Deflection of gravitational acceleration vector due to Sun or Moon

[Adrian B]

On Wolfram Alpha, I get the following values for gravitational acceleration on Earth due to the sun and moon, respectively:

gravitational constant * mass of sun / (distance to sun)^2 → 0.005991 m/s²
gravitational constant * mass of moon / (distance to moon)^2 → 3.68×10^-5 m/s²

If the Earth's gravitational acceleration vector is 9.8 m/s² toward the Earth's center of mass, I'm trying to determine how much that vector is changed by the presence of the sun or moon. I think I can just say that there's a point on Earth where the Sun's 0.005991 m/s² vector would be orthogonal to the Earth's 9.8 m/s² vector so that I can find the angle between the Earth and Earth+Sun vector as:

Inverse Tan(.005911 / 9.8) → 0.03456°

Q1: Is this correct? It is a larger angle than I was expecting.
Q2: On a tangent, why is the moon primarily responsible for tides when the sun's gravitational acceleration at Earth is greater? Is it because the slope of 1/r² is flatter for the Sun than it is for the moon at the distances involved?

 

[BobG]

You use the gradient of each gravitational field, not the forces, so yes to the second part of Q2.

https://en.wikipedia.org/wiki/Tide#Forces

 

[jbriggs444]

I can find the angle between the Earth and Earth+Sun vector as:

Inverse Tan(.005911 / 9.8) → 0.03456°

Q1: Is this correct? It is a larger angle than I was expecting.

Without checking the math, it looks reasonable. This gives you the angle that a ball would fall if dropped from a height onto the ground at sunrise or sunset -- if the ground were not, itself, accelerating toward the sun. Of course, the ground is accelerating sunward. If you pounded an earth-vertical stick into the ground, the ball would trace out a path parallel to the stick and you would measure an angle of zero.