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Gravitational Physics

Here we lay out Newton's law of gravity and crack open the universe of consequences that spring from it.

Ocean tides

                           

Tides have earned their place in the canon of nature’s awe-inspiring spectacles, and have fascinated philosophers old and new. Among these theorists is noted astronomer Bill O’Reilly who famously declared

“Tide goes in, tide goes out, never a miscommunication. You can’t explain that.”

In fact, here you will.

There exist as many wrong explanations for tides as there are grains of sand on the beach that’s getting covered by the rising tide of this horrible pun, but by focusing on first principles, we can find the correct reasoning.

Tides recur roughly once every twelve hours.

Which of the following effects could possibly be the cause of Earth’s tides?

Hint: think about the timescale of the different effects.

Timing is provisional evidence for, but not proof by any means that the gravitational pull from celestial bodies is responsible for ocean tides. To really address the question, we must look at the gravitational field of celestial bodies in the vicinity of Earth. For the next few questions, we're going to frame things in terms of the influence of an object of particular beauty, a celestial donut.

Fig: Illustration of the gravitational attraction of the celestial donut in vicinity of Earth (<strong>not</strong> drawn to scale).

Fig: Illustration of the gravitational attraction of the celestial donut in vicinity of Earth (not drawn to scale).

Drawing the gravitational influence of the donut may seem to immediately refute our hypothesis. In this view, the water on the near side of the Earth is pulled toward the donut, but so is the water on the other side. In fact, everything is pulled toward the donut! How can we have two outward tides?

As damning as this may seem, all is not lost. Recall, Earth is in stable orbit around the Sun and thus the distance \(d\) does not shrink under the pull of \(F_\textrm{celestial}.\) This suggests that we're missing something.

In reality, there is of course no celestial donut, but there are the Moon and Sun.

The relationship between Moon and Earth is often portrayed as one in which the Moon orbits in a perfect circle about a stationary Earth. But the gravitational interaction of the Moon with Earth actually causes Earth to perform a small orbit, as shown below, around their mutual center of mass (dashed grey line).

Earth's slight oscillation about the center of mass of the Earth-Moon system.

Earth's slight oscillation about the center of mass of the Earth-Moon system.

Calculate the magnitude of Earth's acceleration, \(a_\textrm{Earth},\) toward the Moon in \(\si[per-mode=symbol]{\meter\per\second\squared}.\)

Details:

  • the distance from Earth to Moon is \(d = \SI{384.4e6}{\meter}.\)
  • \(M_\textrm{Moon}=\SI{7.35e22}{\kilo\gram}.\)

To cut through the morass, we just have to switch perspectives to the non-inertial frame of Earth. To keep things simple, we're going to first focus on points along the line connecting the centers of the celestial body and Earth.

Suppose the acceleration of the water at the near and far side in the inertial reference frame is \(a(d\mp R_E).\) We can transform these into the frame of Earth according to the Galilean transformation \[a \rightarrow a^\prime = a - a_\textrm{Earth}.\]

Find the sum of accelerations at either side, \[a^\prime_\textrm{near} + a^\prime_\textrm{far}.\]

Details:

  • \(d = \SI{4e8}{\meter}\)
  • \(R_\textrm{E} = \SI{6.371e6}{\meter}\)
  • the mass of the celestial donut = \(\SI{e22}{\kilo\gram}\)

As you just showed, the pull from the donut at the near and far side in the frame of the Earth is very nearly equal and opposite.

This is due to the fact that the gravitational pull from faraway celestial bodies decreases linearly in the vicinity of Earth, and therefore pull the water on the near side with force \(F(d) + \Delta F_\textrm{near}\), the Earth with force \(F(d),\) and the water on the far side with force \(F(d) + \Delta F_\textrm{far}\) yielding the relative forces \[\begin{align} F_\textrm{near} &\approx -\Delta F \\ F_\textrm{center} &= 0 \\ F_\textrm{far} &\approx +\Delta F \\ \end{align}\] Therefore, the effect of a celestial body is to pull the ocean away from Earth's surface at both the near AND far side.

The water level at a given location on Earth will go from maximum to minimum approximately once every 12 hours as The Earth rotates under the tidal bulge.

Note: the Earth rotates under the dual bulge once every \(\SI{24}{\hour},\) and on the longer term, the Moon orbits the Earth once every \(\approx\SI{27.3}{\day},\) dragging the tidal bulges along with it.

Now that we have the insight, let's make things quantitative.

Suppose you’re looking at the Earth from an inertial frame in the presence of a celestial body (above).

Write down an expression for the pull from the celestial body \(a_\textrm{celestial}(h) = F_\textrm{celestial}(h)/m\) at position \(h\) relative to the center of Earth (\(r = d_\textrm{celestial} + h\)).

Take a Taylor expansion about the center of Earth and keep the first two terms. You should find \[a(h) = \overbrace{G\frac{M_\textrm{celestial}}{d_\textrm{celestial}^2}}^{\textrm{constant}} + \overbrace{\Delta a(h)}^{\textrm{correction}}\] What is \(\Delta a(h)\)?

Repeating the Galilean transformation, we see that \(F_\text{near/far} = \pm \Delta F\) holds to linear order in \(h.\) Our calculation applies equally to all celestial bodies, including the Sun and Moon.

Based on your last calculation, which gravitational interaction do you expect to have a greater effect on Earth's water?

