Some math about the satellite that orbited the Earth during
the Tiahuanaco calender time period.
Prior to Noah's flood around 10,900 BC there were evidently
290 days in an Earth year. After Noah's flood it appears
there were 360 days in an Earth year till 701 BC. After 701
BC there were 365 and 1/4 days in an Earth year. It also
seems Mars used to pass very close to the Earth every few years during this time. The
book The Calendar of Tiahuanaco is by H.S. Bellamy and P. Allan. The main part of the
Calendar itself is the gateway which weighs around ten tons, made from hard rock
which is around 12 and 1/2 feet wide, by ten feet tall and 1 and 1/2 feet thick. The
Calendar has many carvings on it. Tiahuanaco is now around 12,596 above sea level
but when it was built it was evidently near sea level. Tiahuanaco is 16 degrees
and 27 minutes South and 68 degrees and 41 minutes West. Several scientist spent
years attempting to understand the meanings of the different symbols or carvings
on the Calendar of Tiahuanaco.
By the way Copyright 2025 by Wayne Mckellips. Last
updated 19 Nov 2025. Please copy and share but do not
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According to the Tiahuanaco calender in Bolivia when it was made there were 290 days
in an Earth year. Also at that time there was a satellite orbiting the Earth a little
faster then the Earth rotates. So it was moving faster than a geosynchrous orbit. It
may have been composed of mostly ice. It may also have been smaller than our moon. A
geosyncherous orbit would put the center of the satellite around 43,623 km from the
center of the Earth.
Roche's Limit
Roche's Limit basically says when a large satellite gets to close to say a planet,
the gravity of the planet will pull the satellite into pieces. I read if the satellite
was greater then 200 miles in diameter it could be pulled apart by roche's limit. But I
haven't been able to verify that 200 mile number yet. So
how close can a satellite get to say the Earth before the gravity of the
Earth will pull it apart? First you take the density of the Earth which is 5513 kg/m3.
Divide that by the density of the satellite. The density of our moon is 3346 kg/m3.
Density of Mars is 3934 kg/m3. Density of Mimas a mostly icy satellite of Jupiter is
1100 kg/m3. Density of Titan another satellite of Jupiter is 1600 kg/m3. Density of an
satellite made entirely of ice would probably be 920 kg/m3. So let's divide 5513 by 920.
We get 5.99. Now we need to determine the cube root of 5.99 by calculator to get 1.816.
Now take 1.816 times the radius of the Earth which is 6378 km to get 11,582 km. Lastly
take 11582 times 2.44 for a Fluid Satellite to get 28,260 km. Or take 11582 times 1.26
for a Rigid Satellite to get 14,593 km. It seems an icy satellite would be tore apart
by Earth's gravity when it's center was between around 28,260 km and 14,593 km from the
Earth's Center. Of course, this does not include variables like the satellite's tensile
strength.
Mimas, a satellite of Jupiter, with a density of 1100 kg/m3 would be tore apart when it
was between 26,612 and 13,745 km away from the Earth. Titan with a density of 1600 would
be tore apart when it was between 23,505 and 12,138 km away from the Earth. The moon with
a density of 3346 would be pulled apart between 18,381 and 9,492 km away from the Earth.
The sites I've looked at all say around 9,500 km. So it appears satellites like the moon
and Mars are considered Rigid for this exercise. However, it seems an icy satellite would
be considered more liquid. Mars density is 3934. It's destruction would occur between 17,415
and 8,993 km away from the Earth.
Mass of moon is 7.347 e22 kg. Radius of moon is 1,737 km. Density of ice is 919 kg/m3.
Surface area of whole Earth is 5.1 X 10 to the 8 km2, or 5.101 X 10 to the 11 m2. The
sea level at Dogger Littoral between Britain and the coast was about 120 meters lower
before the Younger Dryas around 13,000 years ago.
What was the volume of the icy Satellite in m3?
The last ice age around 13,000 years ago covered around 8% of the Earth's surface area.
The entire surface area of the Earth, 5.101 exponent 11 m2 times 8% or .08 equals 4.0808
exponent 10 m2. It seems the average ice depth was around 1,000 meters. So let's take
4.0808 exponent 10 m2 times 1,000 meters to get a volume of 4.0808 exponent 13 m3.
