Worksheet 14.2, Problem 1

Draw a circle representing the Earth (mass M_o), with 8 equally-spaced point masses, m, placed around the circumference. Also draw the Moon with mass M_c to the side of the Earth. In the following, do each item pictorially, with vectors showing the relative strengths of various forces at each point. Don’t worry about the exact geometry, trig and algebra. I just want you to think about and draw force vectors qualitatively, at least initially.

(a)  What is the gravitational force due to the Moon, F_c,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, F_c, due to the Moon? Draw these vectors at each point. 


 

(c)  What is the force difference, ∆F, between each point and Earth’s center? This is the tidal force. 


 In order to get the force difference, we subtract the vector of the central force from the vector of the force at each point and are left with the following (where green arrows represent the net force).

 

(d)  What will this do to the ocean located at each point? 


At each point, the ocean will rise slightly in the direction of the net force (green arrows from the previous problem). As you can see, this will cause high tides on both the near and far sides of the Earth and low tides in between.

(e)  How many tides are experienced each day at a given location located along the Moon’s orbital plane? 


Each given location located along the moon's orbital plane experiences 2 low tides and 2 high tides. This is because the Earth's rotation causes the moon to "cross over" a given location on the Earth two times. The first time it is near or overhead, the location will experience a high tide. As it moves away, the location will experience a low tide. However, once the moon "reaches" the opposite side of the Earth, the location will once again experience a high tide and will then experience a low tide when the moon moves 90 degrees away in the Earth's rotation.

(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:

ΔF=2GmMmoonr3Δr

(HINT: recall that

limx→∞f(x+Δx)−f(x)Δx=ddxf(x)≈Δf(x)Δx

and Δr<<r.

We know that:

\[\frac{d}{dx}f(x)\:\approx\:\frac{\Delta f(x)}{\Delta x}\]

We can then rewrite this with variables that pertain to our specific situation:

\[\frac{d}{dr}F(r)\:\approx\:\frac{\Delta F(r)}{\Delta r}\]

\[\Delta F\:=\:\frac{dF(r)}{dr}\Delta r\]

We know that F(r) is equal to:

\[F(r)\:=\:-\frac{GM_{moon}m}{r^{2}}\]

Taking the derivative and substituting into the equation for \(\Delta F\) we see that:

\[\Delta F\:=\:\frac{2GM_{moon}m}{r^{3}}\Delta r\]

(g)  Compare the magnitude of the tidal force ∆F_c caused by the Moon to ∆F_o 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? 


Using the equation above, we know that:

\[\Delta F_{moon}\:=\:\frac{2GM_{moon}m}{r_{moon}^{3}}\Delta r\]

\[\Delta F_{Sun}\:=\:\frac{2GM_{sun}m}{r_{sun}^{3}}\Delta r\]

Thus, we can see that:

\[\frac{\Delta F_{moon}}{\Delta F_{sun}}\:=\:\frac{M_{moon} r_{sun}^{3}}{M_{sun} r_{moon}^{3}}\]

If we plug actual values into this equation, we see that the ratio is about 2.35. This means that the tidal force from the moon is twice as strong as the tidal force from the Sun. Even though the Sun is much more massive than the moon, it is also much farther away so it doesn't have a larger tidal force.

(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)? 


From the previous part of the problem, we can infer that:

\[\frac{\Delta F_{moon}}{\Delta F_{Jupiter}}\:=\:\frac{M_{moon} r_{Jupiter}^{3}}{M_{Jupiter} r_{moon}^{3}}\]

We can then plug our known values for the radii and mass of both the Moon and Jupiter, and can see that this ratio is equal to about \(1.4\:\times\:10^{5}\). This implies that the tidal force from the moon is about 140,000 times stronger than the tidal force from Jupiter even when Jupiter is on its closest approach.