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Let's consider the universal
gravitational law a little bit

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more.

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Let's consider two
objects in space.

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Let's say this is the sun,
and we have the earth here

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or the earth and the moon.

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And of course, they're
orbiting each other.

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So we can pick a coordinate
system that goes radially.

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And so we're going to have
an r hat direction here,

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and we're going to call this the
r hat direction between objects

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1 and 2.

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What forces are acting
on this little moon here?

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Well, it's the gravitational
force going inward.

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It's force of 1, 2,
F1 2, on object 2,

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due to the interaction
between bodies 1 and 2.

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For that, we can write down the
universal gravitational law,

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F1 2 equals minus G, the
gravitational constant, m1,

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m2 of r1 2 squared.

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That one is the distance between
the two objects times r1 2 hat.

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And the minus goes, actually,
with the unit vector

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here, because the force goes
in the opposite direction

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from our r hat.

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Let's now consider here
is the earth again.

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And we're going to move the
moon or a little moon rock right

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to the surface of the earth.

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And we want to now calculate
and consider what kind of force

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this act on this
moon rock, and what

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is the gravitational
acceleration that this moon

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rock on the surface of
the earth is experiencing?

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So we have the earth.

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Earth has one earth radius,
and it has an earth mass.

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And our moon rock
has the mass m.

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And we know, from this exercise
here already, that, of course,

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this gravitational force
is acting on our moon rock

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as well.

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That hasn't changed.

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What we are now
considering in addition

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is that this moon rock
is also experiencing

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a gravitational acceleration
due to this force,

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and that goes inward as well.

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So it is experiencing an mg.

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And we know that that is
the same as the magnitude

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of this force here.

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So we can equate that with G,
and then we have the earth mass

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and the mass of the moon rock
times the distance squared,

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so an earth radius squared.

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And from that, we
already see that a, we

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can cancel out the small
m, so the moon rock,

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and we get to g here.

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So we can calculate the
gravitational acceleration,

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which is capital G earth mass
over earth radius squared.

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So if we have this
kind of information,

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we can determine the
gravitational acceleration.

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And of course, it will
change, depending on which

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object we are considering.

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It would be different if
we plug in the solar mass

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and the solar radius or the
moon mass and the moon radius,

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if we consider an
astronaut standing here

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on the moon's surface.

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Now let's put some numbers
into this equation.

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So we have g is
capital G. That's

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the gravitational constant.

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We have 6.67, 10 to 11, and
then we have Newton and 1

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over kilogram squared
and mass squared times

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the earth mass, 5.97
10 to 24 kilograms.

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And then we have
to divide this over

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through the earth
radius, 6.37 10 to the 6.

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And we have to square that, and
we have to square the meters.

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If we calculate this, we get to
9.81 meter per second squared.

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And surely you have
seen this number before.

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This number can
either be calculated,

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if you know capital G, the
gravitational constant,

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or you can determine that
gravitational acceleration

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through an experiment.

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And together with the earth
mass and the earth radius,

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you can actually calculate the
gravitational constant there.