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As a matter of fact, we will prove this theorem, employing gravitational field concept for a spherical mass like that of Earth. For the time being, we consider Earth and apple as particles, based on the Newton’s shell theory. In that case, the distance between Apple and center of Earth is equal to the radius of Earth i.e 6400 km.
The magnitude of gravitational force between terrestrial objects is too small to experience. A general question that arises in the mind of a beginner is “why do not we experience this force between, say, a book and pencil?” The underlying fact is that gravitational force is indeed a very small force for masses that we deal with in our immediate surrounding - except Earth.
We can appreciate this fact by calculating force of gravitation between two particle masses of 1 kg each, which are 1 m apart :
$$\Rightarrow F=\frac{6.67X{10}^{-11}X1X1}{{1}^{2}}=6.67X{10}^{-11}\phantom{\rule{1em}{0ex}}N$$
This is too insignificant a force to manifest against bigger forces like force of gravitation due to Earth, friction, force due to atmospheric pressure, wind etc.
Evidently, this is the small value of “G”, which renders force of gravitation so small for terrestrial objects. Gravitation plays visible and significant role, where masses are significant like that of planets including our Earth, stars and such other massive aggregation, including “black holes” with extraordinary gravitational force to hold back even light. This is the reason, we experience gravitational force of Earth, but we do not experience gravitational force due to a building or any such structures on Earth.
Newton’s law of gravitation provides with expression of gravitational force between two bodies. Here, gravitational force is a vector. However, force vector is expressed in terms of quantities, which are not vectors. The linear distance between two masses, appearing in the denominator of the expression, can have either of two directions from one to another point mass.
Even if, we refer the linear distance between two particles to a reference direction, the vector appears in the denominator and is, then, squared also. In order to express gravitational force in vector form, therefore, we shall consider a unit vector in the reference direction and use the same to denote the direction of force as:
$${\mathbf{F}}_{12}=\frac{G{m}_{1}{m}_{2}\stackrel{\u2038}{\mathbf{r}}}{{r}^{2}}$$
$${\mathbf{F}}_{21}=-\frac{G{m}_{1}{m}_{2}\stackrel{\u2038}{\mathbf{r}}}{{r}^{2}}$$
Note that we need to put a negative sign before the second expression to make the direction consistent with the direction of gravitational force of attraction. We can easily infer that sign in the expression actually depends on the choice of reference direction.
Gravitation force is a vector quantity. The net force of gravitation on a particle is equal to resultant of forces due to all other particles. This is also known as “superposition principle”, according to which net effect is sum of individual effects. Mathematically,
$$\Rightarrow \mathbf{F}=\Sigma {\mathbf{F}}_{i}$$
Here, F is the net force due to other particles 1, 2, 3, and so on.
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