Tidal Acceleration from a Distant Body

A distant body does not pull evenly on a large ball: the near side feels slightly more of its gravity than the centre does. That unevenness is the tidal acceleration, and it is what lifts the oceans on the near and far sides of the Earth at once. The formula $a = 2GMR/d^{3}$ is the difference between the body's gravity at the near side and at the centre, kept to first order in $R/d$. Here $M$ is the mass of the body that pulls, $d$ its distance from the centre of the ball, $R$ the radius of the ball and $G$ the gravitational constant.

The cube of the distance is what matters. Take the Moon's mass of $7.34\times10^{22}$ kg at a distance of $3.84\times10^{8}$ m and the Earth's radius of $6.37\times10^{6}$ m: the formula gives $1.1022\times10^{-6}$ m/s². For the Sun, $1.99\times10^{30}$ kg at $1.50\times10^{11}$ m, it gives $5.014\times10^{-7}$ m/s². The Sun's tidal pull is therefore a little under half the Moon's, though it pulls far harder, because the cube of the distance shrinks it.

The page is checked against a published table of tidal accelerations for the Earth, which prints $1.1\times10^{-6}$ m/s² for the Moon and $0.5\times10^{-6}$ m/s² for the Sun. Those printed values have two significant figures for the Moon and one for the Sun, so they check the page only that closely: within 4.5 percent for the Moon and 10 percent for the Sun. That is enough to catch a missing 2 or a wrong power of the distance, but not a small error. Treat it as a first-order approximation to a difference in gravity, not a precision tool.

The formula keeps only the first term. The exact difference, $GM/(d-R)^{2} - GM/d^{2}$, is larger by about $1.5\,R/d$, which is 2.4 percent for the Moon and nothing noticeable for the Sun. It gives the pull along the line to the body; across that line the tide squeezes, with half the size. The sea's real tide follows the shape of the basins and the water's own response, so this is the sky's share of it, not a tide height. As a round-number check, $M = 10^{22}$ kg, $R = 10^{6}$ m and $d = 10^{8}$ m give $1.33486\times10^{-6}$ m/s², and read backwards give $d = 10^{8}$ m again.