Main Menu - Is There a Dynamic Ether? (Page 6) - Appendix Menu
Average Gravity
The acceleration we call gravity varies with the altitude above the surface of the celestial body (attracting mass). As an object falls, it passes through these various magnitudes of gravity. In falling, it spends more time in the zones of lower gravity than it does in the zones of higher gravity. So the average gravity that an object experiences as it falls cannot be found by a simple linear equation. Instead, there are two non-linear equations used to find the average gravity experienced by an object in free-fall.
re = radius of the earth (for example)
ra = distance from the center of the earth
to a point above its surface
gave = average gravity experienced
by an object when falling
1. Gave = Bge + (1 - B)ga, B = 1/[(ra/re) + 1]
2. Gave = [(ve - va)/H] [(ve + va)/2]
H = (ra - re)
Derivation of Escape Velocity (explanation)
Simple potential energy, "Ep", may be the product of an object's
weight and its distance, "d", above the surface of the earth. This product
is expressed here in English measurement as foot-pounds. A pound is
actually a force, "F", rather than a mass unit, "m", and is equal to the
acceleration of gravity, "g", multiplied by the mass of the object which is
expressed in slugs.
So: Ep = Fd = mgd.
Kinetic energy, "Ek, for an object that has been dropped is equal
to the Ep which existed before it was dropped.
Ek = mv2/2 where v is velocity when an object impacts
upon the earth. Ek at impact is the way the object discharges
what was its Ep.
Gravity is a "force" (acceleration) which extends to infinity (or for as far as the universe extends).
If we were to cause an object to escape from our gravity funnel, it would have to leave the earth's surface with a velocity at least equal to that found in its kinetic energy equation, were it to fall from an infinite distance to the planetary surface. In other words, gravity extends to infinity so the maximum kinetic energy a body can have, due to the earth's gravity alone, is what would be given to it if it were to fall to earth from an infinite distance.
In the absence of any other force, an object at an infinite distance from Earth would not be subjected to Earth's gravity, so the gravitational force from Earth would be equal to zero at this distance. Once having been given a very small shove toward Earth, an object would begin to accelerate along with its surrounding nether that also started at zero velocity toward Earth.
So the nether which passes through the earth's surface should have the same kinetic energy as an object dropped from an infinite distance. Since the two Eks are equal, the velocity found for the object's Ek should be the same velocity as that of nether passing through the earth's surface.
So Ep will be equal to the weight (ma) of the object multiplied by an infinite distance. Ek will be equal to the square of its instantaneous velocity multiplied by its Mass, "m", divided by two. In the process of discovering the velocity, both "m" and the infinite distance divide out and are no longer part of the equation.
The equation we use with the energy equality is that used to find the average gravity, "gave", experienced by an object falling from an infinite distance. Bear in mind that the average gravity varies with the velocity experienced during each part of the fall. The time experienced while at lower gravity will be greater than the time experienced while at higher gravity. Thus, we find that gave is a function of B.
Page 2
Page 1 - Page 3 - Page 4 - Page 5
Main Menu - Is There a Dynamic Ether? (Page 6) - Appendix Menu
Derivation of Escape Velocity (equation)
We know that for an object in space not to be affected by earth's gravity, it must be at an infinite distance, ra, from the earth's surface, re. To fall to earth, this object must be pushed slightly. It will eventually strike the earth with an impact velocity, ve, which can be found because we know that its kinetic energy, Ek, on impact is equal to its original potential energy, Ep. Average gravity during the fall is gave. Gravity at the earth's surface is ge. Gravity at the infinitely remote distance is ga. F is force.
1. B = 1/[(ra/re) + 1], from average gravity equation one.
2. When ra is infinite, ra/re is so great that "+1" is negligible.
3. So: B = 1/(ra/re) = re/ra
4. gave = Bge + (1 - B)ga. From average gravity equation one.
5. At distance ra, ga is zero.
6. So: gave = Bge = (re/ra)ge
7. Ek = Ep, Ek = mve2/2, F = ma or mgave, and Ek = Fra. These are known physics equations.
8. mve2/2 = Fra. Substituting.
9. mve2/2 = mgavera. Substituting.
10. ve2/2 = gavera. Dividing both sides by m.
11. ve2 = 2gavera. Multiplying both sides by 2.
12. ve2 = 2(Bge)ra. Substituting.
13. ve2 = 2[(re/ra)ge]ra. Substituting.
14. ve2 = 2rege. Dividing out like terms.
15. ve = (2rege)1/2. Taking the square root of both sides.
The energy needed for a rocket to escape from Earth is the same as its kinetic energy from falling from an escape distance, so ve is its escape velocity. Since the rocket must have fallen at the same rate as the nether surrounding it would have fallen, the velocity of the nether at the earth's surface would also be ve.
The equation ve = (2rege)1/2
can be generalized as
v = (2rg)1/2 or g = v 2/2r.
So the escape velocity for any celestial body at any point is the same as the instantaneous velocity of the nether moving past at that point.
v = (2Gm/r) 1/2 where G is the gravitational constant.
The gravitational constant comes from the equation:
F = Gm1m2/r 2 where F is force, m1 is the mass of one attracting body, m2 is the mass of a second attracting body, and r is the distance between the two bodies.
