WEBVTT
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The little prince is sitting
on his little planet,
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and he's watching
the planets go by.
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And so suddenly, he's
seeing three of them.
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One, two, three.
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And he wonders, hmm, what
would the center of mass
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of these three planets be?
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So let's calculate it.
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We have one planet
here that's three times
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the mass of this guy.
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And then this one
has half of m1.
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And I've written this
up there already.
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And the center of
mass, when we want
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to determine its coordinate--
it is a coordinate
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that depends on this
coordinate system,
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and this origin
here-- then we need
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to have the total
mass of the system,
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because it's a
mass-weighted coordinate.
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And so the total mass
is going to be 4.5 m1.
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And if we want to
calculate this position
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function of the
center of mass, Rcm,
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that is the mass weight here.
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And then we need the
sum of all of our masses
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times our distances.
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We're going to sum here
over j from 1 to n.
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And what that means is
we need to now write out
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the sum for our three planets.
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And we need to give
this a radius here,
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so the r1 would be
going from here to here.
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r2 goes from here to there.
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And r3 from here to here.
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You need to write them
out in the i hat direction
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and in the j hat direction.
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Let's add that here.
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And then sum it all up
and calculate our r.
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So let's write this out.
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We have m1, and we're
going to have 35 and 10.
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35 i hat plus 10 j hat plus m2.
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It's going to be 5 and 20.
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5 and 20.
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And then m3, we have 40 and 30.
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And what we can do--
oh, and of course that
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has to be divided
by my system mass.
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And what we can do is
we can concentrate first
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on the x component, and
then on the y component.
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Maybe we'll just continue here.
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So we're going to have-- and
we can plug it in all the m's.
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We'll do it for the
x component first.
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We're going to have m1, and
then we have 35 i hat plus here
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we're going to have 1/2 m2.
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That is 3m1 5 plus-- and
for m3, we have 0.5 m1 40.
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that's in the i hat direction.
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And we'll have to divide
that over our system mass.
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And then we do the same
for the y component.
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So m1, 10 plus 3. m1,
20 plus 0.5 m1 30 j hat.
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And again, we have to divide
this over our system mass.
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So this boils down to-- hang on.
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Let me redo this again.
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Let me actually look
at the answer first.
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What do I have here?
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70 and 85, OK.
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So this boils to m1
over the system mass.
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And we have 70 in the i hat
plus 85 in the j hat direction.
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So the 70 comes from this
term, the 80 comes from this.
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And I put it back together, and
now we need to plug in this one
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here.
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And so we will get in the end
of that Rcm equals m over 4.5 m.
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So the m goes away, and we
have a factor of 1/4.5 here.
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We'll divide this
through, and we're
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going to have 15.5 in the
i hat direction, and 18.9
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in the j hat direction.
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All right, so let's see where
this fits on our graph here.
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So 15 in the i hat
is somewhere here.
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And 19 is almost
20, so it's going
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to be here, so about there.
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So this is my Rcm,
and this here is
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my center of mass of the system
of these three little planets.
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Of course, we used approximate
math here for all the planets.
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But if we look at
the real numbers,
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imagine that this
would be Earth,
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and it has one Earth mass.
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And if this were Saturn,
it would have something
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like 318 Earth masses.
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And if this is Pluto, it would
have 0.0025 Earth masses.
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You will see that Saturn
really holds all the weight.
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And if we were to do this
calculation with these numbers
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here, then our Rcm would-- and
keeping this coordinate system
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in the arrangement
of the planets,
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then it would go right
into-- if here's the center,
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it would go right next to
the center right over here,
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because Saturn just weighs so,
so, so much more than Pluto
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and Earth together.