We can describe how strong or weak a gravitational field is by stating its gravitational field strength.
We are used to this idea for objects on or near the Earth’s surface. The gravitational field strength is the familiar quantity g. Its value is approximately $9.8\,m\,{s^{ - 2}}$. The weight of a body of mass m is mg.
To make the meaning of g clearer, we should write it as $9.8\,N\,k{g^{ - 1}}$. That is, each 1 kg of mass experiences a gravitational force of $9.8 N$.
The gravitational field strength g at any point in a gravitational field is defined as follows:
The gravitational field strength at a point is the gravitational force exerted per unit mass on a small object placed at that point.
This can be written as an equation:

$g = \frac{F}{m}$

where F is the gravitational force on the object and m is the mass of the object. Gravitational field strength has units of $N\,k{g^{ - 1}}$. This is equivalent to $m\,{s^{ - 2}}$.
We can use the definition to determine the gravitational field strength for a point (or spherical) mass. The force between two point masses is given by:

$F = \frac{{GMm}}{{{r^2}}}$

So, the gravitational field strength g due to the mass M at a distance of r from its centre is:

$\begin{array}{l}
g = \frac{F}{m}\\
 = \frac{{GMm}}{{{r^2}m}}\\
 \Rightarrow \frac{{GM\cancel{m}}}{{{r^2}\cancel{m}}}\\
 = \frac{{GM}}{{{r^2}}}
\end{array}$

Since force is a vector quantity, it follows that gravitational field strength is also a vector. We need to give its direction as well as its magnitude in order to specify it completely. The field strength g is not a constant; it decreases as the distance r increases. The field strength obeys an inverse square law with distance. The field strength will decrease by a factor of four when the distance from the centre is doubled. Close to the Earth’s surface, the magnitude of g is about $9.81\,N\,k{g^{ - 1}}$. Even if you climbed Mount Everest, which is $8.85 km$ high, the field strength will only decrease by $0.3\% $.
So the gravitational field strength g at a point depends on the mass M of the body causing the field, and the distance r from its centre (see Worked example 1).
Gravitational field strength g also has units $m\,{s^{ - 2}}$; it is an acceleration. Another name for g is ‘acceleration of free fall’. Any object that falls freely in a gravitational field has this acceleration, approximately $9.8\,m\,{s^{ - 2}}$ near the Earth’s surface. In Chapter 2, you learned about different ways to determine an experimental value for g, the local gravitational field strength.

We can now see in more detail why the Earth’s gravitational field may be considered to be uniform near a planet’s surface. Consider a planet of radius R, then, at its surface, the gravitational field strength, ,$g = \frac{{GM}}{{{R^2}}}$ where G is the universal gravitational constant.
If we move up from the surface, a distance ΔR, the new gravitational field strength, ${g_{new}} = \frac{{GM}}{{{{(R + \Delta R)}^2}}}$.
The percentage change in R is $\frac{{\Delta R}}{R} \times 100\% $ and, using a similar logic to that used when dealing with uncertainties, the percentage change in the gravitational field strength is $2 \times \frac{{\Delta R}}{R} \times 100\% $.

Looking at this from a different viewpoint, if we refer back to Figure 17.6 and the spreading of the field lines, as we go further from the centre of mass we see that a change in height increases the area subtended by those lines. The area subtended increases by the square of the distance and thus moving a distance $\Delta R$ from the surface of the Earth would give a percentage increase in the area subtended by the field lines of $2 \times \frac{{\Delta R}}{R} \times 100\% $. Following similar logic to the worked example, we can see that a $10 km$ vertical movement from the Earth’s surface leads to a percentage increase in area of $16\% $, which shows that over this distance the lines in Figure 17.4 (a rise of less than 10 metres) are very nearly parallel and the field is very nearly uniform.