Acceleration Due to Gravity of the Earth

Consider an object of mass $m$ resting at the Earth's surface, a distance $R$ (the Earth's own radius) from the Earth's centre, and treat the Earth, for this purpose, as though its entire mass $M$ were concentrated at a single point at its centre (a valid simplification for a body that is very nearly spherically symmetric, which Newton himself proved rigorously using integral calculus). By Newton's law of gravitation, the Earth pulls this object toward its centre with a force

By Newton's second law, this same force also equals $mg$, where $g$ is exactly the familiar acceleration due to gravity used throughout ordinary mechanics problems. Equating the two expressions for $F$ and cancelling the one factor of $m$ common to both sides,

This single equation is the real origin of the number $g \approx 9.8\ \text{m/s}^2$ used throughout mechanics: $g$ is not some separate, independently defined constant of nature at all, but simply what Newton's inverse-square law of gravitation works out to, for an object sitting specifically at the Earth's own surface, a fixed distance $R$ from the Earth's centre. Two consequences follow immediately from this: first, $g$ must depend only on the Earth's own mass $M$ and radius $R$, and not at all on the mass $m$ of the falling object itself -- exactly why, famously, a heavy stone and a light feather fall with the very same acceleration in a vacuum. Second, since $g = GM/R^2$ holds only for an object exactly at distance $R$ from the Earth's centre, any object at a genuinely different distance -- higher up, or lower down -- experiences a correspondingly different value of $g$, which is exactly the subject of the next two sections. …