The Formula of Gravitational Force

The gravitational acceleration experienced by every objects on the earth surface is approximately 9,8 m/s2.  the acceleration of the moon towards the earth can be calculated from the moon centripetal acceleration as follows.

$a_{s}=\frac{v^{2}}{R}=\frac{\left(\frac{2\pi R}{T}\right)^{2}}{R}=\frac{4\pi^{2}R}{T^{2}}$

Where :

R = radius of moon orbit (3.8 x $10^{5}$ m )

T = Period of moon  ( 27.3 days = 2.36 x $10^{6}$ s )

So, the acceleration of gravitational is can determine as follows :

$a_{s}=\frac{4.(3,14)^{2}.(3,84\times10^{8})}{(2,36\times10^{6})^{2}}=0,0027\textrm{ }\frac{m}{s^{2}}$

The ratio of the acceleration of the moon toward the earth and that of the objects on the earth surface is expressed as follows :

$g_{B}=\frac{1}{3600}g_{b}$

Where :

$g_{B}$ = moon acceleration toward the earth ( m/s²)

$g_{b}$ = objects acceleration toward the earth (m /s² )

It means that the moon acceleration towards the earth is approximately $\frac{1}{3600}$ time of object’s gravitational acceleration on the earth surface. The distance of the moon from center of the earth or the radius of the moon orbit is $3,84\times10^{8}$ m and the distance of the earth surface from the center of the earth or the radius of the earth is $6,4\times10^{6}$ m. if we compare the distance of moon from the center of the earth it the distance of earth surface from the  center of earth then we will obtained the following value.

$\tfrac{\textrm{The distance of moon to the center of earth}}{\textrm{The distance of earth surface to the center of earth}}=\tfrac{3.84\times10^{8}}{6.4\times10^{6}}=60$

Thus, the ratio of both distances is 60. It means that distance of the moon from the center of earth is 60 times of the distance of earth surface from the center of the earth its center.

Based on the calculation, newton concluded that there is a relationship between objects distance from the earth and the magnitude of the earth gravitational force acting on that object. Newton stated that the magnitude of gravitational force acting on an object ($F_{g}$) is inversely proportional to the square distance ( r ) from the center of the earth.

$F_{grav}\thicksim\frac{1}{r^{2}}$

Based on the calculation, obtained that the moon lie at 60 times of object distance on the earth surface. It means that the moon experiences the earth gravitational force of $\frac{1}{60^{2}}=\frac{1}{3600}\times$ the earth gravitational force experienced by objects on the earth surface.

Newton realized that the magnitude of the gravitational force on an object not only depends on objects distance, but also on object mass. 

 

Based on newton’s third law, if the earth act gravitational force acting on an object (such as the moon) , then the object will act a force on the earth with equal in magnitude but opposite in direction.

The magnitude of gravitational force acting between the earth an object must be directly proportional to the earth an object mass and inversely proportional to the square of the distance sparating the center of the earth and that of the object.

Newton concluded that the magnitude of gravitational force acting between earth and the object can be expressed as follows :

$F_{g}\thicksim\tfrac{m_{E}.m_{o}}{r^{2}}$

Where :

$m_{E}$ = the earth mass ( kg)

$m_{o}$ = the object mass ( kg )

r = the distance of the object from the earth center ( m)

the equation above is generally valid for two objects which attract each other with gravitational attractive force, so that the equation above becomes :

$F_{g}\thicksim\frac{m_{1}.m_{2}}{r^{2}}$

Where :

$F_{g}$ = gravitational force (N)

$m_{1}$ = the mass of the object 1 (kg)

$m_{2}$ = the mass of the object 2 (kg)

r = the distance between the objects (m)

based on his observation, newton provided the law of universal gravitation, that states “ the gravitational force between two objects is the attractive force which its magnitude is directly proportional to the mass of each object and inversely proportional to the square of the distance between their centers” . mathematically, the law can be expressed as follows :

$F_{g}\thicksim G\tfrac{m_{1}.m_{2}}{r^{2}}$

 where :

G = the universal gravitational constant