Tidal force

Remarks

For $\norm{x} \ll d$, a Taylor expansion gives $$a(x) \approx \frac{GM}{d^3}\big( 3 (x \cdot e)\, e - x \big).$$ Example: at $x = r e$, the point facing $M$, this gives $a \approx 2GMr/d^3$ pointing towards $M$; at $x = -re$ it gives $2GMr/d^3$ pointing away from $M$; at $x \perp e$ it gives $GMr/d^3$ towards the centre. For the Moon acting on the Earth, $2GMr/d^3 \approx 1.1 \cdot 10^{-6}\ \mathrm{m/s^2}$, about $10^{-7}$ of surface gravity. Non-example: the full lunar attraction $GM(d\,e - x)/\norm{d\,e - x}^3$ is not the tidal acceleration. Its value at the centre of mass is a uniform pull that accelerates the whole Earth without deforming it, and only the difference from that value is tidal. The tidal acceleration is the gradient of the tide-generating potential.