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Can you see this too?

  1. Jan 10, 2009 #1
    Hello, I've been wracking my feeble brain trying to find, or derive, an equation that lets one compute under what conditions can an observer on the ground, who has an obstruction of the horizon, see an object high above the Earth, for talks sake, a rocket.
    I tried to draw a diagram to show you what the problem is:

    How high above sea level does the rocket need to be before the over sized man can see it?

    Obviously the height of the rocket and the angle between the persons eyes and the tip of the mountain/obstruction are important values but how to incorporate them is beyond my current ability.
    I understand that one can normally find if it is possible for one to see a satellite/rocket high above sea level by adding the results of the following equations.

    (1) the distance to the horizon for the observer on board the rocket
    (2) the distance to the horizon for the ground observer.

    s= rp(cos^-1)(rp/(rp + h)

    s= curved surface of the Earth below the observer.
    rp= radius of planet
    h= height of observer about sea level.
    Note, your Calculator must be on ''radians''.

    The below link has some info on what I'm trying to compute under the heading'' Optical adjustments and objects above the horizon'' but I have been unable to factor in an obstruction of the horizon for the ground observer.

    Secondly I've been attempting to find a method of converting a flares candela output
    to the astronomical apparent Magnitude scale.
    I've done some searching for the conversion and I found* that 1 candle power =0.981 candela. I think I need to translate candela into lumen and then lumen into lumen/m^2 (lux) and lastly lux into apparent magnitude but I'm not sure on this?

    Let's say I wanted to find out how bright a M112A1 flare would seem to be if it was 10 km above my head at night and ignited. For example, how would an amateur astronomer who saw it classify it on the apparent magnitude scale. Using the apparent magnitude scale would they say it was brighter than Vega, Magnitude of 0, or brighter than Sirius, mag of -1.47.

    The M112A1 weighing 227g emits 120 million candle power
    120,000,000 candle power= 117,720,000 candela

    ¬''A light source that uniformly radiates one candela in all directions radiates a total of 4π lumens. If the source were partially covered by an ideal absorbing hemisphere, that system would radiate half as much luminous flux—only 2π lumens,''

    I will assume that the flare is covered by a hemisphere and I will multiply 2 pi by 117,720,000, which equals 739,655,950 lumen, now I'm completely unsure on this but would I now have to factor in the area of the ground illuminated by the flare and divide the number of lumen by this area to get the lux? if this is the case I need to revise the equation below, yes?

    apparent brightness= L/4(pi)(r^2)
    r = distance from the source to observer
    L =luminosity measured in Lumen
    Brightness is measured in lux.

    **As for converting Lux into apparent magnitude, I'm even more clueless.

    I'd appreciate any helpful pointers or comments.

    I found this usual converter but I'm not exactly sure what units I should be using:

    ** http://www.bautforum.com/archive/index.php/t-32223.html (this source has an ambiguous formula for converting lux to apparent magnitude so I'm not sure on how to do this)
    There's also this site with a useful apparent mag--> lux table
    http://stjarnhimlen.se/comp/radfaq.html but no formula.

    http://zebu.uoregon.edu/~soper/Light/luminosity.html (on this site they use radiometry units and not photometry units so the light is total EM not apparent(preceived light).
    ¬ http://en.wikipedia.org/wiki/Luminosity
  2. jcsd
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