Diagram of the force from a celestial body along the line connecting Earth's center and that of the body, calculated in the frame of Earth.

Diagram of the force from a celestial body along the line connecting Earth's center and that of the body, calculated in the frame of Earth.

Details \[\begin{array}{|c|c|}\hline \textbf{Quantity} & \textbf{Value} \\ \hline M_\textrm{Sun} & \SI{1.99e30}{\kilo\gram} \\ \hline d_\textrm{Sun} & \SI{1.45e11}{\meter} \\ \hline M_\textrm{Moon} & \SI{7.35e22}{\kilo\gram} \\ \hline d_\textrm{Moon} & \SI{3.84e8}{\meter} \\ \hline \end{array}\]

In order to predict the variation in water level between high and low tide, we’re going to use an argument used in Newton’s historical development. Suppose we dig a thin well from the near side of Earth all the way to the center, and then out again to the side, and let it fill with ocean water.

Once things settled down, the water would adopt an equilibrium surface, with the near and far points at the height \(h_\textrm{high}\) and the side points at the height \(h_\textrm{low} < h_\textrm{high}.\) If the surface of the water is at equilibrium, then we should expect the pressure at the bottom of each well to be equal, or else water would flow until it were.

Find the pressure at the bottom of the well by calculating along the nearside well.

Details and a hint:

  • Leave integrals of \(g_\textrm{Earth}(h)\) unevaluated. In the choices, \(I_H\) represents \(\rho\int_0^{R_E+h_\textrm{high}}g_\textrm{Earth}(h)\,dh.\) We will see same integral appears in the side well calculation.
  • You can modify your approach from the Atmosphere Thickness quiz in light of the forces acting here.

It is tempting to think that along the sidewell, the only relevant force is the pull of gravity, but there is actually a tidal contribution we need to take into account there as well.

Because the pull from the moon is not perfectly parallel with the center of the Earth, it has a small component inward toward Earth's center, which enhances the pull from gravity on the water at the surface.

Find an expression for the pull from the Moon along the sidewell.

Hint: construct a right triangle between the Earth, the Moon, and the sidewell.

Find an equivalent expression for the pressure at the bottom of the well, by calculating along the sidewell.


Set this expression equal to your result for the nearwell and solve for \(h_\textrm{near} - h_\textrm{side},\) the daily variation in water level for a location on Earth.

Hint: \(\left(R_\textrm{Earth} + h_\textrm{low}\right)^2 \approx \left(R_\textrm{Earth} + h_\textrm{high}\right)^2 \approx R_\textrm{Earth}^2.\)

In actual fact, the Sun and Moon both contribute to the variation in tides on Earth, with the Sun contributing about half as much as the Moon as you found earlier.

When the Sun and Moon are aligned (either at Full Moon or New Moon), their tidal effects are additive, and we find higher highs and lower lows.

At Full or New Moon, the tidal accelerations due to the Moon and Sun add constructively, leading to a more dramatic tidal surge.

At Full or New Moon, the tidal accelerations due to the Moon and Sun add constructively, leading to a more dramatic tidal surge.

Conversely, at Quarter Moons, their effects are interfering and we find lower highs and higher lows.

At Quarter Moon, the tidal accelerations due to the Moon and Sun interfere, leading to a tide that's much reduced.

At Quarter Moon, the tidal accelerations due to the Moon and Sun interfere, leading to a tide that's much reduced.

All the analysis of the preceding panes has been linear in the acceleration terms, which means that we can perform a simple superposition to find their joint effect, as you'll do in the next problem.

Use your result from the last problem to calculate the daily variation in water level (in \(\si{\centi\meter}\)) during a Full or New Moon.

\[\begin{array}{|c|c|}\hline \textbf{Quantity} & \textbf{Value} \\ \hline M_\textrm{Sun} & \SI{1.99e30}{\kilo\gram} \\ \hline d_\textrm{Sun} & \SI{1.45e11}{\meter} \\ \hline M_\textrm{Moon} & \SI{7.35e22}{\kilo\gram} \\ \hline d_\textrm{Moon} & \SI{3.84e8}{\meter} \\ \hline \end{array}\] Hint: the model is linear with respect to accelerations.

In this quiz you constructed a simple model for the tides as influenced by the gravitational attraction of nearby celestial bodies like the Sun and Moon. As you showed, these are caused by the so-called tidal acceleration inherent to any force that decreases with distance, causing some parts of a body to be pulled differently from others.

The approach we took here is an example of what's called an equilibrium model because we assume that the ocean water has enough time to adjust to the heights required by the tidal acceleration. This is equivalent to assuming that the water instantly snaps into place, or that the rotation of Earth is extremely slow compared to the dynamics of water. In reality things are more complex in that it takes actual time for water to move, water has inertia, and it experiences drag in interacting with the ocean floor.

More seriously, our model treats the ocean floor like a sphere of uniform radius, ignoring its varying topography as well as the presence of continents. Thus, while we expect our predictions to hold in the open ocean where the effects of land are less important, to accurately capture the effect from the varying sea floor, we need to adopt a more sophisticated modeling strategy.

Note: Though we are subject tidal acceleration from the Earth, the effect is not noticeable, its magnitude amounts to just one part in ten million compared to the gravitational field at the surface. On the inside of a black hole, tidal accelerations are typically big enough to rip objects apart, resulting in the so-called "spaghettification" of matter and astronauts in movies.

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