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 of Motion.

Proof

From Acceleration Due to Gravity, we have that:

$g = \dfrac {G M} {r^2}$

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 universal 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.

$\blacksquare$

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.

Sources

• 1966: Isaac Asimov: Understanding Physics ... (previous) ... (next): $\text {I}$: Motion, Sound and Heat: Chapter $2$: Falling Bodies: Free Fall