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This is not a new argument but I wanted to see what sort of counter-arguments are produced here. According to Wikipedia, the Sun's surface gravity is 28.02 times earth surface gravity, and the moon's surface gravity is 0.1654 times earth surface gravity.

Assumption set 1:

The angular diameter or apparent diameter of an object is:

2*arcsin(diameter-actual/2/distance to object)

For the sun and moon, these are very nearly the same (during an eclipse, the sun and moon appear to be almost exactly the same size), so

diameter-sun/2/distance to sun = diameter-moon/2/distance to moon

or

distance to sun / distance to moon = diameter-sun / diameter-moon

Assumption set 2:

SurfaceGravitySun = GravitationalParameterSun/HalfSunsDiameterSquared

SurfaceGravityMoon = GravitationalParameterMoon/HalfMoonsDiameterSquared

SunsGravityForceAtEarth = GravitationalParameterSun/DistanceToSunSquared

MoonsGravityForceAtEarth = GravitationalParameterMoon/DistanceToMoonSquared

where the GravatationalParameter is the Gravitational Constant multiplied by the mass of the Sun or Moon.

So...

SurfaceGravitySun * HalfSunsDiameterSquared = DistanceToSunSquared * SunsGravityForceAtEarth

and

SurfaceGravityMoon * HalfMoonsDiameterSquared = DistanceToMoonSquared * MoonsGravityForceAtEarth

If we divide the equations we get equation 1:

SurfaceGravitySun * HalfSunsDiameterSquared / SurfaceGravityMoon / HalfMoonsDiameterSquared =

DistanceToSunSquared*SunsGravityForceAtEarth / DistanceToMoonSquared / MoonsGravityForceAtEarth

Going back to Assumption set 1,

DistanceToSun/DistanceToMoon = DiamenterSun/DiameterMoon

which we can easily make into equation 2:

HalfDistanceToSunSquared/HalfDistanceToMoonSquared = HalfSunsDiameterSquared/HalfMoonsDiameterSquared

So now substitution of equation 2 into 1 yields...

SurfaceGravitySun * HalfDistanceToSunSquared / SurfaceGravityMoon / HalfDistanceToMoonSquared = 
DistanceToSunSquared * SunsGravityForceAtEarth / DistanceToMoonSquared / MoonsGravityForceAtEarth

Removing and cancelling the Half's on the left side means the distance terms completely cancel, leaving equation 3:

SurfaceGravitySun / SurfaceGravityMoon = SunsGravityForceAtEarth / MoonsGravityForceAtEarth

Assumption set 3:

Lunar tides are stronger than solar tides, therefore SunsGravityForceAtEarth < MoonsGravityForce at earth.

Therefore

SurfaceGravityMoon > SurfaceGravitySun

Something's not right here, obviously. So what's wrong with the argument?

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    $\begingroup$ Tidal forces go like the rate of change of gravitational forces, which makes them vary with the cube of the distance rather than the square. That's why the Moon wins: it's much closer. Previously. $\endgroup$
    – rob ♦
    Commented Jul 2, 2017 at 4:39
  • $\begingroup$ Well that's a head scratcher. $\endgroup$
    – Isaac Bolinger
    Commented Jul 2, 2017 at 4:46
  • $\begingroup$ If it's not clear, we might hash it out over in Physics Chat. $\endgroup$
    – rob ♦
    Commented Jul 2, 2017 at 4:48
  • $\begingroup$ I read the other answer, I see the problem must be resolved in the tidal physics. Thank you $\endgroup$
    – Isaac Bolinger
    Commented Jul 2, 2017 at 4:52

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