(a) What is the gravitational force due to the Moon, F⃗L,cen, on a point at the center of the Earth? Recall that vectors have both a magnitude (arrow length) and a direction (arrow head).
(b) What is the force vector on each point mass, F⃗L, due to the Moon? Draw these vectors at each point.
(c) What is the force difference, ∆F⃗, between each point and Earth’s center? This is the tidal force.
(d) What will this do to the ocean located at each point?
(e) How many tides are experienced each day at a given location located along the Moon’s orbital plane?
(f) Okay, now we will use some math. For the two points located at the nearest and farthest points from the Moon, which are separated by a distance ∆r compared to the Earth-Moon distance r, show that the force difference is given by
\[\Delta F = \frac{2GmM_L}{r^3}\Delta r\]
(HINT: recall that
\[lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} = \frac{d}{dx}f(x) \approx \frac{\Delta f(x)}{\Delta x}\]
(g) Compare the magnitude of the tidal force ∆FK caused by the Moon to ∆F@ caused by the Sun. Which is stronger and by how much? What happens when the Moon and the Sun are on the same side of the Earth?
(b) What is the force vector on each point mass, F⃗L, due to the Moon? Draw these vectors at each point.
(c) What is the force difference, ∆F⃗, between each point and Earth’s center? This is the tidal force.
(d) What will this do to the ocean located at each point?
(e) How many tides are experienced each day at a given location located along the Moon’s orbital plane?
(f) Okay, now we will use some math. For the two points located at the nearest and farthest points from the Moon, which are separated by a distance ∆r compared to the Earth-Moon distance r, show that the force difference is given by
\[\Delta F = \frac{2GmM_L}{r^3}\Delta r\]
(HINT: recall that
\[lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} = \frac{d}{dx}f(x) \approx \frac{\Delta f(x)}{\Delta x}\]
(g) Compare the magnitude of the tidal force ∆FK caused by the Moon to ∆F@ caused by the Sun. Which is stronger and by how much? What happens when the Moon and the Sun are on the same side of the Earth?
(h) How does the magnitude of ∆F caused by the Moon compare to the tidal force caused by Jupiter during its closest approach to the Earth (r = 4 AU)?
(a) What is the gravitational force due to the Moon, F⃗L,cen, on a point at the center of the Earth? Recall that vectors have both a magnitude (arrow length) and a direction (arrow head).
This is pretty simple. We've dealt with gravity before.
\[F_g=\frac{GMm}{r^2}\]
\(G\) is Newton's gravity constant, \(M_E\) is the mass of Earth, \(M_L\) is the mass of our moon (Luna), and \(r\) is the distance between them.
\[F_g=\frac{GM_L M_E}{r^2}\]
(b) What is the force vector on each point mass, F⃗L, due to the Moon? Draw these vectors at each point.
Let's let each point have mass \(m\). Also, let's let \(d\) be the distance from each point to the center of the moon.
\[F_g=\frac{GM_L m}{d^2}\]
\(d\) will vary for each point, but its safe to say that the further away from the far east point on our diagram of Earth we measure from, the greater \(d\) becomes. Due to this, our vectors are approximated and their differences are greatly exaggerated.
(c) What is the force difference, ∆F⃗, between each point and Earth’s center? This is the tidal force.
We now subtract the force of each point upon the earth's center from the moon's force vector (the yellow vector) yielding the following diagram:
The math here is mainly boring 2D algebra, so I will omit it in favor of qualitative explanation.
At the 0 degree point, we subtract the yellow vector format the black, decreasing it's length, but resulting in the red vector towards the moon.
At the 45 and -45 points vector subtraction calculus a bit of the horizontal component, but he vertical component is intact, yielding a net force towards the moon and towards the Earth-Moon orbital plane.
At the 90 and -90 points, the force on the point is almost equal to the force on the center, so the horizontal components cancel leaving only a vertical component towards the center of the Earth.
