from airglow is V = 22 per square arcsecond at a high-altitude observatory on a moonless night; in excellent seeing conditions, the image of a star will be about 0.7 arc-seconds across with an area of 0.4 square arc-seconds, and so the emission from airglow over the area of the image corresponds to about V = 23. This gives the number of photons from airglow, :

The signal-to-noise for an ideal groundbased observation with a telescope of area (ignoring losses and detector noise), arising from Poisson statistics, is only:

If we assume a 10 m diameter ideal ground-based telescope and an unresolved star: every second, over a patch the size of the seeing-enlarged image of the star, 35 photons arrive from the star and 3500 from air-glow. So, over an hour, roughly photons arrive from the air-glow, and approximately arrive from the source; so the S/N ratio is about :

We can compare this with "real" answers from exposure time calculators. For an 8 m unit Very Large Telescope telescope, according to the FORS exposure time calculator you need 40 hours of observing time to reach V = 28, while the 2.4 m Hubble only takes 4 hours according to the ACS exposure time calculator. A hypothetical 8 m Hubble telescope would take about 30 minutes.

It should be clear from this calculation that reducing the view field size can make fainter objects more detectable against the airglow; unfortunately, adaptive optics techniques that reduce the diameter of the view field of an Earth-based telescope by an order of magnitude only as yet work in the infrared, where the sky is much brighter. A space telescope isn't restricted by the view field, since they are not impacted by airglow.