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  • Part II: Fountains of the Great Deep
    • The Hydroplate Theory: An Overview
    • The Origin of Ocean Trenches
    • Liquefaction: The Origin of Strata and Layered Fossils
    • The Origin of the Grand Canyon
    • The Origin of Limestone
    • Frozen Mammoths
    • The Origin of Comets
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  • Part III: Frequently Asked Questions
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This is the online edition of In the Beginning: Compelling Evidence for Creation and the Flood
(7th Edition) by Dr. Walt Brown. The online version of the book is designed to be read online.
A PDF version or hardbound print version may be ordered.
Copyright © 1995–2008, Center for Scientific Creation. All rights reserved.

Click here to order the hardbound print edition of this online book.

[ Technical Notes > How Long Would It Take the Moon to Recede from Earth to Its Present Position? ]

How Long Would It Take the Moon to Recede from Earth to Its Present Position?

Evolutionists believe that (1) the Earth and Moon are 4.5 billion years old, and (2) with enough time bacteria will change into people. We have all heard some evolutionists say, “Given enough time, anything can happen.” This simplistic attitude overlooks two things. First, most conceivable events will not happen, because they would violate well-established laws of science.1 Second, if 4.5 billion years have elapsed, many things should have occurred that obviously have not. Rather than time being “the hero of the plot,” as one prominent evolutionist stated,2 immense amounts of time cause problems for evolution, as you will now see.

Most dating techniques, including the majority that indicate young ages, make the three basic assumptions given on page 32. The following dating technique has few, if any, major assumptions. It relies basically on only the law of gravity and one undisputed and frequently repeated measurement. We will look at the forces causing the Moon to spiral farther and farther away from Earth. Then we will see that this spiraling action could not have been happening for the length of time evolutionists say that the Earth and Moon have been around.

It will be shown that if the Moon began orbiting very near the Earth, it would move to its present position in only 1.2 billion years. Stated another way, if we could run time backwards, in 1.2 billion years the Moon would be so close to Earth that ocean tides would sweep over all mountains. Astronomers who are aware of this problem call it “the lunar crisis.”3  Notice that this conclusion does not say that the Earth-Moon system is 1.2 billion years old; it only says that the Earth-Moon system must be less than 1.2 billion years old. Had the Moon begun orbiting Earth slightly inside the Moon’s present orbit, its age would be much less. Obviously, something is wrong with either the law of gravity or evolutionists’ belief that the Earth-Moon system is 4.5 billion years old. Most astute people would place their confidence in the law of gravity, which has been verified by countless experiments.

What causes tides?  If the Moon’s gravity attracted equally every particle in and on Earth, there would be no tides. Tides are caused by slight differences in the Moon’s gravitational forces throughout Earth.4 As shown in Figure 181, the Moon pulls more on ocean particle A, directly under the Moon, than it does the center of Earth, C, because A is closer to the Moon. Therefore, A, pulled with slightly more force, moves proportionally farther toward the Moon than C, creating a tidal bulge. Likewise, water particle B, on the far side of Earth, is pulled with slightly less force than C. This difference pulls Earth away from B, creating the far tidal bulge.

tide1.jpg Image Thumbnail

Figure 181: Why the Moon Produces Tides on Earth.

How does the height of ocean tides relate to the Earth-Moon separation distance (R)?  According to Newton’s law of gravitation, the Moon’s gravitational force pulls on Earth’s center of mass (C) with a force proportional to 1/R2. Water particle A directly under the Moon is one Earth radius (r) closer, so it is pulled by a force proportional to 1/(R-r)2.  The difference between these forces is proportional to

tnmoon01.jpg Image Thumbnail

Because r is much less than R, the numerator on the right is almost 2rR and its denominator is almost R4. Therefore, the force difference producing tides and tide heights is approximately proportional to

tnmoon02.jpg Image Thumbnail

Because Earth’s radius (r) is constant, we can conclude that the height of the tides is proportional to 1/R3. For example, if the Earth-Moon distance suddenly doubled, tides caused by the Moon would be only 1/8 as high.5

How do tides affect the Moon’s orbit and the Earth’s spin rate?  Surprisingly, the tidal bulges do not line up directly under the Moon as shown in Figure 181. This is because the spinning Earth carries the bulges out of alignment as shown in Figure 182. If Earth spun faster in the past, as we will see, the misalignment would have been even greater.

