This force is the reason you feel yourself pulled towards the Earth.

The magnitude (size) of the gravitational force increases when the masses of the objects increase, and it decreases as their separation increases.

The gravitational force that you experience when standing on the Earth's surface depends on your mass, the mass of the Earth, and on the separation between you and the mass of the Earth.

For a large (approximately) spherical object like the Earth, it is the separation between you and the centre of the Earth that is relevant, and this separation is essentially the same for all objects close to the Earth's surface. See centre of mass.

So, the force of gravity you experience due to the Earth (your weight - w) is proportional to your own mass (m).

w = mg

'g' is the gravitational field strength - the force with which gravity pulls each kilogram of mass.

This is roughly the same value at the surface of the Earth at all points.

It is 9.81 N/kg (to 3 s.f.).

It actually increases by about 0.5% between the Equator and the poles.

There are two reasons for this; one is the slightly non-spherical shape of the Earth and the other is the Earth's rotation.

However, we usually ignore this small variation and take the value of the acceleration due to gravity near to the Earth's surface to be 9.81 m s^{−2}.

Let's consider the motion of an object of mass m that is falling towards the Earth - for example, an apple falling from a tree.

Newton's second law of motion states that the acceleration a of the object is related to the unbalanced force F acting on it by the equation:

F = ma

If we assume that the force due to air resistance is very small, and can be neglected, so that the object falls freely, acted on by only the gravitational force g_{}. we can see that:

F = w

so, ma = mg

and therefore

a = g

acceleration due to gravity = g - which on Earth is roughly 10 ms^{-2}