|Feature Article Supplement- February 2001|
|by Do-While Jones|
|(G) Gravitational constant||6.67E-11||nt-m2/Kg2|
|(Me) Earth mass||5.98E+24||Kg|
|(Mm) Moon mass||7.35E+22||Kg|
Table 2 (below) shows calculated values for five distances between the Earth and Moon. Column A shows what the values would be if the distance between the Earth and Moon was ¼ of what it is now. Column B shows the values for ½ the current distance. Column C contains the calculations for the current distance. Columns D and E calculate what the values would be for 2 and 4 times the current distance, respectively.
|A (¼)||B (½)||C (1)||D (2)||E (4)|
|(Rm) Radius to the moon||9.63E+07||1.93E+08||3.85E+08||7.70E+08||1.54E+09||meters|
|Om/Vm/sec per day||(Pm) Period||3.44||9.72||27.49||77.76||219.95||days|
|½*Mm*Vm2||(Em) Kinetic energy||1.52E+29||7.62E+28||3.81E+28||1.90E+28||9.52E+27||nt-m|
|Mm*Vm*Rm||(Lm) Angular Momentum||1.44E+34||2.04E+34||2.88E+34||4.07E+34||5.76E+34||Kg*m2/sec|
The reason that F changes more rapidly than Rm is that there is a 1/Rm2 term in the equation. Gravitational force varies inversely proportional to the square of the distance. Four squared is 16, so changing the distance by a factor of 4 changes the force by a factor of 16. This means that as the Moon gets closer to the Earth, the tides produced will be much greater.
Velocity of the Moon
We need to know how fast the Moon orbits the Earth as the distance between them changes. The centripetal force required to keep an object of mass M orbiting at a speed V at a distance R is F=MV2/R. Since we know F, M, and R, and want to know V, we use a little algebra to rearrange the equation to V=(FR/M)½ . Row 5 of Table 2 computes the velocity of the moon at the five distances.
Looking at Row 5 you can see that as the radius increases, the speed decreases. That is because the mass of the Moon (Mm) is constant. The only things that change are F and Rm. Even though Rm is increasing, we have already seen that F decreases faster (at a rate of 1/R2.) The net result is that V decreases at a rate of 1/R½.
Just to be sure, let’s do a sanity check on the equation. Table 2 row 5 column C, says that the Moon is moving at 1018 meters/sec. How can we tell if that is true or not? Here is a simple check. The Moon’s orbit is not a perfect circle, but it is pretty close. Let’s see what we get if we assume the Moon has a circular orbit. If the radius of a circle is R, then its circumference is 2*pi*R. So, the distance the Moon travels in one compete orbit is 2.42x109 meters (row 6, column C). If the Moon really is going 1018 meters/sec, it will take 2.42x109 meters / 1018 meters/sec = 2.38x106 seconds to go all the way around. That is 27.49 days, as shown in row 7, column C. The actual period of the Moon is 27.32 days. The 0.6% error is due to the fact that the Moon’s orbit isn’t perfectly circular. This should give us confidence that the calculation of the force, velocity, and period are correct.
In row 8 of Table 2 we have computed the kinetic energy of the moon. It shows that decreasing the radius by a factor of 4 increases kinetic energy by a factor of 4. In other words, the kinetic energy of the Moon is inversely proportional to its distance from the Earth.
The equations show that as the Moon gets farther away from the Earth, it loses kinetic energy.
It turns out, however, that the mistake largely canceled itself out, so it does not change the results enough to change the final conclusion is concerned. The Moon doesn’t orbit the Earth as fast as we had originally calculated. But, the flip side of our incorrect assumption that the Earth would not have slowed down significantly in the last 2 billion years resulted in the conclusion that days were 24 hours long back then. Shortly we will show the calculations of the change in the Earth’s speed that would be necessary to compensate for the change in angular momentum of the Moon (assuming that moment of inertia of the Earth remains constant). When we do, you will see that it turns out that the days were shorter back then. So, although the Moon was only going slower than we originally calculated, the Earth was spinning faster, so the error almost completely canceled out. There is a revised graph at the end of the main essay.
Since the Moon weighs a whole lot more than a figure skater, we can’t assume its angular momentum stays the same. We have to consider the angular momentum of the Earth-Moon system. The amount of angular momentum gained by the Moon will have to be offset by an equal amount lost by the Earth. So, we have to calculate the change of angular momentum of the Earth.
In the 179 years from 1820 to 1999, the angular rotation (usually represented by lower-case omega, but we use W because some web browsers might not support Greek letters) of the Earth decreased from W = 2 Pi radians per 86,400 second to W = 2 Pi radians per 86,400.002 seconds. If you try to compute (2Pi/86,400)-(2Pi/86,400.002) you may or may not get the correct answer, depending upon the number of digits of precision you pocket calculator uses. It is a classic example of the “small difference between large numbers problem” that is well known to numerical analysts. The best way to solve this problem is to use calculus. But, since this essay is written for the general public, many of whom don’t do calculus, we will show you how to solve it using algebra.