Let’s think of Earth as composed of two parts: a spherical portion (gray in Figure 182) and the tidal bulges—both water and solid tides.6  Gs is the gravitational force the Moon feels from the spherical portion of Earth. Because Gs is aligned with the centers of Earth and Moon, it does not alter the Moon’s orbit. However, the near tidal bulge, because it is offset, pulls the Moon in a direction shown by Gn, with a tangential component, Fn, in the direction of the Moon’s orbital motion. Fn accelerates the Moon in the direction it is moving, flinging it into an increasingly larger orbit. The far tidal bulge has an opposite but slightly weaker effect—weaker because it is farther from the Moon. The far bulge produces a gravitational force, Gf, and a retarding force on the Moon, Ff. The net strength of this accelerating force is (Fn - Ff). It can also be thought of as a thrust pushing the Moon tangential to its orbit, moving the Moon farther from Earth. This accelerating force allows us to calculate an upper limit on the age of the Moon. Today’s recession rate has been precisely measured at 3.82 cm/yr,7 but as you will see, it was faster in the past.

Conversely, the Moon’s net gravitational pull applies an equal and opposite force on Earth’s tidal bulges, slowing Earth’s spin. In other words, the Earth spun slightly faster in the past.

tide2.jpg Image Thumbnail

Figure 182: Rotated Tidal Bulges.

How does (Fn  -  Ff) relate to the Earth-Moon separation distance (R)?  Using similar triangles, 

tnmoon33.gif Image Thumbnail

where y is the misalignment distance of each tidal bulge, m is the Moon’s mass, mb is the mass of each tidal bulge, and G is the gravitational constant.  Solving for (Fn - Ff)

 tnmoon04.jpg Image Thumbnail

Equation 1b showed that the mass of a tidal bulge, mb, is approximately proportional to 1/R3, that is

tnmoon05.jpg Image Thumbnail

where C1 is the constant of proportionality. Therefore

tnmoon06.jpg Image Thumbnail

The velocity of the Moon (or any body in a circular orbit) is

tnmoon07.jpg Image Thumbnail

where M is Earth’s mass (or the mass of the central body).

Differentiating both sides with respect to time (t) and solving for tnmoon08.jpg Image Thumbnail gives

tnmoon09.jpg Image Thumbnail

Because the Moon’s tangential acceleration, tnmoon10.jpg Image Thumbnail , is equal to tnmoon20.jpg Image Thumbnail, which is known from equation (2)

tnmoon11.jpg Image Thumbnail

The slight displacement of the tidal bulge (y), as mentioned earlier, is proportional to the difference in the Earth’s spin rate (w) and the Moon’s angular velocity (wL). In other words,

tnmoon12.jpg Image Thumbnail

Substituting (4) into (3) and replacing the product of all constants by C gives

tnmoon13.jpg Image Thumbnail

C is found by using today’s values (subscript t)

tnmoon14.jpg Image Thumbnail

Kepler’s third law shows how (w - wL) varies with R:

tnmoon15.jpg Image Thumbnail

Applying the law of conservation of angular momentum gives

tnmoon16.jpg Image Thumbnail

where the constant L is the angular momentum of the Earth-Moon system, and P is Earth’s polar moment of inertia.  Combining (7) and (8) gives

tnmoon17.jpg Image Thumbnail

Substituting (6), (7), and (9) into (5) gives us the final equation. Because it has no closed-form solution, it will be solved by numerical iteration. The steps begin by setting the clock to zero and R to its present value of 384,400 km. Then time is stepped backwards in small increments (dt) until the centers of the Moon and Earth are only 15,000 km apart. Had this happened, ocean tides would have steadily grown to a ridiculous 12.8 km (8 miles) high and left marks on Earth that would be—but obviously are not—visible.8

tnmoon18.jpg Image Thumbnail

The QuickBasic program that solves this system of equations (shown on page 413) gives 1.2 billion years as the upper limit for the age of the Moon. (If the Moon began moving away from Earth 1.2 billion years ago, the Earth would have rotated once every 4.9 hours.)

