**5. Gravitational attraction**

Newton's law states that in case of two masses (m1) and (m1) separated by a distance (r) between them, then these two masses attract each other's by force (F).

$$F = \frac{Gm\_1m\_2}{r^2} \tag{19}$$

G = Universal gravitational constant

If The mass of the earth (M), and the radius of the earth (r), then the weight (w) of a body with a mass (m) on the surface of the earth equal to:

$$W = mg = \frac{GMm}{r^2} \tag{20}$$

The above equation used only for non-rotating earth.

$$\mathbf{g} = \frac{GM}{r^2} \tag{21}$$

(g) usually expressed as a weight of a unit mass or earth attraction.

The gravity attraction of the earth varied from point to point on the earth surface, due to that the radius of the earth (r) (which is not constant everywhere on the earth) in addition to the centrifugal force create duo to the rotation of the earth, (**Figure 3**).

**Figure 3.** *The variation of centrifugal force value with latitude.*

The gravity attraction (g) can be measured with high accuracy and the size of the earth can be determined using the astronomical geodetic studies. And as the universal gravitational constant (G) is known, the earth mass can calculated easily.

The gravity attraction can be measured using a pendulum method with accuracy of 5 ppm, while the relative variation in gravity can be measured using special high accurate gravity meter instruments (composed of springs system) with an accuracy of 10�<sup>9</sup> .

The following equation used on assumption of non-rotated earth

$$F = \frac{GMm}{r^2} \tag{22}$$

But the earth is a rotated body. It is rotate around the long axis. This rotation created a centrifugal force (a) varied relatively with the variation of the radius of the circular shape plotted by a rotated points on the earth surface.

$$\mathfrak{a} = rw^2 \tag{23}$$

Where ω is the angular velocity T is the period

$$
\alpha = \frac{2\lambda}{T} \tag{24}
$$

Then

$$\mathbf{a} = 4\pi^2 \,\mathrm{r/T^2} \tag{25}$$

So

α value at the equator is 3.4 gal

$$a = 3.4cm/\sec^2 = \frac{\text{g}}{289} \tag{26}$$

For the other latitude circle of the earth which symbolized by Ø, the radius (r) replaced by

## r cos Ø

then the angular acceleration become

α cos Ø

This acceleration have two components: The vertical component is

$$\mathfrak{a}\text{ }\cos^2\mathfrak{D}$$

This component reach its maximum value at the equator (3.4 gal), while its value at the pole is zero which is the minimum value.

The horizontal component is

 $\alpha$   $\cos \mathcal{O} \text{ sin } \mathcal{O},$ 

The horizontal component reach its maximum value (0.5 gal) at the latitude Ø = 45°, while it reach its minimum value (zero) at the equator and poles.

*Gravity Field Theory DOI: http://dx.doi.org/10.5772/intechopen.99959*

The total gravity force which is determine the weight of anybody on the earth during a free fall represented as a resultant of the gravitational attraction of the earth and the centrifugal force at certain point on the earth.

Tenth thousands of gravity measurements on the earth surface indicate that g values at poles is greater than g values at equator by about (1/189) of the total gravity value. Where the centrifugal force caused a difference between the equator and poles by about (1/289), while the increase in radius at the equator by (21 km.) more than pole caused a variation by about (1/547).

1/289 = 0.003460207 Centrifugal force effect.

1/547 = 0.00182815 Earth radius difference effect

!/189 = 1/289 + 1/547

1/189 = 0.0052884 The difference in gravity between the equator and poles relative to the total gravity value.
