View Full Version : Homework help :(
Jenisi
09-06-2005, 12:39 PM
I can't figure out how to do this damn conversion. Imagine that the Sun were the size of a basketball (or any similar sized, but common, sphere of your choosing, e.g. volley ball, soccer ball, etc), and then scale the sizes and distances to Mercury, Earth, the Earth's Moon, Mars, Jupiter and Pluto. You may use whatever units you like, but you must be consistent.
Well the sun is 1,400,000 km
Mercury = 4,880 km
Earth = 12,756 km
Earth's moon = 3,474 km
mars 6,794 km
Jupiter = 142,984 km
Pluto = 2,300 km
39,370 inches in a kilometer
Anyone wanna lead me in the right direction?
Jenisi
09-06-2005, 12:42 PM
Would it be easy to just do a 1" = 92,888 km or something like that?
Detri
09-06-2005, 12:46 PM
Are you making a solar system out of golf/tennis/volleyballs or what?
Jorddyn
09-06-2005, 12:49 PM
Originally posted by Jenisi
Nevermind I think I figured out how to do it
Do yourself a favor and skip the metric to English conversions, though, unless it is part of the assignment.
Jorddyn
Jenisi
09-06-2005, 12:50 PM
Well I figured if i shrink the model down. Basketball is 9" and if 1 in = 155,000 that would be right for the sun. Now I'm just trying to figure out how to scale down the other sizes
Jenisi
09-06-2005, 12:51 PM
Should if I take Mercury which is 4880 and divide 155555 by it. I get 32. So would it be Mercury is 1/32th of an inch?
Celephais
09-06-2005, 12:56 PM
Sun = 9"
Mercury = 0.0313"
Earth = 0.082"
Moon = 0.0223"
Mars = 0.0436"
Jupy = 0.919"
Pluto = 0.0147"
Jorddyn
09-06-2005, 12:56 PM
Originally posted by Jenisi
Should if I take Mercury which is 4880 and divide 155555 by it. I get 32. So would it be Mercury is 1/32th of an inch?
Your conversion is:
9" = 1,400,000km
1" = 155,555 km
So, solving for the diameter of Mercury:
x = 4880km
x = 4880km * (1"/155555km)
x = .03137"
Jorddyn
Jorddyn
09-06-2005, 12:57 PM
Originally posted by Celephais
Sun = 9"
Mercury = 0.0313"
Earth = 0.082"
Moon = 0.0223"
Mars = 0.0436"
Jupy = 0.919"
Pluto = 0.0147"
<poke> She has to know HOW to do it, not just the answers.
Jorddyn, could be a teacher :D
Jenisi
09-06-2005, 12:58 PM
Hehe, now I have to convert the distances.
Celephais
09-06-2005, 01:01 PM
Basically take your frame of reference, (9") divide by the actual (1,400,000) take that crazy small fraction (6.428 E-6) and multiply all your other actual sizes by it.
Take that same number and multiply all your distances by it. Voila.
Edit: Damn smileys.
[Edited on 9-6-2005 by Celephais]
Terminator X
09-06-2005, 01:06 PM
I had a chemistry teacher who used to make us write out an entire equation when doing conversion factors and would take points away on quizes and exams if we didn't show the work of multiplicative identity properties, moving the decimal, writing out scientific notation even when it the final product was xy(10^0) feh.
Gigantuous
09-06-2005, 01:25 PM
I slept through chemistry. :P
Jenisi
09-07-2005, 11:05 PM
Anyone have an analemma and know what degree latitude the sun was on Sept 6th?
Yeah, this analemma's been a pain in my ass, literally, for the past two weeks.
Latitude was 48.
Jenisi
09-07-2005, 11:07 PM
central
Jenisi
09-07-2005, 11:09 PM
thanks peam
I hope someone sees this before she edits it.
Celephais
09-07-2005, 11:10 PM
Yeah I dont' know what I'm talking about...
I did find this out though:
SUNPOS: Longitude:-70 11 0.00, Latitude: 32 45 22.00 9/6/2005
Sunrise: 05:19 AM, Azimuth 82.9 degrees from North
Sunset: 05:57 PM, Azimuth 277.1 degrees from North
Civil Twilight - AM (For VFR pilots - 6 degrees below horizon) : 04:54 AM
Civil Twilight - PM (For VFR pilots - 6 degrees below horizon: 06:22 PM
Solar noon: 11:38:22 AM, Azimuth: 180 0 0.00, Altitude: 63 10 44.93
Eq of time: 2.37 minutes This is the amount that sun position varies from average because earth's equator and orbit do not lie in the same plane and the because the earth's orbit is slightly elliptical.
Declination: 5 56 6.93 degrees(D M S). The angle of the sun today if observed from the equator (because the earth's axis is tilted relative to it's orbit)
Jenisi
09-07-2005, 11:53 PM
k it was 6 degree's north not 48
Latrinsorm
09-08-2005, 12:24 AM
If you took like one of those time-lapse photographs, and somehow managed to prevent the sky from clouding over, you'd see an analemma. It's the variations in the sun's position caused by the elliptical orbit and tilt of the earth.
http://www.analemma.com/Pages/framesPage.html
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