Gravity at Earth's Surface

Physical Law
The acceleration due to gravity at the surface of Earth is approximately given by:
 * $$g = 9.8 \ \mathrm {N} \ \mathrm{kg}^{-1}$$

Note that this is equivalent to
 * $$g = 9.8 \ \mathrm {m} \ \mathrm{s}^{-2}$$

as an acceleration can be defined as the force per unit mass from Newton's Second Law.

Proof
From Acceleration Due to Gravity, we have that:


 * $$g = \frac {G M} {r^2}$$

where:
 * $$g$$ is the acceleration on the body caused by the gravitational field of Earth;
 * $$M$$ is the mass of Earth, approximately $$5.9736 \times 10^{24}\,\mathrm{kg}$$;
 * $$G$$ is the gravitational constant, approximately $$6.674 \times 10^{-11} \, \mathrm N \, \mathrm m^2 \, \mathrm{kg}^{-2}$$;
 * $$r$$ is the radius of Earth, approximately $$6.371 \times 10^{6}\,\mathrm{m}$$.

Hence the result, by direct numerical calculation.

Note that the value of $$r$$ is significantly variable, depending on where you are on Earth.

For fuller detail, see Gravity of Earth at Wikipedia.