Since the force (F) on a body of mass m in a gravitational field of a body of mass M = GMm/r2 you can see that the force per unit mass is given by F/m. So:

Gravitational Field Intensity (EG) = (g) = F/m = GM/r2

the units for EG are Nkg-1 or since F = ma, F/m = a and so EG may be expressed in ms-2

### Uniform field and radial field

A uniform gravitational field is one where the field lines are always the same distance apart - this is almost exactly true close to the Earth's surface (Figure 1(a)).However if we move back and look at the planet from a distance the field lines clearly radiate outwards (Figure 1(b)), getting further apart as the distance from the Earth increases.

When viewed from an even greater distance the complete field can be seen (as shown in Figure 1(c)).

Such a field is called a radial field - the field intensity (g) decreasing with distance.

The separation of the field lines gives an indication of the strength of the field if they are close together the field intensity is high and of they are far apart it is low.

Diagram 1(d) shows the distortion of the **gravitational field lines** by high- density rock. This was most important for the Apollo Moon landings where NASA discovered concentrations of massive rock below the lunar surface. The resulting variation in the gravitational acceleration at that point would have affected the approach of the lunar lander.

### Acceleration of a satellite in orbit and the value of g at that height

The acceleration of a satellite in orbit is equal to the value of the gravitational field at that height. Since g can be expressed as either Nkg-1 or ms-2 you can see that it can be thought of as an acceleration.Therefore:

g = GM/r2 = a = v2/r for a satellite moving at velocity v in an orbit of radius r.

**Example problems**

A satellite orbits a planet in an orbit or radius 6x106 m where the gravitational field intensity is 0.3 Nkg-1. (See Figure 2)

Calculate :

(b) the velocity in orbit

(c) the mass of the planet (using G = 6.67x10-11 Nm2kg-2)

(a) centripetal acceleration = g = 0.3 ms-2

(b) velocity in orbit a = v2/r

therefore v = (ar)1/2 = (0.3x6x106)1/2 = 1340 ms-1

(c) Using g = GM/r2 M = gr2/G = 1.6 x 1023 kg

### Change in g with height

This is minimal until satellite orbit heights are reached. In fact for a height of some 200 km above the Earth's surface the g value has decreased from 9.8 ms-2 to around 7.5 ms-2.### Comparison between gravitational fields and electric fields

(a) Dependent on BOTH the masses or charges(b) Proportional to the inverse square of their separation

(c) Gravitational fields are ALWAYS attractive

(d) Electric fields can be attractive or repulsive

(e) Electric fields are affected by the intervening medium