Yes,it is a perfectly good explanation,but only qualitative analysis, not to mention the quantitative calculation about the moon away from the Earth (3 cm per year).

Correction; the evidence is quaNTitative. The angular momentum of the Earth is changing (measured) and has ended up in the Moon's motion.

Are you suggesting that all this is some sort of State Secret? Is there a conspiracy to suppress what could be a very interesting topic - if it were actually needed to explain what we have observed?

" one stone hit four birds"

The tidal force is composed of a normal component (universal gravitation) and a tangential component (vorticity force). Exactly, the vorticity force enables the moon’s angular momentum to get away from the earth. The earth rotation rate gets lower because of the conservation of angular momentum in the process.

Quantitative Calculation:

Applicable formula: tangential component [vorticity force] Ft = kGMmωCosα/ r2

Basic Data:

Cosα≈ 1

Basic Data:

Mass of the earth Me = 6×1024 kg，Moon-earth distance r = 38.4×107 m

mgh = m v2 /2 ①

v = aT ②

Average velocity of the moon in a week: v = ＝2v/π ③

Angular velocity difference: q = 1-1/30 = 29/30

v = 2aT/π, ω = 1/T

a = KGωM /r2 ④

mg = GMm/r2 ⑤

Daily departing distance of the moon from the earth: h = 2GK2 Mq/π2 r2

= 2×6.67×10-11×0.16×6×1024×29

(38.4×107)2×3.142×30

= 8.51×10-5m

Yearly departing distance of the moon from the earth: 8.51×10-5×365= 0.031m

The results are identical with the measured values.

The vortex theory is also applicable to the earth-moon system. The earth-rotation vorticity force acts on the moon. The moon receives exactly what the earth loses because of the conservation of angular momentum. The counterforce from the moon may cause great waves on the earth surface.

对地球离日等三问题之预判

鉴于六年前运用 涡旋力公式解决 月亮离开地球 问题，现用其中经验公式

h=2GK^2Mq/π^2r^2 对地球离日， 火卫一 Phobos 轨道下降， 火卫二 Deimos 轨道上升等类似问题一并处理并作为对三问题之预判。

calculate the earth are gradually moving away from sun:

cosа≈1

Basic Data:

Me =1.98×10^30kg, sun-earth distance r=1.5×10^11M

mgh=mv － ^2/2 ①

v=at ②

Average velocity of the earth in a week: v － =1/π∫vsinωtdωt= 2v/π ③

Angular velocity difference: q=1-25/365=0.93

v － =2at/π; ω=1/T

a=KGMω/r^2 ④

mg=GMm/r^2 ⑤

25Daily departing distance of the earth from the sun:

h=2GK^2Mq/π^2r^2

=2×6.67×10^-11×0.16×1.98×10^30×0.93/3.14^2×[1.5

×10^11] ^2

=1.77×10^-4 m

It's not much – just 0.25 cm per year

refrence： Why is the Earth moving away from the sun?

§ 18:22 01 June 2009 by Kelly Beatty, SkyandTelescope.com

§ For similar stories, visit the Solar System Topic Guide

calculate about Phobos approaching Mars 轨道下降：

Basic Data: M=6.6×10^23 kg, 火卫一 —— 火星距离 Phobos -Mars distance r=9.345×10^6 m

cycle= 24.62 h， 火卫一公转Phobos cycle： 7.66 h

Angular velocity difference: q=1-24.62/7.66=-2.21 ，

Phobos approaching Mars perday

h=2GK^2Mq/π^2r^2

=2×6.67×10^-11×0.16×6.6×10^23×-2.21/3.14^2×[9.345×10^6] ^2

=-3.6×10^-2 m

Phobos approaching Mars 12 . 8 m peryear

calculate about Deimos moving away from Mars：

Basic Data: M=6.6×10^23 kg, 火卫二Deimos —— 火星Mars distance距离 r=2.3459×10^7 m

Mars cycle: day ， Deimos cycle ： 1.26244day

Angular velocity difference: q=1-1/1.26244=0.2078

Deimos moving away from Mars perday:

h=2GK^2Mq/π^2r^2

=2×6.67×10^-11×0.16×6.6×10^23×0.2078/3.14^2×[2.3459×10^7] ^2

=5 。 39 ×10^-4 m

Deimos moving away from Mars 0 . 19 m peryear

refre:

sharma

由于火卫一的轨道周期是比火星天短，潮汐减速是在减少约 20 米（ 66 英尺），每世纪率其轨道半径。

Because Phobos' orbital period is shorter than a Martian day, tidal deceleration is decreasing its orbital radius at the rate of about 20 metres (66 ft) per century.