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

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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

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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Sherica
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Sherica
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Tamia
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China
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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