W is the angular rotation of the Earth. dW is the small difference in angular rotation we want to compute. Let D represent the number of seconds in a day in 1820. Let dD represent the difference in the length of a day. Symbolically, we want to compute dW = (2Pi/D) - (2Pi/(D + dD). We can factor out 2Pi to get dW = 2Pi*(1/D - 1/(D + dD)). Convert both fractions to common denominator to get dW = 2Pi*((D + dD)/(D*(D+dD) - D/(D*(D+dD)). The two Ds in the numerator cancel out, leaving dW = 2Pi*(dD/(D*D+dD)). At this point we recognize that, as far as a calculator is concerned D+dD = D when dD is very small. So, dW = (2Pi/D2)dD, which is what we would have gotten if we had used calculus in the first place. Since D is 86,400 seconds, and dD is 0.002 seconds, dW is -1.68x10-12 radians/sec (in 179 years). That’s slow! An object spinning that slowly would only turn 0.00000000009.6 degrees in 179 years. But that’s fast enough to affect our calculations.
Angular momentum can also be expressed as a moment of inertia multiplied by an angular velocity. So, the change in angular momentum will be the moment of inertia multiplied by the change in angular velocity.
|Le = Ie*We|
|dLe = Ie*dWe|
All we have to do is multiply -1.68x10-12 radians/sec by the moment of inertia of the Earth to determine how much angular momentum it has lost. This is easier said than done, because we don't know what the moment of inertia of the Earth is. We are going to have to calculate it, based on conservation of angular momentum.
We know that Lm=Mm*Vm*Rm. We know that Vm=(Rm*F/Mm)½. So we can substitute that value for Vm in the equation for Lm. We also know F=G*Mm*Me/Rm2, so we can substitute that value for F in the equation for Lm. After algebraic simplification we get an equation for Lm as a function of constants and Rm.
|Lm = Mm*(G*Me*Rm)½|
We could do the same algebraic trick to figure out dL/dR as we did to compute dW/dD, but you saw what a pain that was. So, let's just use calculus and get the answer directly.
|dLm/dRm = ½*Mm*(G*Me/Rm)½|
Plug in the values for the mass of the Moon, G, mass of the Earth, and current distance to the Moon to get dLm/dRm, and multiply that by the 7.16 meters the Moon receded in 179 years, and you will find that the Moon’s angular momentum increased by 2.68x1026 Kg.m2/sec from 1820 to 1999.
That means the Earth’s angular momentum must have decreased by 2.68x1026 Kg.m2/sec during that same period. Since the change in the Earth’s angular velocity was -1.68x10-12 radians/sec during that time period, the Earth’s moment of inertia must be 15.9x1037 Kg.m2.
|Ie = 15.9x1037 Kg.m2|
Let's do a sanity check to see if this is reasonable. The moment of inertia of a solid sphere is I=(2/5)*MR2, an equation which can be found in any physics or math book that deals with moments of inertia. Therefore, a sphere with uniform density having a mass and volume equal to the mass and volume of the Earth would have a moment of inertia of 9.71x1037 Kg.m2. The Earth isn’t a perfect sphere, nor is it solid, nor does it have uniform density. So, we would not expect the Earth to have exactly this moment of inertia. We would, however, expect it to be reasonably close to this value.
Our calculated value of 15.9x1037 Kg.m2 isn't too far from the estimated value of 9.71x1037 Kg.m2 that we obtained using the solid Earth model, so we feel reasonably sure that it is correct.
We saw previously that Le, the angular moment of the Earth, is equal to Ie*We, but we didn't know Ie at the time. Now we know the moment of inertia of the Earth must be 15.9x1037 Kg.m2. The Earth turned 2*pi radians in 86400 seconds (in 1820), so its angular velocity was 7.27x10-5 rad/sec. Multiplying these two values together we discover that the Earth's current angular momentum is 1.16x1034 Kg.m2/sec.
|Current Angular Momentum of the Moon||2.88E+34 Kg.m2/sec|
|Current Angular Momentum of the Earth||1.16E+34 Kg.m2/sec|
|Total Angular Momentum||4.04E+34 Kg.m2/sec|
The sum of the angular momentum of the Moon and the Earth must remain constant. Therefore, we can use the Lm angular momentum values from Table 2 to figure out what the angular moment of the Earth must be if the Moon is at 1/4, 1/2, and 1 times its current distance. Knowing the Earth's angular momentum we can calculate its angular velocity and kinetic energy (assuming the Earth's moment of inertia doesn't change too much). From the angular velocity we can compute the number of seconds in a day, which can be converted to hours by dividing by 60 twice.
