From: Jan Wee <jwee@mail.arc.nasa.gov>
Subject: Challenge Question #2 rebuttal
Date: Sun, 10 Nov 1996 09:29:43 -0600
Dear discuss-lfm members,
I received another perspective on Week #2 Challenge Question's answer from high
school senior Philip Gressman. His comments are below for all to see.
Jan Wee, discuss-lfm co-moderator
PS -- On another matter -- the student worksheets are co-packaged and
shipped with the print copy of the Teacher's Guide (when you order it
from PTK for $10.00). You will not find the student activity sheets
included in our online version at this time.
*************************************************************************
>From Philip:
I am a senior at Ava High School in Ava, Missouri. I received a copy of the
problem through my physics teacher, Mr. Verl Smith. I read the given
explanation, yet found it somewhat mathematically vague. I decided to
derive an equation so that I could see the results for myself.
(Being an earthling, I found it difficult to picture such an immense
canyon.) I first used calculus and went through several messy steps to arrive
at the formula, but I re-derived it in the method described and found its
simplicity to be preferable.
I really enjoy getting into the mathematical guts of science because it helps my
mathematical head picture things as well as "manipulate" the universe.
The solution given to last week's challenge question was a bit
misleading. While it is possible that one could stand on one side of the
Valles Marineris and be completely unable to see the other side, the
chances of this occuring are very slim. In fact, if the part of the
canyon in question is more than approx. 1.5 km deep, the other side will
be visible. This arrives from writing an equation for this effect.
C B
\ A /
D\---|---/E
\ | / (mOC) = (mOB) = 3393 km
\ | / (DOE) is a sector of Mars
\|/
O
(In the above diagram, OC and OB are two normal Martian radii and OA is
the distance from the center of Mars to the floor of the hypothetical
canyon floor.Here CD is the depth of the canyon such that (mOA)+(mCD) =
radius of Mars) Ideally, an observer standing at C would cease to be
able to see point B when the line of sight is tangent to A. This occurs
when (mOB)*cos(AOB) = (mOA). Knowing by elementary geometry that DE, the
width of the canyon, is equal to (mOA)*angleCOB, all angles in radians.
Then, (AOB) = arccos[(mOA)/(mOB)], and thus the width of the canyon sould
be more than 2*(mOA)*arccos[(mOA)/(mOB)]. Upon evaulation of this
quantity, it should be noted that the necessary width of the canyon
exceeds 200 km when the depth of the canyon exceeds 1.5 km. According to
the data we discovered on the Internet, 200 km is the maximum width of
the Valles Marineris. So while it is true that the canyon _could_ be
unimpressive, chances are that it would still impress most of us.