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The little prince lives
on his asteroid B-612.

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And he really likes
to watch the stars.

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And what he really wishes
for is to watch the stars

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while floating, to have
the best possible view

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and to just immerse
himself in the stars.

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And so the little
prince has to think

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if there is a position
in space with respect

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to other celestial bodies where
the gravitational force would

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just cancel, so
that he can float.

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Let's look at that.

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So there is the planet
of the businessman.

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We're going to give
it the mass m1.

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And there is also the small
planet from the lamplighter

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floating around with a mass m2.

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And we know that gravitational
forces act radially.

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So we can already see here that
if we put the asteroid B-612

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somewhere here, then maybe
we find a constellation where

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gravitational forces cancel.

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So let's start by placing the
little asteroid over here.

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And what is this body
experiencing in terms

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of gravitational forces?

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Well, it's going to experience
a gravitational force of object

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1.

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So we have object 1 or
F1m, due to the interaction

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between object 1-- so
the businessman planet

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and our little asteroid here.

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And then we also, of course,
have the same direction

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here due to the
interaction of object

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2 and our little asteroid,
so that would be F2m.

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And you can clearly see that
if we tally up these forces,

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they're not going to add to 0.

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So this is not a good location
for the little prince's

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asteroid to be.

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The same would be true
if we put it out here.

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It will experience
those same forces again.

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What do we have here--
1m and 2m just going

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in the opposite direction.

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But you have guessed it already.

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We can find a spot here in the
middle where one force goes

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in this direction.

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So this is F2m.

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And actually we should give
those forces direction.

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And the other body
exerts a force

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going in the other direction.

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And so we can see
that we now just need

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to find that exact position
here, in between these two

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objects with two
different masses, where

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the little prince can free
float while watching the stars.

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We can say already that if
these two masses were the same,

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it would obviously
be in the middle.

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But because they
are not the same,

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we have to calculate
what that distance is.

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And so we're going to
label this distance here x.

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And we're going to say
that the two bodies here

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are a distance d apart.

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All right.

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How are we going
to go about this?

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Well, we have to apply the
universal law of gravitation

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in our F equals ma analysis.

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Well, before we started with
our F equals ma analysis,

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we actually have to pick a
coordinate origin and a unit

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vector.

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And so let's place the
coordinate origin in here.

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And let's have i hat
go in this direction.

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And we're going to label this
position A, and position B,

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and position C.
These are our three

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options for asteroid placement.

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And well, let's
look at position B.

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And well, we have
the universal law

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of gravitation between the
mass of the object here.

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And we'll have to
describe both components.

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So first this one
F1m and then F2m.

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And if we start with F1m, that
goes in the negative i hat

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direction.

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So minus Gmm1 on the distances x
squared in the i hat direction.

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And then the other one goes
in the plus i hat direction.

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And we have Gmm2.

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And now we have d
minus x gives us

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this portion-- d minus x squared
also in the i hat direction.

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And that needs to add up to 0,
because that is what we want,

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right?

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If it adds up to 0, then we
have no gravitational forces

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acting on the asteroid.

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So now we need to
solve this for x here.

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And in the first step, you
see that actually G and m

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will fall out here.

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And then we're
left with m1 over x

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squared minus plus m2
over d minus x squared.

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And we can write
that as m2 x squared

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minus m2 d minus x squared.

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And what you see here
is that this will

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turn into a quadratic equation.

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And if we do a few
steps of arithmetic,

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and then write down
the general solution

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to this quadratic equation,
we will find this here.

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x equals 2dm1 plus minus 2dm1
squared minus 4m1 minus m2m1d

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squared, and then the square
root of that over m1 minus m2.

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And actually we
need two of those.

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So we have this quadratic
equation here-- well,

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the solution.

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And we now need to consider
one more thing-- namely,

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this equation here is
only valid if-- well,

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can I write this here--
if x is between 0 and d.

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Right?

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In position C, my
x is larger than d,

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which means this-- sorry,
this force here flips sign,

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and then this
would be different.

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So this plus here refers
to x being less than d.

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And we have to decide
which of the two signs

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here gives us the
correct, which will

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fulfills this requirement here.

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So this simplifies to x
equals dm1 plus minus m1m2

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and here we have
the square root.

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And then here, we have another
bracket over m1 minus m2.

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And we need to get
this term here smaller

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than 1, which will be
smaller than 1 if the x is

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between 0 and d.

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And as it turns out,
that is indeed true

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if we use the minus sign here.