From flux to AlbedoPosted by Daddy-o Mar 21, 2013 10:49AM
In posts below we have discussed how to best investigate colour differences. Here we saw that sky brightness and exposure time problems can be detected.
Using selected good images in B and V we found the pairs that also were close in time, and generated B-V images. We noted (discussed here) that B-V on the BS is not always near the published value of 0.92, even in images selected for not having (obvious) exposure time problems. We wonder if the value 0.92 is more of a classical photometer value? That the colour of the whole BS on average is 0.92? Perhaps - but we also wonder if the reflectivity of the Moon has a phase dependence so that the B-V colour, even if a BS average, is lunar phase dependent?
Here, we choose to bring the B-V value of the BS in our selected images to 0.92 in order to study which values we get for B-V on the DS. This in an attempt to see if we can discuss what the B-V color of the DS, and therefore of Earth is.
For a range in lunar phases equivalent to illuminated fraction from 35% to 50% we plot the B-V values of a slice across the lunar disc, through the centre and 40 rows wide. We average over the rows:
We have aligned the images by the deep cut, which corresponds to the BS/DS border - the terminator. On the right of this we have the BS and on the left the DS.
We see that the BS is level. We have offset the image values so that the BSs are near 0.92 (by eye). We see that the DS has a slope. We see some level differences in these slopes but the slopes themselves are fairly similar. For one of the profiles the B-V reaches as low as 0.5ish, but there is still a slope.
On the basis of that I think we ought to say that "B-V for earthshine is at or below 0.5 in absolute terms". Better may be to say that "the DS B-V is 0.42 below the BS B-V level."
The B-V os sunlight is 0.656 ± 0.005 (Chris has measured a similar value of 0.64). If we also know that the sunlight on the Moon appears to us to have B-V of 0.92 then we can infer that one reflection from the Moon reddens the sunlight by (0.92-0.656)=0.264. The Sun also shines on Earth (lucky us!) and that light has its colour altered by the Earth - when it strikes the Moon and comes back to us it has B-V of 0.5 or less. Knowing what one reflection off the Moon does to the B-V color we infer that B-V of Earthshine is 0.25 or less. Franklin performed UBV photometry [with a 'diaphragm' of diameter near 1 arc minute - it means he observed areas on the DS, preventing the BS light to enter the photometer, but not BS-scattered light from optical elements before the photometer] on the Earthshine and found that B-V 'for Earth' was 0.17 below B-V for the Sun - this implies that B-V for Earth is 0.47. We have a value (from our lowest value) slightly above that.
1) The above slope is rather straight. The halo itself is there because the halo from B and V have not cancelled. This must be telling us something about the scattered light halos? Probably that it is linear in a lin-log plot - which is what we use using the 'BBSOlog' method. What would Ve1-VE2 images look like? We cannot tell since the halo is not (primarily) due to atmospheric scattering and Mette knows no rule for how scattering in lenses depends on wavelength. We have to try, before we know if Vegetation edge data can be extracted in this way.
2) We also see that we have no images without a B-V gradient across the DS. If we had had an SKE we might hope to achieve 'clean' B-V images. But we don't. So we can't.
3) All is not lost - when we fit images profiles directly we compensate for the halo and can therefore extract actual albedo data that does not depend on the presence of the halo.
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