Acceleration Due to Gravity/Earth's Surface

Physical Law
Let a body $B$ be situated in a uniform gravitational field $M$ given rise to by a body $P$.

When the body $P$ of mass $M$ is the Earth, and the body $B$ of mass $m$ is located at or near its surface, it is usual to use $g$ for the quantity $\dfrac {G M} {r^2}$.

Therefore the force on $B$ can be expressed as:
 * $F = m g$

Its value is approximately:
 * $9.8 \mathrm m \ \mathrm s^{-2}$ in SI units
 * $32 \mathrm f \ \mathrm s^{-2}$ in FPS units.