Re: Challenge to Jim Scotti
Article: <[email protected]>
Subject: Re: Challenge to Jim Scotti
Date: 26 May 1998 14:32:45 GMT
Since no one else took this on, and the Zetas are obviously trying to
make a point that human theories of gravity and orbital mechanics
cannot be put together, I'll take a stab at it.
In article <[email protected]> the Zetas write:
> In article <[email protected]>
> M.C. Harrison writes.
>> F=M1*M2/r^2 where M1 is one mass, M2 is the
>> other mass, and r is the separation of the masses.
>
> Per your fellow, M.C. Harrison, the force of gravity is equal to
> the force BETWEEN two objects, and one must take them both
> into consideration. Take the weight of a square foot of granite,
> on the surface of your Earth, and use that as a constant for mass,
> as some constant must exist.
I located in a book that the Moon is 3,476 km in diameter, is 400,000
km from the Earth, orbits in a sidereal month in 27.32 days, and the
Earth is 13,528 km in diameter
VOLUME: Not having a handy formula for calculaing volumn (I'm no
mathematician), I did as follows to calculate volume. I recall
somewhere the argument being made that if the 12th Planet was 4 times
the diameter of the Earth, that it's volume would be 68 times that of
Earth. Conversely then, if the Moon has 1/4 the Earth's diameter, it's
volume would be 1/68th that of the Earth, correct? Likewise, I
calculated the volumn of an quarter of either sphere to be slightly
more than half the volumn of a square formed by the diameter of that
sphere, right? Going on the light side. If the Earth is 13,528 km in
diameter, then 1/2 the volume formed by a cube of this is
12,378,578,000,000,000 cubic kilometers, right? And the volume for the
Moon would be 182,037,910,000,000 cubic kilometers.
MASS: To give, as the Zetas suggested, a mass to the Earth and Moon
that the readership can relate to, we are using the weight of granite
at the surface of the Earth (since, as they said, there must be SOME
constant used. My Britannica states that granite is 2.5-2.7 g/cm3
(cubic centimeter). It also states that a metric ton is 1,000,000
grams. Taking the lighter weight of 2.5 g/cm3, then, a cubic kilometer
of granite would be 250,000 grams or .25 metric tons (1/4 of a metric
ton). The Earth, using the constant of a weight on its surface for
mass, would have a mass of 30,946,445,000,000,000 metric tons, and the
Moon 455,094,770,000,000 metric tons.
FORCE of GRAVITY: Using the given of the Moon being 400,000 km from
Earth, on average, and using the given formula of F=M1*M2/r^2, we find
F=30,946,445,000,000,000 * 455,094,770,000 / 400,000^2 or
F=88,022,131,000,000 metric tons. The force of gravity between the Earth
and Moon, then, at the distance the Moon is from the Earth, is equal to
the weight of all those metric tons, trillions of them. Now we find
what keeps the Moon from crashing down to the Earth, weighing what it
does.
In article <[email protected]> the Zetas write:
> In article <[email protected]>
> M.C. Harrison writes.
>> V=v+At, where V is the new velocity and v is the old
>> velocity, A is the acceleration and t is the number of
>> seconds that have passed.
>
> Now we deal with velocity, which you can calculate nicely as
> you have your Moon at a determinate distance from the Earth,
> and it rounds in a short ellipse every 28 days or so. What is
> that speed, in miles per hour, something the common man can
> relate to! Or is your math not REAL and meant to reflect the
> world around you. Do you only deal in abstracts?
Websters defines pi as 3.14 something times the radius, so given the
Moon is some 400,000 km from the Earth, then it travels 1,256,000 km in
27.32 days, or 45,974 km in a day, or 32 km a minute.
In article <[email protected]> the Zetas write:
> Now, if we state, as many have argued, that the Moon is
> FALLING toward Earth, and this is accelerating its rate
> rather than slowing it, then A is SOMETHING ... This
> would mean, for the Moon to maintain an even keel, that
> the force of its existing velocity must be a something that
> is equal to F, the force of gravity.
Given V=v+At, for the Moon to maintain a steady pace, At must be the
sum of it's deceleration due to it's FALL toward Earth, combined with
the forward momentum of the Moon moving in a straight line, which could
be considered the acceleration.
V=v+At
32 km =32 km + (A-D)t where t=a minute.
Here I'm boggling, as the pull AWAY from Earth is not 32 km/minutes,
since the angle is mostly parallel or tangent to the orbit. The pull
AWAY from Earth would be slight, a fraction of 90 degrees, say only 9
degrees at most, 1/10 of the full pull away, or about 9 trillion metric
tons of Moon compared to 88 million metric tons of Moon wanting to fall
toward Earth. So .. we're saying that the equivalent of 88 trillion
metric tons of Moon, speeding along at a rate of 32 km per minute,
would lift away from the Earth some 3.2 km during that minute, and this
is enough to balance out a fall to Earth?
How fast would 88 trillion metric tons fall to Earth, at any speed?
In article <[email protected]> the Zetas write:
> Pardon us for placing your math together in one lump, a
> situation that is avoided, because it simply DOES NOT
> WORK!