Also, sea levels have risen around 120 meters since the last ice age. Around 70% of the
Earth's surface is oceans. So let's take the entire surface area of the Earth, 5.101
exponent 11 m2 times 70% or .70 to get 3.5707 exponent 11 m2. So let's take 3.5707 exponent
11 m2 times 120 meters to get a volume of 4.28484 exponent 13 m3.
I was surprised to find the above two answers in the above two paragraphs were almost the same.
So let's go with a volume of the icy satellite of 4.18 exponent 13 m3 for the rest of our
calculations.
What was the possible diameter of the somewhat icy Satellite?
Now, if the somewhat icy satellite had a volume of 4.18 exponent 13 m3 what was it's diameter.
Fortunately, this website can do the calculations for us. Plugging in a volume of 4.18
exponent 13 m3 I get a diameter of 26.76 miles or 43 km.
What was the possible mass of the somewhat icy satellite?
If the satellite was the density of Mimas, a mostly icy satellite of Jupiter it would have been
1100 kg/m3. So let's take 1100 kg/m3 times a volume of 4.18 exponent 13 m3 to get a mass of
4.598 exponent 16 kg. It seems the icy satellite must have weighted around 4.598 exponent 16 kg.
How many seconds did it take for the Satellite to orbit the Earth?
According to the Tiahuanaco calender in one 290 day year the Satellite orbited the Earth
between 447 and 449 times. The satellite would appear in the west and disappear in the
east moving faster than the Earth was rotating ccw or from east to west as observed
from the North Star. The Earth orbited 290 times in one Tiahuanaco Earth year. However,
during that Earth year the satellite made between 447 and 449 orbits around the Earth.
There are 86,164 seconds in a sidereal day. 447 days divided by 290 days is 1.542. Then
divide the number of seconds in a sidereal day 86,164 by 1.542 to find there were 55,878
seconds in the satellites orbital period when there were 447 days in a year. 449 days
divided by 290 days is 1.548. Now divide the number of seconds in a sidereal day 86,164
by 1.548 to find there were 55,661 seconds in the satellites orbital period when there
were 449 days in a year. It's possibles the satellite got closer to the Earth, and thus
faster, as they were making the Tiahuanaco calender.
How far from the Earth's center was the center of the Satellite as it orbited the Earth?
Distance cubed = {[(mass body one plus the mass of body two) times the universal
gravitational constant] times (orbital period squared)} divided by [(pi squared) times four].
For this formula orbital period is in seconds. Mass is in kilograms. Distance is the meters
from the center of one body to the center of the second body. According to a nasa web site at
https://en.wikipedia.org/wiki/Sidereal_time there are 86,164 seconds in a sidereal earth
day. I used 5.972 e 24 for the Earth's mass, and 4.598 exponent 16 kg as the satellite's mass.
Mass of Earth plus mass of the satellite is 5,972,000,045,980,000,000,000,000.
Times gravitational constant 6.67390 exponent -11, times orbital period squared.
To find orbital period you first divide 447 by 290 to get 1.542. Then divide the
number of seconds in a sidereal day 86,164 by 1.542 to find there were 55,878 seconds
in the satellites orbital period. Now square 55,878 seconds to get 3,122,350,884.
Multiply the previous number by 3,122,350,884 orbital period squared. Then divide
that product by [ (pi squared) times four] or 39.478 to get the cube of the radius.
Now using a computer you can find the radius was 31,535 km when there were 447
satellite orbits in a Tiahuanaco year. However, when there were 449 satellite orbits
in the perhaps later years the distance from the center of the Earth to the center
of the satellite was around 31,508 km. So I believe something happened to bring the
icy satellite even closer to the Earth so that it broke up and as Donald Patten
speculated the pieces picked up a magnetic charge and came in mostly over the North
and South magnetic poles.
Could the break up of the icy satellite explain the rapidly frozen mammoths found
in Siberia and Alaska?
I think so. Mammoths had to eat lots of food just to survive. A lot of food has been
found in their stomachs. It appears they went from a warm climate with lots of
vegetation to suddenly being entombed in icy.
In 1908 on 30 June, the Tunguska asteroid came apart over Siberia while it was between
5 to 10 km or 3 to 6 miles above the Earth. It is estimated to have been only 50 to 60
meters or 160 to 200 feet in diameter. Yet its destructive power as it was pulled apart
flattened around 80 million trees covering over 830 square miles.