Two complicated effects were neglected that would further reduce this upper limit for the Moon’s age.9

1. Evolutionists believe that the Earth formed by gravitational accretion of smaller bodies. If so, the impacts would have left a molten Earth. The Earth, throughout its history, would have been less rigid than it is today. Therefore, tidal bulges would have been larger, causing the Moon to spiral away from the Earth even faster than we calculated here.

2. Internal friction from tidal stretching of the solid Earth reduces Earth’s spin velocity. A greater value for w in the past would have increased the tidal misalignment and the Moon’s recession over what we assumed above. This would have been especially severe if the Earth had been less rigid in the past.

Incorporating these effects into the above analysis would make the upper limit on the Moon’s age even less than 1.2 billion years.

One might argue that 1.2 billion years ago the Moon was captured by the Earth or blasted from the Earth by an extraterrestrial collision.10 These events would have placed the Moon in a very elongated orbit. Today, Earth’s Moon and most of the almost 200 other known moons in the solar system are in nearly circular orbits.11 So many circular, or nearly circular, orbits are difficult to explain with any rigor.12 Therefore, it is highly unlikely that the Moon (1) was captured, (2) was blasted from Earth by an extraterrestrial collision, or (3) somehow began orbiting Earth 1.2 billion years ago.  Its orbit is too circular. (Other problems with evolutionary theories on the Moon’s origin are discussed under “Origin of the Moon” on page 26.)

Besides mountain-eroding tides, what other implications would a 1.2-billion-year-old Moon have for organic evolution and the age of Earth? Evolutionists claim that certain fossils are 2.8–3.5 billion years old. Had the Moon begun orbiting Earth 1.2 billion years ago, such fossils would have been pulverized by the havoc of gigantic tides. Evidently, the Moon did not originate near Earth. This further reduces the maximum age of the Moon.

All other dating techniques must assume how fast the dating clock has always ticked and the clock’s initial setting. For example, radiometric techniques assume, with much less certainty, that each radioactive isotope has a constant half-life. This analysis on the Moon’s recession assumes that only the law of gravity has been constant. Neither assumption can be proven, but there is no doubt which assumptions scientists would favor. If Newton’s law of gravitation did not hold in the past, our scientific foundations would crumble. However, if the Moon is less than 1.2 billion years old, a few evolutionary preconceptions must be discarded.  But that’s progress.

PROGRAM

DEFDBL A–Z  ‘DOUBLE PRECISION
dt = 1  ‘TIME INCREMENT (yr)
G = 6.64E-08  ‘THE GRAVITATIONAL CONSTANT (km3 gm-1 yr-2)
LOP = 13486.23  ‘ANGULAR MOMENTUM OF EARTH-MOON SYSTEM / P (1/yr)
ME = 5.97E+27  ‘MASS OF THE EARTH (gm)
mm = 7.35E+25  ‘MASS OF THE MOON (gm)
P = 8.068E+34  ‘EARTH’S POLAR MOMENT OF INERTIA (gm km2)
R = 384400  ‘TODAY’S EARTH-MOON SEPARATION DISTANCE (km)
Rdot = 0.0000382 ‘TODAY’S RATE OF CHANGE OF R (km/yr)
w = 2301.22  ‘TODAY’S ANGULAR VELOCITY OF THE EARTH’S SPIN (rad/yr)
wL = 83.993  ‘TODAY’S ANGULAR VELOCITY OF THE MOON’S ROTATION (rad/yr)
t = 0  ‘TIME, THE NUMBER OF YEARS AGO (yr)

a = SQR(G * (ME + mm))
b = ME * mm * SQR(G / (ME + mm)) / P
C = Rdot * R ^ 5.5 / (w - wL) ‘FROM (6)

‘marching solution begins

DO
R = R - (C * (w - wL) / R^5.5) * dt ‘FROM (5)
IF R < 15000 THEN LPRINT “The upper limit on the Moon’s age is”; t; “years.”: END
w = LOP - b * SQR(R) ‘FROM (9)
wL= a * R ^ -1.5 ‘FROM (7)
t = t + dt
LOOP

OUTPUT

The upper limit on the Moon’s age is 1,198,032,532 years.

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