Hi Praveen, Going on a long trip and need to know what film to buy? I did some quick, back of the envelope calculations and came up with this: based upon the Earth being 10 time higher albedo than the moon, 13.5 times greater cross section area and 8 times closer, I calculate that the Earth would be 9400 times as bright. 10,000 times would mean exactly 10 magnitudes (2.511886...etc raised to the 10th power) I would guess that, depending on the phase of the earth beneath you, it would be 10 mag.s brighter than the moon. If it were full (you standing between the sun and the Earth), it would be mag. -22 or -23 (bright - take fast film!) More quick guesses: noon time sun gives 0.002 moles photons/m2/s on Earth = about 1.2 x 10^21 photons, the reflection up to geosynch (at full-earth condition) would be 50 to 100 times less = about 20,000,000,000,000,000,000 photons/m2/s or about 20 photons per square nanometer per second (a lot). To get it back on topic - nearby geosats would be impressively bright, and worth a few pictures too! -Tyler Quoting Praveen Bharadhwaj <psbad1@sancharnet.in>: > Pls. forgive the slightly off-topic nature of the question, but what would > be the earth's magnitude value as viewed from a geosynchronous orbit? > > More precisely, can someone point me to where I can get information on this > question: > How many photons per second would be captured from the earth with a given > aperture, in a given slice of the visible/IR band, when using a given FOV? > Thanks. > > -- Praveen > > ----------------------------------------------------------------- > To unsubscribe from SeeSat-L, send a message with 'unsubscribe' > in the SUBJECT to SeeSat-L-request@satobs.org > List archived at http://www.satobs.org/seesat/seesatindex.html > > ----------------------------------------------------------------- To unsubscribe from SeeSat-L, send a message with 'unsubscribe' in the SUBJECT to SeeSat-L-request@satobs.org List archived at http://www.satobs.org/seesat/seesatindex.html
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