I think the universe is self-winding.

However, Newton's laws of motion are not consistent with this intuition.
Newton realised that according to his mechanics God's clock-work universe
was winding down so God would need to intervene periodically to wind things
up. Leibniz mocked Newton: If God was a perfect clock maker why would he
have mad such an imperfect clock?

Then Laplace declared God was irrelevant to physics and eventually the
winding down of the universe became enshrined as a law of nature -- the
second law of thermodynamics.


Harry

David Jonsson wrote:

> Hi
> 
> Check this. 
> 
> David
> 
> ------------------------------
> 
> _*A simple and approximate method to determine the gravitational
> torque between two astronomical bodies.*_
> 
> _Introduction_
> Bodies in space rotate slower the farther away from each other they are.
> 
> Look at
> Sun                 Mercury          Venus
> almost still       fast                not that fast
> 
> Also look at
> Sun                 planet's left side            planet's right side
> almost still        wants to move fast        wants to move not that fast
> 
> The left side of the planet want to move faster than the right side.
> The reason is the same as in the case with several planets. If the
> left side experiences a higher speed than the right we get an angular
> momentum or a torque on the planet. The torque contributes to the
> planets rotation around its own axis. The planets' rotation around
> their axises is a mystery since this motion causes a lot of viscous
> heat production in the planet slowing it down. The tidal force on the
> planet would also cause it to slow down. Despite this the planets
> continue to rotate. Lets see if the gravitational torque can explain
> why.
> 
> _Calculation of the torque_
> One way is to view how the position of the sun changes as gravity is
> experienced in the front versus the back of the planet. Assume that
> gravity moves at the speed of light, as many say. (There are evidence
> against this showing it moves instantaneously. This is currently
> ignored.)
> 
> Say the Sun is in horizontal position when viewed from the central of
> Earth. This means the Sun will be a little above the horizon when
> viewed from the equator (say it is midsummer's day for simplicity). On
> midnight on the other side of earth the gravity of the Sun will be
> experienced an equal distance below the horizon.
> 
> This little difference in experiencing the Sun will cause a little
> tangential force on earth.
> 
> 
> Earth at t1
> Sun                                                         Earth at t0
> Earth at t-1
> 
> 
> The torque on earth is symmetrical in a plane through its axis. Lets
> calculate the effect from that plane to the further end and multiply
> by two.
> Some values
> R = distance between Sun and Earth
> r = distance from Earths center
> ft = tangential force on Earth
> fs = Suns force on Earth
> d = displacement of Earth seen from non rotating frame on earth.
> vs = Earths speed relative the Sun
> te = time elapsed since gravity passed Earths center = r/c
> re = Earths radius
> 
> ft is proportional to d/R so
> ft=fs*d/R (1)
> The displacement d is determined by Earths speed and the time in takes
> gravity to pass Earth so
> d=vs*te=vs/c*r (2)
> 
> (2) in (1) makes
> ft=fs*vs*te/R=fs*vs/c*r/R
> 
> Suns force fs on a an element dm of Earth mass
> df=G Ms dm / R²
> and the tangential component
> dft=G Ms dm / R² *vs/c*r/R = vs/c r G Ms dm / R³
> 
> Earths mass per unit length
> dm/dr = rho pi (re²-r²)
> 
> The torque is
> T= r f
> dT = r dft = r vs/c r G Ms dm / R³ = r² vs/c G Ms rho pi (re² - r²) dr / R³
> T=2 vs/c G Ms rho pi / R³ integral (r=0 to re) r² (re² - r²) dr =
> T=2 vs/c G Ms rho pi / R³  (r=0 to re of) (re² r³/3 - r^5/5) =
> T=4 vs/c G Ms rho pi / R³ re^5/15
> 
> This formula is general for any two bodies. The power is P = T*omega
> where omega is the angular velocity of Earth around its axis. Lets
> calculate it to see if it corresponds with the observed value of
> Earths heat production 500 W/m² equal to 2.5*10^17 watts.
> 
> Omega for Earth in radians per seconds is 2 pi / 86400
> For earth P = T omega =
> 4*10^8/365/86400/3/10^8*6.7*10^-11*2*10^30*5515*3.14/(1.5*10^11)^3*(6.3*10^6)^
> 5/15*2*3.14/86400
> = 1.4*10^12 watt = 1 400 gigawatt
> 
> Far too little by a factor 10 000. Maybe I did something wrong? Is
> gravitational torque really this small? Please correct me in that
> case.
> 
> David
> 


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