WEBVTT
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This is Giancoli Answers with Mr. Dychko.
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We have that the centripetal force
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on the satellite is provided by gravity.
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So, *m v squared* over *r* equals
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*g* times mass of the satellite times
mass of the Earth over
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distance from the center of the Earth
to the satellite squared.
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Now, the distance from the
center of the Earth
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is gonna be the radius of the Earth
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plus this distance above the
surface of the Earth.
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So, we'll call that *delta r 1* for the
satellite 1, the close one,
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and that's 7500 kilometers,
above the surface.
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And then *delta r 2* is 15000 kilometers,
above the surface.
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And we can do some algebra
on this to solve for
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what the speed should be.
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We can multiply both sides by *r*,
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and that turns this into
*r* to the power of 1.
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And divide both sides by
mass of the satellite
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and then take the square root of
both sides to solve for *v*.
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And so we have *v*, for the
close satellite, is
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square root *G*
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Well, it looks like we made
a slight mistake here
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but it doesn't actually affect the answer
which is interesting.
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So, this mass of the satellite disappears
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because it ends up canceling down here.
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Anyway.
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So, we have this is correct.
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We have, in the first case,
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the speed of the satellite is
square root *G*
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mass of the Earth divided by
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the total distance from the center of
the Earth and the satellite.
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So that's radius of the Earth plus
this distance above the surface.
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And then, for satellite 2, it's the
same idea with *delta r 2*.
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And then taking the ratio
of those velocities,
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you go *v 1* divided by *v 2*.
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We can think of the square root as,
on the top and on the bottom,
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separately and then it becomes more clear
that this thing cancels with this thing
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although neither of them contain the mass
of the satellite as it turns out but either way.
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So, we are doing this division
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of a fraction is the same as multiplying
by its reciprocal
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is what else is going on here.
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So, dividing by *v 2* is the same
as multiplying by
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the reciprocal of *v 2*.
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So multiplying by the denominator
put on top and the
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numerator is put on the bottom.
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And then these square root *G* mass of
the Earth factors, cancel
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and we are left with the
ratio of the velocities
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is the same as the square root of
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the distance from the center of the Earth,
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for each of these.
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So square root of
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6.38 times 10 to the 6 metersâ€”radius of
the Earthâ€”plus
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the distance above the surface
of the Earth for the
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far satellite, which is 15000 times
10 to the 3 meters
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divided by 6.38 times 10 to the 6
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plus 7500 times 10 to the 3 meters
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and that gives 1.2.
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So, the close satellite is
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1.2 times as fast as the far satellite
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because we could rewrite this as
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*v 1* equals 1.2 times *v 2*.
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And so there's an algebraic way
of saying that,
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or arithmetic way of saying that
*v 1* is faster than *v 2*.
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But you can also think of it
in a different way.
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You can say the close satellite
has to be faster because
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since the acceleration due to gravity
is greater here, closer to the Earth,
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but it means, it has more
of a tendency to
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go down; it's falling faster,
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and so it has to scoot to the side faster
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in order to avoid colliding with the Earth.
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That's another way to think about it.