From twirl-8@juno.com Sat Mar 17 13:27:43 2001
Date: Sun, 25 Feb 2001 14:58:37 -0600
From: Ian S Speir (twirl-8@juno.com)
Subject: Fake Moon Landings

Dear Sir or Madam,

I am writing to you in regards to your raising the question concerning
the truth of NASA's landing on the moon. Many of the sites I have visited
on this subject have offered no solid, scientific proof outside of their
interpretation of photographs that would convince anyone that the moon
landings were faked. Some even claim hearsay as evidence; others show
their ignorance for the physical laws of the universe when proposing
ideas. In this letter to you I wish to conduct a scientific experiment
that will prove conclusively to some extent that the moon landings were
faked or were a reality.


First, a small physics lesson. The total distance a body travels is
defined by the equation

d = vt

where d is distance, v is velocity (speed), and t is time. For example,
if a car travels at a rate of 30 m/s for 1 hour, the total distance
traveled is

d = (30 m/s)(3600 s)
d = 108,000 m


Now, this equation works only under the following two conditions: 1)
velocity, v, is constant (i.e., that rate of 30 m/s was exactly that for
the full 3600 seconds), or 2) velocity, v, is an average velocity.
Objects in a gravitational field behave differently because they are
being accelerated when gravity's constant force acts upon them.
Therefore, a different equation must be used to describe their
motion--which, by the way, is called projectile motion for objects in a
gravitational field. This equation is as follows (the symbol "&" has been
substituted for the Greek symbol Delta, which, when used before a
variable in an equation means "change of"):

&d = v.t + .5a(&t)^2

In other words, the change of displacement (or distance), &d, is equal to
the product of the initial velocity, v., and the time, t, PLUS one-half
the product of the rate of acceleration and the change of time, &t,
squared. 


For example, let's say that we are standing atop a cliff and we drop a
heavy rock off the cliff. We can calculate with this equation how far the
rock has fallen for a certain amount of time. Let's say the rock has
fallen for a time of 3 seconds--how far will it have fallen? We insert
our values into the equation:

v. (initial velocity) = 0  --> (This quantity is zero because we did not
throw the rock down the 	cliff with some velocity. We simply dropped it,
which means its velocity right before it 	began to fall was zero.)
t = 3 s 
a = 9.8 m/s^2  --> (This is Earth's acceleration due to gravity, a
constant.)
&t = 3 s

&d = (0)(3 s) + (.5)(9.8 m/s^2)(3 s)^2
&d = 0 + (4.9 m/s^2)(9 s^2)
&d = 44.1 m

So, the rock has fallen 44.1 meters down the cliff, below us.


---------------------
THE EXPERIMENT
---------------------

All right, now that that's behind us and understood, allow me to explain
my experiment. Many-- probably most--of you have seen video footage of
astronauts on the moon. There are certain clips where we see the
astronauts jump into the air. Using our equation we can prove
conclusively if the astronauts are truly jumping on the lunar surface.

If we modify our equation by algebra, we can solve for the acceleration:

&d = v.t + .5a(&t)^2
&d - v.t = .5a(&t)^2
2(&d - v.t) = a(&t)^2
(2&d - 2v.t)/(&t)^2 = a

In other words, the rate of acceleration is equal to the difference of
twice the distance and twice the product of the initial velocity and time
. . . all divided by the change of time squared.

So, if we can insert values into the equation, we can solve for the rate
of acceleration. Now, I have found the moon's acceleration due to gravity
to be 1.6888 m/s^2. I obtained this quantity by the following method. A
video entitled "Man in Space" put out by TIME Magazine in 1989 said that
the space suits worn by the moon-walking astronauts weighed 383 pounds on
earth and about 66 pounds on the moon. Dividing the first number by the
second yields 5.803, and earth's gravity is then that many times greater
than the moon's. So, is we take earth's acceleration due to gravity of
9.8 m/s^2 and divide by 5.803, we get 1.6888 m/s^2. We may then call this
the accepted value of the moon's acceleration due to gravity.

The value can also be obtained using

a = (2&d - 2v.t)/(&t)^2

Let's assume we watch a man on the moon jump into the air. At the peak of
his jump, we measure the distance he has jumped off the ground. Then,
just before he begins to descend, we start our clock. We stop the clock,
of course, when the astronaut again makes contact with the ground. If the
experiment is conducted in this way, we can modify our equation.

Since we measured the time from the peak of his jump to his touching of
the ground, we can say that the astronaut's initial velocity, v., was
zero--just like in our example with the cliff and the rock above.
Therefore the entire -2v.t term goes to zero. The equation then will look
like this:

a = 2&d/(&t)^2


There you have it: the equation. The laws of physics are undeniable. I am
asking you to conduct the experiment described in the previous two
paragraphs using the video footage available from NASA. Of course, the
distance the astronaut jumps will have to be measured in inches first
since we are using the TV screen. Then, it must be compared to the height
in inches of the astronaut. A ratio can be obtained. For example, the
'naut might jump 2 inches in the air and may be 9 inches tall on the TV
screen. If in real life the astronaut is 5'8" (68 inches) tall, then the
following ratio could be determined:

2 inches is to x inches as 9 inches is to 68 inches,  OR
9x = 136
x = 15.1 inches

So the astronaut jumped 15 inches into the air. To insert into the
equation, this must be converted to meters, which would end up being
.3775 m. The time, of course, would not be that difficult to measure. 

I am conducting this experiment myself. However, I wanted to employ the
minds and measurements of others to attempt to eliminate the amount of
error present in this experiment. I ask this of you: that you conduct
this experiment and send me the measurements you used (i.e., the distance
and the time) and your results from the equation. I am hoping that enough
people will respond to this e-mail and we can lay to rest the mystery
surrounding the lunar landings. 


Sincerely and with high hopes,
Ian S. Speir