Why the Moon is Receding

This article proposes a model of the earth moon system not advocated by mainstream scientists. Do not copy it as you may get an 'F' on your term paper as well as be ridiculed by friends.

It is generally believed that the recession of the moon is due to the friction caused by the tidal bulge of the earth being carried forward of the moon by the earth's rotation. The bulge causes the earth's rotation to slow down and the loss energy is transferred to the moon, causing it to speed up and therefore recede.

In this article, the angular momentum transfer of an earth moon closed system is calculated using data from the Time Service Dept.^[1], U.S. Naval Observatory and NASA’s Moon Fact Sheet^[2]. The study concludes that tidal friction alone cannot explain the current recession rate. However, with an added recession described in the Hubble’s expansion model, the sum more closely matches the observed expansion of 3.8 cm/year.

We first focus on the recession of the moon due to loss in angular momentum of the earth. The amount of angular momentum gained by the orbital moon will have to be offset by an equal amount lost by the Earth.

Let ω be the angular velocity of the earth. According to the Time Service Dept., U.S. Naval Observatory, in the 188 years from 1820 to 2008, the angular rotation of the Earth decreased from ω = 2 π radians per 86,400 second to ω = 2 π radians per 86,400.002 seconds. Let T_e be the time for the earth to complete one revolution, we have

dω_e = (2 π /T_e²)dT_e

Substituting data from the U.S. Naval Observatory, we have

dω_e ~ -1.683 x 10^-12 radians/sec per 188 years or -8.95 x 10^-15 radians/sec per year

Earth’s moment of inertia can be derived directly from NASA’s Moon Fact Sheet: I/MR² = 0.3308, M_e = 5.9736 x 10²⁴ kg and R_e = 6371 km

I_e = 0.3308 x 5.9736 x 10²⁴ x (6371000) ² = 8.021 x 10 ³⁷ Kg.m²

The change in the earth’s angular momentum (dL_e) is the product of moment of inertia (I_e) and the change in angular velocity (dω_e).

dL_e = I_e dω_e

Substituting above values yields

dL_e = I_e dω_e = 8.021 x 10³⁷ dω_e

dL_e ~ -7.18 x 10²³ Kg.m²/sec per year

As the earth slows down, this change in angular momentum is transferred to the orbital angular momentum (L_T) of the moon

dL_T = -dL_e = 8.021 x 10³⁷ dω_e = 7.18 x 10²³ Kg.m²/sec per year

Let M_m be the mass of the moon, V_o be the mean orbital velocity and R_o be the distance from the moon to the earth, we have

L_T = M_mV_o R_o

and its derivative is

dL_T = M_mV_o dR_o + M_mR_o dV_o

Since orbital velocity can be expressed in terms of gravitational constant (G), orbital radius (R_o) and earth mass (M_e)

V_o = (G M_e)^0.5 R_o^-0.5

Its rate of change is

dV_o = -0.5 (G M_e)^0.5 R_o^-1.5 dR_o

dL_T can be rewritten as

dL_T = M_mV_o dR_o - 0.5 (G M_e)^0.5 R_o^-1.5 dR_o

The equation can be rearranged as

dR_o = dL_T / ( M_mV_o – 0.5 (G M_e)^0.5 R_o^-1.5 )

We can get the moon mass, moon's orbital velocity and the earth mass from NASA’s Moon Fact Sheet

M_m = 0.07349 x 10²⁴ kg; V_o = 1023 m/s; M_e = 5.9736 x 10²⁴ kg

From NASA's "Measuring the Moon's Distance"^[3], the center to center distance of the moon to the earth is R_o = 385,000,000 m

Gravitational constant G = 6.673 × 10^-11 m³ kg^-1 s^-2

Previously we derived dL_T = -dL_e = 8.021 x 10³⁷ dω_e ~ 7.18 x 10²³ Kg m^-2/sec per year

Substitutions of values to dR_o = dL_T / ( M_mV_o – 0.5 (G M_e)^0.5 R_o^-1.5 ) yield

dR_m = 2.12 x 10¹² dω_e

The recession due to tide and angular momentum transfer is approximately

ΔR_o = 2.12 x 10¹² x 8.95 x 10^-15 = 0.019 m /year

According to NASA’s Moon Fact Sheet, the recessional rate of the moon from the earth is 0.038 m/year. As the angular momentum transfer only account for 50% of the recessional rate. There are at least 2 reasons for this under-estimate:

1) I made a mistake and the result is off by a factor of 2. If you see my err, kindly email me at CreationWord (at) live (dot) com.

2) Some other force contributes to the recession. Let's go down this route.

Taking the Hubble constant H = 2.3×10^-18 s^-1 and the distance from the moon to the earth as 385,000 km, the cosmic recessional velocity dR_c as calculated by the expansion model of the cosmos is

dR_c = H R_o

And the recessional distance is

ΔR_c ~ H R_o Δt

The recession per year (31,556,930 seconds) is

ΔR_c ~ 385,000,000 x 2.3×10^-18 x 31,556,930 = 0.028 m /year

The sum of recessions due to cosmic expansion and angular momentum transfer is

ΣΔR = ΔR_c + ΔR_o ~ 0.028 + 0.019 = 0.047 m / year

This value is 24% higher than the observed recessional rate of 0.038 m per year and is closer than that from discounting the expansion effect. Assuming the above measurements and calculations are precise, accurate and correct, a model in which the recession of the moon from the earth is due to both the cosmic expansion and the transfer of the earth’s angular momentum does yield a better fit than one in which only one factor is considered.

Acknowledgements

Reference to the following sources does not imply their endorsement or agreement with the views and opinions expressed in this article.

¹ LEAP SECONDS. Time Service Dept., U.S. Naval Observatory, Washington, DC, http://tycho.usno.navy.mil/leapsec.html, 31 December 2008.
² Dr. David R. Williams. Moon Fact Sheet. NASA Goddard Space Flight Center, http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html, 02 February 2010.
³ Measuring the Moon's Distance. NASA Goddard Space Flight Center, http://eclipse.gsfc.nasa.gov/SEhelp/ApolloLaser.html, 2005 July 11.

I am not a career astrophysicist. This and other articles on the universe stem from my attempts to reach the Creator via examining the creation. I hope you find something useful as you read them judiciously. Thanks. D. N. Pham.