|A (¼)||B (½)||C (1)|
|(Rm) Radius to the moon||9.63E+07||1.93E+08||3.85E+08||meters|
|L-Lm||(Le) Angular momentum||2.60E+34||2.00E+34||1.16E+34||Kg*m2/sec|
|Le/Ie||(We) Angular velocity||1.63E-04||1.26E-04||7.27E-05||rad/sec|
|0.5*Ie*W*W||(Ee) Energy of Earth||2.12E+30||1.26E+30||4.21E+29||Kg*m2/sec|
|2*pi/We/60/60||Length of a Day||10.7||13.9||24.0||hours|
What we really care about is kinetic energy (Em) as a function of radius to the Moon (Rm), and the change in energy with change in radius. So, we can substitute the values for Vm and F into the equation for Em. That makes it easy to take the derivative. When we do that, here is what we get:
|Em = ½*G*Me*Mm/Rm|
|dEm/dRm = -½*G*Me*Mm/Rm2|
Table 2 tells us that the current force of attraction between the Earth and the Moon is 1.98x1020 newtons. That force squared multiplied by some efficiency constant must produce the 1.12x1016 joules of energy transfer. Therefore, the efficiency constant, K, must be 2.86x10-25 joules per newton squared.
|K = 2.86x10-25 joules/nt2|
Now we have all the pieces. We can put them in a very wide spreadsheet and let the spreadsheet figure it all out.
The result is in Table 4, which we have put on a separate page for several reasons.
In the first column, we have let the Moon's radius go from its current value to one-tenth of its current value, in one-tenth steps. The second column differs from the first column only in that we have shown the actual value in meters, rather than a fraction of the current value. In other words, 3.47x108 meters is 0.9 times the current radius. Since these values were produced by an Excel spreadsheet, the value is shown using Excel exponential notation, which uses "E" to indicate raised to a power.
The third column is just 2 pi times the radius in column 2.
The fourth column contains the gravitational force at that separation. It is calculated using the equation in row 4 of Table 2.
The fifth column contains the velocity of the Moon at that separation. It is calculated using the equation in row 5 of Table 2.
The sixth column calculates the number of times the Moon will circle the Earth in one year at that radius. It is obtained by dividing the circumference by the velocity. This tells how many seconds it takes for one orbit. Divide it by 60 twice and 24 once to convert the number of seconds into days. Finally, divide 365.25 days by the number of days it takes for one orbit to determine the number of orbits per year.
The seventh column calculates the amount of kinetic energy change per meter of change in the radius. It is calculated using the derivative found in the section titled, Energy Loss per Year. Since we really want to calculate how much the radius will change as the result of energy transfer, the eighth column contains the reciprocal of the seventh column (simply for convenience).
The ninth column computes the angular momentum of the Moon at that radius and velocity. It uses the equation from row 9 of table 2.
The tenth column subtracts the value of column 9 from the constant total angular momentum of the Earth-Moon system to determine the portion of angular momentum that is the result of the Earth's rotation.
The eleventh column computes the angular velocity of the Earth from its moment of inertia and angular momentum.
The twelfth column computes seconds in a day by dividing 2 pi radians by the angular velocity in column 11. Then it divides by 60 twice to convert seconds to minutes, and minutes to hours.
Since we prefer to compute motion in terms of years that consist of 365.25 twenty-four hour days, we need to compute how many Earth rotations there are in a standard year. This is done by multiplying 365.25 by 24 hours, then dividing by the number of hours in a day from column 12.
The fourteenth column computes the number of tide cycles per standard-length year (i.e. 1820). There is one tide cycle per day, with the loss of cycle per month. So, column 14 is equal to column 13 minus column 6.
In the fifteenth column we have computed the energy per tide cycle using the efficiency constant, K (2.86x10-25 joules per newton squared) and the gravitational force from column 4.
Multiply the tide cycles per year (from column 14) by the energy per cycle (from column 15) to get the energy transfer per year in column 16.
In column 17 we finally start to get to the problem we really want to solve. We want to calculate how far the Moon will move per year at various distances. Column 8 tells how far the Moon moves per joule of energy transferred. Column 16 tells the amount of energy transferred per year at that distance. So, Column 17 contains the product of column 8 and column 16.
The final four columns require some explanation because it is difficult to capture in a simple column heading.
Columns 18 and 19 tell the minimum and maximum times it would take the Moon to go from 90% of today's separation to 100% of today's separation. This is where the "grocery store estimates" come it. Row 2 (counting the header) of column 17 computed the rate of recession to be 0.0398 meters/year when the separation is 100% of today's distance. Row 3 of column 17 tells us that the recession rate would have to be 0.0554 meters/year. So the average rate must be between 0.0398 and 0.0554 meters/year. In column 18 we have determined how long it would have taken for the Moon to go from 90% to 100% of its current radius by dividing the distance by 0.0554 meters per year. This is the minimum time it could have taken. Column 19 makes the same calculation using 0.0398 meters/year. Although we can't compute the actual value, we know it must be between these minimum and maximum values.
Row 2 of columns 18 and 19 tell us that it would take between 695 million to 966 million years to go from 90% to 100% of today's distance. Row 3 of columns 18 and 19 tell us that it would take 492 million to 695 million years to go from 80% to 90%. (Of course, it is no coincidence that 695 million years appears twice. The minimum value of one segment will always be the maximum value of the previous segment because the distance of each 10% segment is the same, and rate used is the same.) Columns 20 and 21 contain the cumulative total of years for each segment. Column 20 says the time to go from 80% to 100% is sum of the minimum number of years to go from 80% to 90% plus the minimum number of years to go from 90% to 100%. Similarly, column 21 contains the sum of the maximum number of years.
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