Please help me out on this or give me a reference.
Thank you.
NS
You can find this in most earth science or astronomy texts. They took two
measures of the time it took the sun to travel. Knowing the distance between
each point in degrees (1 degree equalling
15 miles) and the measures of the two angles, they can compute the actual
distance. But, this distance does vary. Also, the kids should know the
distance in metric.
The follow thru should extend only to the target. Vunch
What do you mean by 1 degree equalling 15 miles? 15 miles where? An 1-degree
arc that has an arclength of 15 miles has a radius of about 860 miles. I
can't think of anything that is 1720 miles in diameter than has any
relevance to this question.
Now, the Earth rotates one degree on it's axis every 15 minutes - is that
what you meant? Again, that has application to finding the Earth's
circumference, but not the distance to the sun.
I don't know if the earliest measurements of the solar distance were done
this way, but by carefully noting the locations of the sun and planets
relative to the stars (which are assumed to be at such a great distance that
they can be treated as fixed) at six month intervals you can back out the
solar and planetary distances pretty easily based on parallax.
Vunch wrote in message <199805091131...@ladder01.news.aol.com>...
First, Erasthones (SP?) used the trick with shadows to find the size of
the Earth. This is pretty well documented, so I won't go into details.
Second, they knew the Earth's shadow fell on the moon during a lunar
eclipse. By extending the arc of that shadow (easy with a compass and a
photo of an eclipse... the latter is fairly accessible o the web) you can
find out how many times bigger than the moon the Earth's shadow is.
Next, consider a solar eclipse. The moon's shadow nearly pinches out to
nothing as it passess from the moon to the Earth, giving a total solar
eclipse over only a very narow swath. Thus, a shadow cast by the sun at a
distance of about 1 A.U. (give or take .25 million miles) loses about a
lunar diameter traveling the distance from the Earth to the moon. Applying
this to the lunar eclipse scenario, we get that the Earth's shadow must be
about one lunar diameter smaller than the Earth itself. Knowing the
diameter of the Earth, solve for the diameter of the moon.
Now for the distance to the moon. A penny held up at about arm's length
just covers the moon. Measure the diameter of the penny and the distance
from the penny to the eye. This gives a triangle that is similar to the
one that has the lunar orbital radius as one side and the diameter of the
moon as another side. Use proportions to calculate the distance to the
moon.
Finally, that triangle incolving the moon's diameter is similar to the
corresponding triangle whose sides are the distance to the Sun and the
solar diameter... Now they somehow used that relationship to find the
distance to the sun. And damned if I can remeber how atthe moment. I'll
look into it...
Jonathan Heiles
jhe...@mhv.net
Poughkeepsie Day School
Poughkeepsie, New York
Try looking up Aristarchus of Samos (c. 310-230 BC)
-mike
Michael Umbricht, Planetarium Coordinator (401) 785-9457 x227
Museum of Natural History & Cormack Planetarium (401) 461-5146 fax
Roger Williams Park, Providence, RI 02908 http://www.osfn.org/museum/