Worksheet 14.2: Tides

Draw a circle representing the Earth (mass M), with 8 equally-spaced point masses, m, placed around the circumference. Also draw the Moon with mass \(M_{moon} \) 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, \( F_{moon,cen} \) on a point at the center of the Earth? Recall that vectors have both a magnitude (arrow length) and a direction (arrow head).

The gravitational force due to the Moon, \( \vec{F_{moon,cen}} \) is given by: \[ F_{moon,cen} = \frac{ G M_{moon} M_{earth} } {r^2} \] This force vector will be in the direction of the earth towards the moon.

(b) What is the force vector on each point mass, \( F_{moon} \), due to the Moon? Draw these vectors at each point.

For each point mass of mass m, the force vector will be: \[ F_{moon} = \frac{ G M_{moon}m } {a^2} \] In this case, a is equal to the distance from each point mass to the moon.

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

The force difference is simply the vector subtraction of the force at Earth's center and the force at each point mass. This would be given by the following expression: \[ \Delta \vec{F} = \vec{F_{moon}} - \vec{F_{earth, center}} \] This produces different force differences for the point masses around the earth, which accounts for different tides in different areas around the globe.

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

This will result in the oceans moving according to the tidal forces. On the far side of the earth, away from the moon, the tidal force will create a low tide. At the point closest to the moon, the tidal forces will create a high tide. And at the points on the top and bottom of the earth, the tidal forces will create a slack tide (in-between).

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

During the earth's rotation, each given location experiences 2 high and 2 low tides a day. During these period, the high tides will occur when the location is facing the moon, while the low tides will occur when the location is pointing away from the moon.

(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 \[ \Delta F = \frac{2GmM_{moon}}{r^3} \Delta r \] To begin, we can compare the forces at each of these two points: \[ F_{near} = \frac{GmM_{moon}}{r^2} \] \[ F_{far} = \frac{GmM_{moon}}{(r + \Delta r)} \] We can express this difference as: \[ \Delta F = F(r + \Delta r) - F(r) \] This is useful because we can divide both sides by \( \Delta r\) to get the derivative definition expression. This then allows us to calculate \( \Delta F \) by taking the derivative of F(r) and then multiplying by \( \Delta r \) in the end. \[ \Delta F = \frac{d}{dr} ( \frac{GmM_{moon}}{r^2} ) \Delta r \] \[ \Delta F = \frac{2GmM_{moon}}{r^3} \Delta r \]

(g) Compare the magnitude of the tidal force \( \Delta F_{moon} \) caused by the Moon to \( \Delta F_{sun} \) 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?

The tidal force generated by the sun is expressed by: \[ \Delta F_{sun} = \frac{2GmM_{sun}}{AU^3} \Delta r \] From this, we can see that the primary differences between this expression, and our expression for \( \Delta F_{moon} \) in part (f) is the sun distance and the sun mass. Since \( \Delta r \) is the same value in both expression, we can set those terms equal to each other, and generate a relationship for the sun and moon forces. \[ \Delta F_{sun} = \frac{r^3}{AU^3} \frac{M_{sun}}{M_{moon}} \Delta F_{moon} \] Inserting known quantities for r, AU, \( M_{sun} \), and  \( M_{moon} \), we get the following relationship: \[ \Delta F_{sun} \approx 0.4 \, \Delta F_{moon} \] or \[ \frac{ \Delta F_{moon} } { \Delta F_{sun} } \approx 2.5 \] This tells us that the tidal force caused by the sun is approximately 2.5 times weaker than the tidal force caused by the moon. This makes sense, since we generally don't consider the sun as a factor in our oceans' 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)?

We can again take the same approach that we followed in part (g): \[ \Delta F_{jupiter} = \frac{2GmM_{jupiter} }{(4AU)^3} \] Now, setting the \( \Delta r \) terms equal, we can solve for \( \Delta F_{jupiter} \) in terms of \( \Delta F_{moon} \): \[ \Delta F_{jupiter} = \frac{r^3}{(4AU)^3} \frac{M_{jupiter}}{M_{moon}} \Delta F_{moon} \] Now, plugging in our known quantities of r, AU, \( M_{jupiter} \), and \( M_{moon} \), we can get our relationship between \( \Delta F_{jupiter} \) and \( \Delta F_{moon} \). \[ \Delta F_{jupiter} \approx 1 \times 10^{-5} \Delta F_{moon} \] \[ \frac{ \Delta F_{moon}}{ \Delta F_{jupiter} } \approx 1 \times 10^5 \]

Acknowledgements: Team EE