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The gravitational force at different places on Earth due to the Sun and the Moon depends on each point's distance from the Sun or Moon, and this variation is what causes the tides. Research the values for the Earth-Moon distance $R_{\mathrm{EM}},$ the Earth-Sun distance $R_{\mathrm{ES}},$ the Moon's mass $M_{\mathrm{M}},$ the Sun's mass, $M_{\mathrm{S}},$ and the Earth's radius $R_{\mathrm{E}}$ . (a) First consider two small pieces of the Earth, each of mass $m,$ one on the side of the Earth nearest the Moon, the other on the side farthest from the Moon. Show that the ratio of the Moon's gravitational forces on these two masses is$$\left(\frac{F_{\text { near }}}{F_{\text { far }}}\right)_{M}=1.0687$$(b) Next consider two small pieces of the Earth, each of mass $m,$ one on the nearest point of Earth to the Sun, the other at the farthest point from the Sun. Show that the ratio of the Sun's gravitational forces on these two masses is$$\left(\frac{F_{\text { near }}}{F_{\text { far }}}\right)_{\mathrm{S}}=1.000171$$(c) Show that the ratio of the Sun's average gravitational force on the Earth compared to that of the Moon's is $$\left(\frac{F_{\mathrm{S}}}{F_{\mathrm{M}}}\right)_{\mathrm{avg}}=178$$Note that the Moon's smaller force varies much more across the Earth's diameter than the Sun's larger force. (d) Estimate the resulting "force difference" (the cause of the tides)$$\Delta F=F_{\text { near }}-F_{\text { far }}=F_{\text { far }}\left(\frac{F_{\text { near }}}{F_{\text { far }}}-1\right) \approx F_{\text { avg }}\left(\frac{F_{\text { near }}}{F_{\text { far }}}-1\right)$$for the Moon and for the Sun. Show that the ratio of the tide-causing force differences due to the Moon compared to the Sun is$$\frac{\Delta F_{\mathrm{M}}}{\Delta F_{\mathrm{S}}} \approx 2.3$$Thus the Moon's influence on tide production is over two times as great as the Sun's.

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a) 1.0687b) 1.000171c) 178d) 2.3

Physics 101 Mechanics

Chapter 6

Gravitation and Newton's Synthesis

Physics Basics

Newton's Laws of Motion

Applying Newton's Laws

Gravitation

Cornell University

University of Michigan - Ann Arbor

University of Washington

University of Sheffield

Lectures

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basically about the gravitational forces of the moon and sun that caused the tides. And so we want to look at, um basically want to consider the gravitational effect of the moon or the sudden on a piece of the earth on the same side as the moon or the sun and on a piece of the earth on the opposite side. And then look at the ratios of these and how much they change. And we're going to discover that the change in the force of the moon is greater than the change in the force from the sun. And so the moon is more responsible for, um for the tides, although not entirely so. First, we want to consider the ratio of the forces from the moon for the near and far sides. So just using the law of universal gravitation, we're going to have the gravitational, constant times, the mass of the moon. And now we're going toe to say that each little point of the earth has some mass m. And what that is is going to matter, because we're going to find the ratio between the near and far sides. And so that'll cancel out So this is going to be the earth moon distance minus the radius of the Earth because we're on this side. So you go from the center of the moon to the center of the earth and then back the radius of the earth and then, similarly here for the far side, Uh, you're going the distance from the center of the moon to the center of the earth and then plus the radius of the Earth. And so this ratio for the moon is going to just be the ratio, these two denominators here. So, like the problem says we should get We get 1.6 at 1.687 now for the same thing for the sun. Everything is going to look exactly the same as it did before. Except we'll have the mass of the sun and the distance from the Earth to the sun. So we're going to get Hey the earth some distance, plus the radius of the earth provided by the Earth Sun distance minus the radius of the earth squared. And so we'd stop and think of moment. The radius severe. There's way smaller than the distance between the Earth and the sun. So we would expect this ratio to be closer to one, because R S plus Ari is almost the same. Azari s and similarly arias minus already is almost the same as our Yes, And in fact, we get 1.0 zero. So we have to go away further out to start getting decimals that are non zero 171 So now I'm part see here we want the average force from the sun divided by the average force from the moon. And so for that, we're going to use the just the distance to the center of the earth from the sun and the moon. So we're going to get the mass of the sun divided by the distance between the earth and the sun squared and the distance between the Earth and the moon squared, divided by the mass of the moon. And, you know, the G and the little am of the piece of earth were considering canceled out. And so this is 178 like it says it should be. So now we want to do is to estimate the sort of forced difference F n minus for both the sun and the moon. This is what causes the tides. And we want to do this for both the moon and the sun and so and then showed that the ratio over the dealt EFS for the moon divided by the sun is about to. So in order to, uh, to find approximately what this is going to be, because if we actually wanted Toa find this, we need to have some hypothetical mass for our piece of the earth. And that depends on how big a piece of the earth is etcetera. And that's not so good, because we don't know, and it shouldn't matter, But then it shows us that weaken. First of all, rewrite this expression like this. We know what this this value is. We just have this minus, uh, one just factored in nfl. And then we're going to say that the force on the far side of the earth is approximately the same as the force at the centre of the earth. So the force difference for the moon provided by the force difference for the sodden This is going to be the average force for the Moon times. Our, uh, we already know that effin over f f for the moon is ah 1.687 So that's one minus that. Are that minus one? This is going to be zero point 0.0 687 And then for the sun, the average force, we have the 1.171 minus one. So this is your go right? 000 171 And we already found this ratio. The F S over fm is 178 so one over that is one over 178. And so this evaluates to 2.3. So the moon's influence the tides is greater than two times that the sun's. Even though the force from the sun is greater, the the actual difference and forests from the sun from one side to the other is about two times a small, which makes sense because there is diameter. Uh is a lot smaller than the distance from the earth to the sun. Then it is, you know, the distance between the earth and moon

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