WEBVTT
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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.