WS 5.2, Problem 2: Tidal Forces

Recall the reading assignment about tides. Draw a circle representing the Earth (mass M$_{\oplus}$), 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).

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

Both (a) and (b) are drawn in the picture below.

(c) What is the force difference, $\Delta$F, between each point and Earth's center? This is the tidal force.

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

At the points where the vectors are pointed away from Earth, the ocean will bulge.  We see this as an increase in sea level, known as high tide.  At the points where the points are pointing in towards the Earth, the ocean is slightly compressed.  We interpret this as low tide, a decrease in sea level.

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

Along the moon's orbital plane, oceans experience exactly two full cycles of high and low tide.  The first high comes when the moon is directly overhead.  As the moon orbits Earth the tides at that original location get lower until it's time for low tide.  Low tide happens with the moon is at a position in its orbit 90$^{_o}$ from the original location.  After that, tide gets higher and higher until the moon is above the spot on the Earth exactly opposite its original location.  There begins the second tide cycle.

(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  $\Delta$r, show that the force difference is given by

$\Delta \mathbf{F}=\frac{2GmM_{moon}}{r^3}\Delta r$

This is the formula for gravitational force according to Newton's law of gravity, found in the reading on tidal waves.

$f(r)=\frac{-GmM_{moon}}{r^2}$

Below, I set up an equality between $\frac{\Delta F}{\Delta r}$ and the derivative of the gravitational force formula to find the force difference.  This equality works because the force difference, or tidal force, is actually the change in force over change in distance, r.  

$\frac{\Delta F(r)}{\Delta r}=\frac{dF f(r)}{dr}=\frac{\left ( \frac{-GmM_{moon}}{(r+\Delta r)^2} \right )-\left ( \frac{-GmM_{moon}}{r^2} \right )}{\Delta r}$

$\frac{\Delta F}{\Delta r}=\frac{2GmM_{moon}}{r^3}$

$\Delta F = \frac{2GmM_{moon}}{r^3}\Delta r$

(g) Compare the magnitude of the tidal force F$_{moon}$ caused by the Moon to F$_{\odot}$  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?

$\frac{\Delta F_{moon}}{\Delta F_{\odot}}=\left ( \frac{2GmM_{moon}\Delta r}{r_{moon}^3} \right )\left ( \frac{r_{\odot}^3}{2GmM_{\odot}\Delta r} \right )=\frac{M_{moon}r_{\odot}^3}{M_{\odot}r_{moon}^3}$

The tidal forces caused by the moon are about 2.25 times stronger than the tidal forces caused by the sun.  This makes sense because even though the sun is a lot more massive than the moon, it's a LOT farther away.  

If the sun and moon were on the same side of the Earth, their tidal forces would add together, creating an extra high tide called Spring Tide.  When the sun and moon happen to align in such a way that they form a 90$^{_o}$ angle with the Earth, we see the lowest tides, called Neap Tides.  

(h) How does the magnitude of $\Delta$F caused by the Moon compare to the tidal force caused by Jupiter during its closest approach to the Earth (r is about 4 AU)?

$\frac{\Delta F_{moon}}{\Delta F_{jup}}= \frac{M_{moon}r_{jup}^3}{M_{jup}r_{moon}^3}= \frac{M_{moon}(4r_{\odot})^3}{M_{jup}r_{moon}^3}$

The tidal forces by the moon on the Earth are 1.5x10$^5$ times stronger than the tidal forces caused by Jupiter.  Jupiter is even lighter and farther away from the Earth than the sun is, so its effect should be really small compared to the moon.