At the 135 and -135 points, the yellow vector is greater than the black, thus it reverses the direction of the resultant, giving us red vectors similar to the 45 points, but in the opposite direction.
At the 180 degree point, there are no vertical forces, and the yellow vector is greater then the back, resulting a a negative red resultant.
(d) What will this do to the ocean located at each point?
This will pull the ocean in the direction of the force vector. In the diagram below, the dashed line represents the highly exaggerated open level at around the Earth.
(e) How many tides are experienced each day at a given location located along the Moon’s orbital plane?
From our diagram, we can see as the Earth rotates, a location on the orbital plane will pass through two high tides and two low tides per day.
(f) Okay, now we will use some math. For the two points located at the nearest and farthest points from the Moon, which are separated by a distance ∆r compared to the Earth-Moon distance r, show that the force difference is given by
\[\Delta F = \frac{2GmM_L}{r^3}\Delta r\]
(HINT: recall that
\[lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} = \frac{d}{dx}f(x) \approx \frac{\Delta f(x)}{\Delta x}\]
\[\Delta F = \frac{2GmM_L}{r^3}\Delta r\]
(HINT: recall that
\[lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} = \frac{d}{dx}f(x) \approx \frac{\Delta f(x)}{\Delta x}\]
Doing this with exact math will be a nightmare. Bottom line is, I don't want to deal with algebra right now. Let's use a slick trick instead. We already have an expression for force on a point:
\[F_g=\frac{GM_L m}{r^2}\]
Lets' take the first derivative of this.
\[\frac{dF_g}{dr}=\frac{-2GM_L m}{r^3}\]
\[dF_g=\frac{-2GM_L m}{r^3}dr\]
We can ignore the negative, since it's just there to tell us the potential is negative.
\[\boxed{\Delta F_g=\frac{2GM_L m}{r^3}\Delta r}\]
That was a lot easier.
(g) Compare the magnitude of the tidal force ∆F caused by the Moon to ∆F caused by the Sun. Which is stronger and by how much? What happens when the Moon and the Sun are on the same side of the Earth?
More math!
Let's set this up as a ratio:
\[\frac{\Delta F_L}{\Delta F_S}=\frac{2GM_Lmr_S^3}{2GM_Smr_L^3}\]
\[\frac{\Delta F_L}{\Delta F_S}=\frac{M_Lr_S^3}{M_Sr_L^3}\]
Let's fill in some numbers from the internet.
\[\frac{\Delta F_L}{\Delta F_S}=\frac{(7.3 \times 10^{25}g)(1.5\times 10^{13}cm)^3}{(2\times 10^{33}g)(3.8\times 10^{10}cm)^3}\]
\[\boxed{\frac{\Delta F_L}{\Delta F_S}=2.2}\]
Therefore the tidal force from the moon is 2.2 times the tidal force from the sun, which is why we pretty much ignore the Sun's tidal force.
If the three spheres align, the forces will all add, resulting in lower low tides and higher high tides.
(h) How does the magnitude of ∆F caused by the Moon compare to the tidal force caused by Jupiter during its closest approach to the Earth (r = 4 AU)?
This is the same as above with different numbers.
\[\frac{\Delta F_L}{\Delta F_J}=\frac{2GM_Lmr_J^3}{2GM_Jmr_L^3}\]
\[\frac{\Delta F_L}{\Delta F_J}=\frac{M_Lr_J^3}{M_Jr_L^3}\]
Let's fill in some numbers from the internet.
\[\frac{\Delta F_L}{\Delta F_J}=\frac{(7.3 \times 10^{25}g)(6\times 10^{13}cm)^3}{(1.9\times 10^{30}g)(3.8\times 10^{10}cm)^3}\]
\[\boxed{\frac{\Delta F_L}{\Delta F_J}=1.5\times 10^5}\]
Therefore the tidal force from the moon is 150000 times the tidal force from Jupiter, which is why we never even bother to consider tJupiter's tidal force.
I worked on his problem with G. Grell and S. Morrison.
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