The moon's orbit is increasing 3.8 cm per year. How much does that warm the earth?

Update: ********
Prefer an answer in watts, or watts per unit area. Thx.
Update 2: ************************** This is conservation of angular momentum and energy. Moon Orbital semi-major Axis = a = 384,399 km da/dt = 3.5 cm/yr m = 7.347 7 x10^22 kg Period = P.m = 27.321 582 days Earth M.e = 5.97 x10^24 kg r.e = 6370 km Rotation earth = P.e = 23 hr, 56 min. Angular momentum... show more **************************
This is conservation of angular momentum and energy.

Moon

Orbital semi-major Axis = a = 384,399 km
da/dt = 3.5 cm/yr
m = 7.347 7 x10^22 kg
Period = P.m = 27.321 582 days

Earth
M.e = 5.97 x10^24 kg
r.e = 6370 km
Rotation earth = P.e = 23 hr, 56 min.

Angular momentum moon
L.m = a *mv
v.m^2/a = G M/a^2
v.m = √ (GM/a)
L.m = m √ (a GM)
dL.m/dt = d L.m/da * da/dt
dL.m/dt = m 1/2 √ ((GM/a) da/dt

The increase in the moon's angular momentum will be offset by a decrease in the Earth's angular momentum.

dL.e/dt + dL.m/dt = 0 ..... Conservation of angular momentum
dLe/dt = - dL.m/dt

L.e = I ω
dL.e/dt = dL.e/dω *dω/dt
dL.e/dt = I *d&omega/dt

dLe/dt = -dL.m/dt
I dω/dt = - m 1/2 √ ((GM/a) da/dt
dω/dt = - m (1/2da/dt) √(GM/a) /I

Earth's rotational energy is
E = 1/2 I ω^2
Update 3: dE/dt = dE/dω *dω/dt dE/dt = I ω * dω/dt dE/dt =( I ω) (- m (1/2da/dt) √(GM/a) /I ) dE/dt = -1/2 mω √(GM/a) da/dt dE/dt = - 3.015x10^12 watts This loss of energy will go into the increase in the moon's orbit and the remainder will be heating of the... show more dE/dt = dE/dω *dω/dt
dE/dt = I ω * dω/dt
dE/dt =( I ω) (- m (1/2da/dt) √(GM/a) /I )
dE/dt = -1/2 mω √(GM/a) da/dt

dE/dt = - 3.015x10^12 watts

This loss of energy will go into the increase in the moon's orbit and the remainder will be heating of the earth. Energy to increase orbit of the moon, per Bekki, below, is

Power = dE/dt = (dE/da)(da/dt)
= (GMm / 2a^2)(da/dt)
= 1.099x10^11 watts

Ergo: the amount heating the earth is roughly 2.9 x10^12 watts

I am at a loss regarding the difference between Wiki, per cite of Captain M, and my answer

Excellent answers by both Captain M and M. Bekki. I'd like to give best answer to both of them. But, I think I'll hold off a day.
Update 4: ***********
Oops,
I used 3.5 cm/yr instead of 3.8 cm per year (3.8 is the correct value)

This increases the power as a result of the earth slowing down to
3.3x^12 Watts, the amount going to increase the moon's orbit to 1.2x10^11 Watts, and the amount heating the earth to 3.1x10^12 Watts
Update 5: *********** I looked at the abstract to the reference cited in Wiki. http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6VGB-3VTY6CC-1&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=d4cf2770df73655388f96f890370568c It says the... show more ***********
I looked at the abstract to the reference cited in Wiki. http://www.sciencedirect.com/science?_ob... It says the 3.75x10^12 watts is from tidal acceleration from astronomical sources. Id. This also includes the sun. The abstract also states that only 2.5x10^12 is from the moon. Id.

I hadn't considered how the sun would also cause a gradual increase in the moon's orbit. So my calculation of 3.3 will be just a tad high for the amount caused solely by the moon. (I do get the 30:1 ratio cited by Wiki for the energy going into increasing the moon's orbit and the energy going into tidal heating on earth, which is good confirmation). ;-)
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