# 0.2 Force, momentum and impulse  (Page 20/35)

 Page 20 / 35

The force exerted by a field of strength $g$ on an object of mass $m$ is given by:

$F=m·g$

This can be re-written in terms of $g$ as:

$g=\frac{F}{m}$

This means that $g$ can be understood to be a measure of force exerted per unit mass.

The force defined in  [link] is known as weight.

Objects in a gravitational field exert forces on each other without touching. The gravitational force is an example of a non-contact force.

Gravity is a force and therefore must be described by a vector - so remember thta gravity has both magnitude and direction.

## Newton's law of universal gravitation

Newton's Law of Universal Gravitation

Every point mass attracts every other point mass by a force directed along the line connecting the two. This force is proportional to the product of the masses and inversely proportional to the square of the distance between them.

The magnitude of the attractive gravitational force between the two point masses, $F$ is given by:

$F=G\frac{{m}_{1}{m}_{2}}{{r}^{2}}$

where: $G$ is the gravitational constant, ${m}_{1}$ is the mass of the first point mass, ${m}_{2}$ is the mass of the second point mass and $r$ is the distance between the two point masses.

Assuming SI units, $F$ is measured in newtons (N), ${m}_{1}$ and ${m}_{2}$ in kilograms (kg), $r$ in meters (m), and the constant $G$ is approximately equal to $6,67×{10}^{-11}N·{m}^{2}·\phantom{\rule{3.33333pt}{0ex}}k{g}^{-2}$ . Remember that this is a force of attraction.

For example, consider a man of mass 80 kg standing 10 m from a woman with a mass of 65 kg. The attractive gravitational force between them would be:

$\begin{array}{ccc}\hfill F& =& G\frac{{m}_{1}{m}_{2}}{{r}^{2}}\hfill \\ & =& \left(6,67×{10}^{-11}\phantom{\rule{0.166667em}{0ex}}\mathrm{N}·{\mathrm{m}}^{2}·\phantom{\rule{3.33333pt}{0ex}}{\mathrm{kg}}^{-2}\right)\left(\frac{\left(80\mathrm{kg}\right)\left(65\mathrm{kg}\right)}{{\left(10\mathrm{m}\right)}^{2}}\right)\hfill \\ & =& 3,47×{10}^{-9}\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\hfill \end{array}$

If the man and woman move to 1 m apart, then the force is:

$\begin{array}{ccc}\hfill F& =& G\frac{{m}_{1}{m}_{2}}{{r}^{2}}\hfill \\ & =& \left(6,67×{10}^{-11}\phantom{\rule{0.166667em}{0ex}}\mathrm{N}·{\mathrm{m}}^{2}·\phantom{\rule{3.33333pt}{0ex}}{\mathrm{kg}}^{-2}\right)\left(\frac{\left(80\mathrm{kg}\right)\left(65\mathrm{kg}\right)}{{\left(1\mathrm{m}\right)}^{2}}\right)\hfill \\ & =& 3,47×{10}^{-7}\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\hfill \end{array}$

As you can see, these forces are very small.

Now consider the gravitational force between the Earth and the Moon. The mass of the Earth is $5,98×{10}^{24}$  kg, the mass of the Moon is $7,35×{10}^{22}$  kg and the Earth and Moon are $3,8×{10}^{8}$  m apart. The gravitational force between the Earth and Moon is:

$\begin{array}{ccc}\hfill F& =& G\frac{{m}_{1}{m}_{2}}{{r}^{2}}\hfill \\ & =& \left(6,67×{10}^{-11}\phantom{\rule{0.166667em}{0ex}}\mathrm{N}·{\mathrm{m}}^{2}·\phantom{\rule{3.33333pt}{0ex}}{\mathrm{kg}}^{-2}\right)\left(\frac{\left(5,98×{10}^{24}\mathrm{kg}\right)\left(7,35×{10}^{22}\mathrm{kg}\right)}{{\left(0,38×{10}^{9}\mathrm{m}\right)}^{2}}\right)\hfill \\ & =& 2,03×{10}^{20}\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\hfill \end{array}$

From this example you can see that the force is very large.

These two examples demonstrate that the greater the masses, the greater the force between them. The $1/{r}^{2}$ factor tells us that the distance between the two bodies plays a role as well. The closer two bodies are, the stronger the gravitational force between them is. We feel the gravitational attraction of the Earth most at the surface since that is the closest we can get to it, but if we were in outer-space, we would barely feel the effect of the Earth's gravity!

Remember that

$F=m·a$

which means that every object on Earth feels the same gravitational acceleration! That means whether you drop a pen or a book (from the same height), they will both take the same length of time to hit the ground... in fact they will be head to head for the entire fall if you drop them at the same time. We can show this easily by using the two equations above (Equations  [link] and [link] ). The force between the Earth (which has the mass ${m}_{e}$ ) and an object of mass ${m}_{o}$ is

$F=\frac{G{m}_{o}{m}_{e}}{{r}^{2}}$

and the acceleration of an object of mass ${m}_{o}$ (in terms of the force acting on it) is

#### Questions & Answers

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research.net
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Introduction about quantum dots in nanotechnology
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there is no specific books for beginners but there is book called principle of nanotechnology
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That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
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s. Reply
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of graphene you mean?
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or in general
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in general
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Graphene has a hexagonal structure
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