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Calculate the magnitude and direction of the net gravitational force on the moon due to the earth and the sun when the moon is in each of the positions shown in Fig. 12.32 . (Note that the figure is not drawn to scale. Assume that the sun is in the plane of the earth-moon orbit, even though this is not actually the case.) Use the data in Appendix F.

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a) $6.30 \times 10^{20}$b)$24.83^{\circ}$c) $2.37 \times 10^{20} \mathrm{N}$

Physics 101 Mechanics

Chapter 12

Gravitation

University of Michigan - Ann Arbor

Hope College

University of Winnipeg

Lectures

03:55

In physics, orbital motion is the motion of an object around another object, which is often a star or planet. Orbital motion is affected by the gravity of the central object, as well as by the resistance of deep space (which is negligible at the distances of most orbits in the Solar System).

03:18

Sir Isaac Newton described the law of universal gravitation in his work "Philosophiæ Naturalis Principia Mathematica" (1687). The law states that every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them.

01:59

(a) What is the magnitude …

04:15

The drawing (not to scale)…

04:00

Calculate the magnitudes o…

03:47

05:00

When the Earth, Moon, and …

02:25

07:13

(II) Find the net force on…

06:45

Using the masses and mean …

05:27

Repeat the previous proble…

02:19

Determine the net force of…

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

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