WEBVTT
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So in this question, we are asked to call
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him the magnitude and direction of the gravitational force on
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the moon. Do to the earth Onda the sun
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in each of the positions that are depicted in for
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your 12.32. And we are to assume that the
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sun is in the plane of the Earth moon orbit
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and use the data in appendix F to go ahead
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and accomplish this. So at first glance, um
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, you know this question looks like it's going to
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be a lot of work because the distance between the
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sun and the moon is different in each of the
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, um in each of the diagrams A, B
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and C Um so it looks like we're going to
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need to calculate three gravitational forces between the moon and
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the sun. Um, what I would say is
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, don't do that, because the the change Anthee
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Ah, distance between the moon and the sun is
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so small as the moon goes around the earth that
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it really doesn't change the the force of gravity between
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the sun and the moon. So what I'm going
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to start off by doing is just calculating the fourth
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from the earth. And then I will calculate the
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force from the sun on the moon, and then
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we will go ahead and, um, just think
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about how they add up together in the different scenarios
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. So the force from the earth on the moon
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is gonna be G um, he, um em
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divided by the separation between the moon and the earth
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squared. And if you look into your appendix F
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, then you can get all of these values,
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so I'll just quickly read them out of here.
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Mass of the Earth is a 5.94 times 10 to
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the 24 free, and the mass of the moon
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is 7.35 times 10 to the 22. And they
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are on Lee a distance of 3.84 times 10 to
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the eight apart. And so once you throw that
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into a calculator, you see that they force from
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the earth is 1.98 times 10 to the 20 and
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now I'm going to do the same thing. But
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I'm going to use the values for the sun.
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So we're calculating the force between the sun and the
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moon. So we're going to use the mass of
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the sun and the moon and thesis aeration between,
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uh, I'm just gonna use the separation between the
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earth and the sun that will give an approximate distance
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between the sun and the moon. Because, as
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I said, those changes that the moon's orbit creates
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in the in the distance um, really doesn't have
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a big effect on this year. So from the
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appendix Ah, we can sub in our values mass
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. The sun is 1.99 times 10 to the 30
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and J the mass of the earth again. Nursery
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mass of the moon should go in there so that
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7.35 times 10 to the 22 and then I'm gonna
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do the distance between the earth and the sun.
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That's 1.5 times 10 to the 11 and that needs
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to be squared one speed. Go ahead and plug
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that into the calculator. We see the force from
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the sun on the moon is going to be 4.34
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times tend to the 20 Newtons. Now these air
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just the magnitude of the forces on the moon.
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Um, the difference between scenarios A, B and
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C in the figure is how these two forces add
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up. Sort of Victoria Lee. Right? So
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let's take a look at that. So in part
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A, we've got the moon on the other side
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of the earth from the sun. So something like
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this. And so the forces from the earth and
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the sun it looks like this. They both point
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in towards, um, the earth innocent. So
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they both point in the downward direction in the way
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this is drawn. Right? So the magnitude,
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um, the magnitudes of the forces here are going
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to add up together. Right? So we've got
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the net force on the moon in this case is
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going to be 1.98 times tend to the 20 plus
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4.34 times 10 to the 20 and so that will
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give us a total of 6.32 times 10 to the
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20 Newtons. And of course, that total force
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is in the direction, um, towards the earth
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and the sun. The moon wants to, um
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the moon isn't experiencing a force towards the earth and
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the sun. Ah, in part B, we're
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a little bit more of a complicated situation because believe
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God's the moon off to the side of the earth
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Kind of the earth sun line. And so the
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earth is pulling the moon towards the laughed. But
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the sun is pulling the moon kind of radial e
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kind of diagonally like this. And so we do
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need to do a little bit of vector work here
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in order to get the correct answer. So first
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of all, let's try to calculate what this angle
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is here. And in order to do that,
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I'm going to create a little triangle between the Earth
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, the moon and the sun, because we know
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what these distances are there 1.5 times 10 to the
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11 and this one over here is the re 10.4
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times 10 88. And so we can use soak
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Ottawa to find this angle here. So I'm gonna
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use, um because we have opposite and adjacent.
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I'm going to use chan. So tan data is
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1.5 times tend to the 11 divided by 3.84 times
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tend to the eight. And so we take the
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tan inverse of this and we get an angle that
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is basically 90 degrees of 89 degrees. Right?
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Because that distance between the earth and the sun is
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so so big that this angle is in reality,
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even though it doesn't look like in the in the
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picture, it's almost a 90 degree angle. So
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we could resolve the, um, the Red Force
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here, the force between the moon and the sun
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into components. Um, it's really not gonna be
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worth it because it basically all points kind of in
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a downward direction, basically all points in the Y
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direction. So what we really have in effect here
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is one force that is, um, pulling the
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moon this way. Ah, from the earth and
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then e force from the sun that is basically pulling
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it, um, completely in that downward direction.
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So in order to get the net force, I'm
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just gonna add thes two forces together. So this
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is just to be clear, this is the force
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from the Earth. This is the force from the
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sun, and this is gonna be our total force
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. So to get the magnitude, we will just
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use Pythagoras theorem. So the force from the earth
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squared, plus you force from the sun squared and
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will take the square root. And so we get
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an answer of 4.77 times 10 to the 20 for
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the magnitude and the direction. Um, we can
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just use this angle here. So what? The
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angle with respect to kind of that horizontal in the
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diagram, Um, again, we could use ah
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, tan. So we've got Santa is the force
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from the sun divided by the force from the earth
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. And when we do bant and take the tan
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inverse, we get Ah, an angle of 65.5
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degrees approximately, um so that's with respect to the
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horizontal in the diagram. Obviously, in space,
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horizontal doesn't mean very much. But, um,
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you know, with respect to the diagram, um
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, the horizontal in the diagram. And then,
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ah, you could also rate that as an angle
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with respect, Teoh the vertical instead. Ah,
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so you just need to do 90 minus state s
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O. It makes an angle of 24 0.5 degrees
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with respect to the vertical in the diagram. Okay
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, so that's the force that is experienced by the
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moon in the second picture and part B. So
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this is the magnitude and this is our angle.
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And then in part, C um, we again
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have some lining up between the earth sun and the
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moon. So we have something that looks like this
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now, So the earth is here, the moon
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is here and the sun would be down here.
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So now we have the force from the earth is
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pulling the moon upwards, and the force from the
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sun is pulling the moon downwards. So we get
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ah, sort of a cancellation, a little bit
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of a cancellation here. So if we label the
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ah up direction as positive, then when we add
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the forces together to get the net force, we
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can put a negative onto the force from the sun
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and that will take into account this sort of cancellation
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here. So I've got 1.98 times 10 to the
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20 newtons, minus 4.34 times 10 to the 20
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Newtons. And so that gives a net of negative
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2.36 ah times 10 to the 20. And we
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could write that as a vector and sad. So
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we can say that the net force is 2.36 times
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10 to the 20 Newton's, um, it down
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again, you know, kind of in the in
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terms of the direction in in the diagram. Ah
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, so this is our final answer for part C
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. So just to recap in this particular problem Ah
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, we found the force between the earth and the
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moon and the force between the moon and the sun
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, and we use those same values for each part
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. But we just added up the vectors, um
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, a little bit differently, because the forces are
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pointing in different directions.