**4. Newton's law and large dimensions mass**

When the dimensions of the source are large, it is necessary to extend the theory. The procedure is to divided the mass into many small elements, and to add the effects of each of these elements, together to measure the gravity effect on certain point.

Because force or acceleration is a vector having both magnitude and direction, it is necessary to resolve the force from each element of mess into its three components (most generally its vertical component and its north–south and east–west components in horizontal plane), before the attraction of the body at any point can be determined.

If we consider the attraction of an irregular laminar body (part of twodimensional sheet) in the xz plane at an external point (p). We first determine the x (horizontal) and Z (vertical) components of acceleration at (p) associated with this attraction.

To do this we divide the plate into (N) small elements of mass, each of area (Δs), if the density (*ρ*n) is uniform within the (nth) elements we express the (x) component of celebration at point (p) due to the attraction of this element as, (**Figure 2**).

laminar body, by divided the mass into small masses (Δs) [1].

$$\mathbf{g}\_{xn} = \frac{G\rho\_n \Delta \mathbf{S}}{r\_n^{\,^2}} \cos \theta = \frac{G\rho\_n \Delta \mathbf{S}}{r\_n^{\,^2}} \frac{X}{r\_n} = \frac{GX}{r\_n^{\,^3}} \rho\_n \Delta \mathbf{S} \tag{9}$$

**Figure 2.** *The gravity determination of two dimensional irregular.*

and

$$\mathbf{g}\_{xn} = \frac{\mathbf{G}\rho\_n \Delta \mathbf{S}}{r\_n^2} \sin \theta = \frac{\mathbf{G}\rho\_n \Delta \mathbf{S}}{r\_n^2}. \frac{Z}{r\_n} = \frac{\mathbf{G}Z}{r\_n^3} \rho\_n \Delta \mathbf{S} \tag{10}$$

Adding the acceleration for all elements

$$\mathbf{g}\_{\mathbf{x}} = \mathbf{G} \sum\_{1}^{N} \frac{X \rho\_{n}}{r\_{n}^{3}} . \boldsymbol{\Delta S} \tag{11}$$

$$\mathbf{g}\_x = \mathbf{G} \sum\_{1}^{N} \frac{\mathbf{Z}\rho\_n}{r\_n^{\ 3}} \Delta \mathbf{S} \tag{12}$$

Where ΔS is very small we can express the two components of acceleration at (p) by the respective integration.

$$\mathbf{g}\_{\mathbf{x}} = G \int \frac{\rho\_{\mathbf{x}} X}{r^3} ds \tag{13}$$

$$\text{gz} = G \int \frac{\rho\_x Z}{r^3} ds \tag{14}$$

Where S = area of body

P = (mass per unit area).

This case can be extended to three dimensional case for a mass per unit volume.

$$\mathbf{g}\_x = G \int \frac{\rho\_x X}{r^3} dv \tag{15}$$

$$\mathbf{g}\_y = \mathbf{G} \int \frac{\rho\_y y}{r^3} dv \tag{16}$$

$$\mathbf{g}\_x = \mathbf{G} \int \frac{\rho Z}{r^3} dv. \tag{17}$$

Where (V) is the volume of the body [1].

In gravity exploration, only the vertical component of force is measured, so that we are normally concerned only with (gz) in determining the attraction at the surface of a buried body.

One of the most important properties of Potential, that it is satisfy Laplace equation anywhere outside the effective gravity mass.

$$\nabla^2 \mathbf{A} = \frac{\partial^2 A}{\partial \mathbf{x}^2} + \frac{\partial^2 A}{\partial \mathbf{y}^2} + \frac{\partial^2 A}{\partial \mathbf{z}^2} = O \tag{18}$$

Laplace equation caused the Ambiguity in gravity and magnetic fields.
