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The motion of
objects in space is

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governed by the
universal law of gravity.

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So let's consider
how this works.

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We have two objects
here, object one and two.

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And that could be two
planets or two asteroids

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or two white dwarfs,
black holes, any kind

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of objects you can imagine.

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And they are mutually
attracted by each other due to

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gravitation.

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So we have a gravitational
force and we label this as F.

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And this is the
force on object one

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due to the interaction between
the bodies one and two.

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And here we have a
force on object two

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due to the interaction
between objects one and two.

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And the objects are separated
by a distance r1, 2.

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Now, we want to derive the
universal law of gravitation.

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How are we going
to go about that?

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Well, Newton figured
out a while ago

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that it is proportional to the
masses of objects one and two.

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So this one has a mass
m2 and this one has m1.

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And it is also proportional
to the square of the distance

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between the two objects.

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Now, we need to do
one more consideration

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and then we can derive this.

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We need to actually--
well, first,

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we need to pick
some kind of origin

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from where we are considering
these two objects to be.

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And this object here goes
from the origin to there

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and we call this r1.

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And then we have
here r2, which also

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means that the distance here
between object one and two

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is r1, 2.

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And actually, we know
from vector decomposition

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that r1, 2 equals r2 minus r1.

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So just this minus this
gives us this distance here.

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Now, if we want to write
down the universal law

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of gravitation, there's a
magnitude component to it

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and we also need a direction.

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And we haven't yet chosen
a coordinate system.

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We could, of course,
choose our usual way

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of placing the i-hat
direction in the x-direction

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and the j-hat coordinate
in the y-direction.

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But when we deal with the
universal law of gravitation,

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it's actually better to
adopt a slightly different

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coordinate system.

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Everything in space
usually orbits one another

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so it's much better to think
in a radial direction rather

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than just normal
Cartesian coordinates.

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And so in this case, we're going
to choose an r-hat vector which

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gives us a radial direction and
we're going to do this here.

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So this is going to be our
r-hat direction on object two.

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And here, we have an
r-hat direction 2, 1.

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One And we're going to
come back to the r-hat unit

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vectors later.

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For now, we can just write here
quickly down the definition

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for a unit factor.

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So our r-hat 1,2 is, of course,
the vector itself, r1, 2,

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divided over the
magnitude of the vector.

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We can write it like this.

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And now we can write down
the gravitational law.

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So the force on object-- we're
going to look at object two.

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The force on object two due to
the interaction between objects

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one and two is proportional to
the mass of the two objects--

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we already said that
in the beginning--

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and the square of the distance
between the two objects.

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But what about the direction?

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The direction here, we're
looking at object two.

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We placed our r-hat unit
vector to point down

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but the force is going in
the opposite direction--

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so in the negative
r1, 2 hat direction.

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So we have to add a minus
here and then our 1, 2 r-hat.

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And as it is the case
with most of these laws,

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it has a
proportionality constant

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and Newton called this capital
G. And G, as we know it today

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from experiment, is
6.67 10 to the minus 11.

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And then in terms of
units, we have Newton.

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Force goes in Newton.

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We have mass.

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This is kilogram squared
and we have meter squared.

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So those are the units.

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And if you plug those in, then
the units of the whole equation

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will work out.

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So let's quickly consider
the force in object one

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to see what's
happening over there.

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So we have F2, 1 equals minus
G m1 m2 over r1, 2 squared.

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And in terms of the
unit vector, we now

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have r2, 1 hat here going.

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So again, this minus goes
with this unit factor

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and that one is pointing here
in the opposite direction

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than our force.

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So that's all good.

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But what we see from this
one here-- and actually

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from our diagram already--
that r1, 2 equals minus r2, 1.

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And so we see from this
then that actually,

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Newton's third law, every
action has an equal and opposite

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reaction, is true for
this little setup here,

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as well, because the forces
are of opposite direction

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and of equal magnitude.