Gravity, g, is usually given by the equation:
g = Gm/r 2 where m is the mass of a planet and r is its radius.
By substituting g for Gm/r2 in the usual equation for escape velocity as follows, the equation becomes the one derived above:
v = (2Gm/r) 1/2
v = (2Gmr/r 2) 1/2
v = [(2r)(Gm/r 2)] 1/2
v = [(2r)(g)] 1/2
v = (2rg) 1/2
The derived equation is a more direct means of showing why escape velocity equals nether velocity at any point within a gravity funnel.
Page 3
Page 1 - Page 2 - Page 4 - Page 5
Main Menu - Is There a Dynamic Ether? (Page 1 - Page 6) - Appendix Menu
Proof for Nether Velocity Equation
Hypothesize two spheres (funnel cross-sections), one above the other, and
use the earth's surface as the lower sphere (called "e") with an upper
sphere (called "a"). We want to find the average gravity between the two
spheres. The difference between Ve and Va is the
increase in nether velocity between the two spheres. Gravity is an increase
in nether velocity which is normally given as an increase per second.
[(ve - va)/H], where H is the distance in feet between
the two spheres, is the increase in nether velocity per foot. To find the
increase in nether velocity per second, we must multiply
[(ve - va)/H] by the average number of feet in one
second.
vave = [(ve + va)/2] gives us the average
number of feet per second.
H = (ra - re) gives us difference in the radii
of the two spheres.
1. gave = [(ve - va)/H][(ve + va)/2] for the two spheres. From average gravity equation two.
2. gave = [(ve - va)/(ra - re)][(ve + va)/2]. Substituting for H.
3. gave = [(ve - va)(ve + va)]/[2(ra - re)]. Rearranging the equation.
4. gave = [ve2 - va2]/[2(ra - re)]. Multiplying term on the right side.
5. 2gave(ra - re) = (ve2 - va2). Dividing both sides by the term on the right.
6. va2/ve2 = re/ra. Each spherical cross-section is an energy level.
7. va2 = ve2(re/ra). Multiplying both sides by the bottom term on the left.
8. 2gave(ra - re) = {ve2 - [ve2(re/ra)]}. Substitution into equation five.
9. 2gave(ra - re) = ve2[1 - (re/ra)]. Removed term from parentheses on right side.
10. 2gave(ra - re) = ve2[(ra - re)/ra]. Multiplied one by bottom term on right side and rearranged.
11. 2gave = ve2/ra. Divided both sides by the same terms in parentheses.
12. 2ragave = ve2. Let H approach and become zero, the ra becomes re and gave becomes ge.
13. 2rege = ve2. Substituting.
14. ve = (2rege) 1/2 and
ge = ve2/2re.
Taking the square root of both sides and reversing sides.
For the general equations:
15. v = (2rg) 1/2 and g = v 2/2r.
Page 4
Page 1 - Page 2 - Page 3 - Page 5
Main Menu - Is There a Dynamic Ether? (Page 1 - Page 6) - Appendix Menu
From the above, we know that nether velocity inward is equal to the escape velocity at any point in a gravity funnel. We may call this velocity "vi". The best form of equation for our use here is:
(1) vi2 = 2rg
For instance, to find the incoming nether velocity toward the sun at our orbital distance from it, "vi" is the velocity of nether moving toward the sun, "r" is our distance from the sun, and "g" is the value of solar gravity where we are.
When an object like the earth is in orbit, it has an orbital velocity which we may call "vo". A planet in orbit about a sun remains in orbit when the centrifugal force of its motion is equal to the centripetal force of the sun's gravity at the planet's orbital distance from the sun. If "Fc" is centrifugal force, "m" is the mass of the earth, "a" is acceleration, and "r" and "g" are the same as given in the preceding paragraph, the equation for centrifugal force is:
Fc = mvo2/r = ma
Gravity is the acceleration that balances the "a" in the equation, so:
mvo2/r = mg and if we take "m" from both sides:
vo2/r = g This can be re-written as:
(2) vo2 = rg
If we divide equation (1) with equation (2):
vi2/vo2 = 2rg/rg = 2
then the equation can be re-written as:
(3) vi = (2vo2)1/2
The sun's mass is approximately 333,000 times the mass of the earth. The approximate average orbital velocity of the earth is 18.5169 miles/second. The approximate average distance of the earth from the sun is 93,000,000 miles. Using these figures, the gravity program I devised using the older equations (which do not include orbital velocity) provides an answer for "vi" of 26.12849 miles/second. Equation (3) shown above provides an answer of 26.18682 miles/second. The input figures are approximations, so the two answers should be slightly different from one another. Of the two, the answer provided by equation (3) should be the most correct if we were to be able to find our precise orbital velocity at any given moment. Perhaps this will be possible in the future if someone has not already discovered a means to do so.
Page 5
Page 1 - Page 2 - Page 3 - Page 4
Main Menu -
Is There a Dynamic Ether? (Page 1 -
Page 6) -
Appendix Menu
[The following has been taken from my book, The Oldest Magic, which
was created from magazine articles published in 1989, 1991, and 1992, with
the addition of several other chapters published in 1995.]
The Role of