I have been interested in this problem for a long time, but I had set it aside without much research. Recently, while reading the section on “Tidal Forces” in Gravitation and Spacetime, I returned to this topic and decided to write something about it. Here, I will not delve deeply into the formal definition of hydrostatic equilibrium; as the name suggests, it is the state of equilibrium reached by a fluid under a specific force field. Hydrostatics tells us:
When hydrostatic equilibrium is reached, the surface of the fluid must be an equipotential surface.
Why is this the case? Let us perform a simple analysis from a mathematical perspective. Considering only the two-dimensional case, if the equation of the equipotential surface is U(x,y)=C, then differentiating both sides gives: 0 = dU = \frac{\partial U}{\partial x}dx + \frac{\partial U}{\partial y}dy = \left(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}\right) \cdot (dx, dy)
This means that the vector \left(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}\right) and the vector (dx, dy) are perpendicular. The former is the force function (gradient of the potential), and the latter is a tangent vector (in three dimensions, it would be a tangent plane). In other words, the net external force must be perpendicular to the fluid surface. Only in this way can an equal and opposite internal force be provided to keep the entire structural system in equilibrium!
This result has quite a few applications in astronomy. Two examples are provided below:
I. Earth’s Oblateness
We have explored this issue before, but the previous investigation was based on an assumption that might not necessarily hold, meaning it lacked a solid physical foundation. Now, we provide a derivation from the perspective of hydrostatic equilibrium.
Wait, some readers might object: the Earth is clearly a solid planet, so why are we talking about fluids? Here, the fluid equilibrium state is actually a state of matter distribution. It is true that the Earth is a solid (rigid body) and does not reach equilibrium as easily as air or liquids. However, the Earth has been formed for over 4 billion years. Research suggests that the Earth’s rotation speed is constantly slowing down, meaning today’s rotation speed is almost a lower bound in its history. Over long periods of motion, under the continuous action of even small centrifugal forces, the Earth’s shape tends toward a state of hydrostatic equilibrium. Using this to estimate the magnitude of oblateness is consistent with reality.
In a simple analysis, a small object on the Earth’s surface is subject to two forces: first, the Earth’s gravity, and second, the “centrifugal force” caused by the Earth’s rotation, as shown in the figure:
The gravitational force on the Earth’s surface can be denoted as mg, and the gravitational potential energy is mgh = mg(r-R), where R is the polar radius of the Earth. The centrifugal force is F = m\omega^2 x, so the centrifugal potential energy is -\frac{1}{2}m\omega^2 x^2. Such an equipotential surface is given by: gh - \frac{1}{2}\omega^2 x^2 = C
When x=0 (at the North or South Pole), h=0, so C=0. Thus, the equation for the Earth’s shape is: g(r-R) - \frac{1}{2}\omega^2 x^2 = 0
At the equator, we have: gh = \frac{1}{2}\omega^2 (R+h)^2 \approx \frac{1}{2}\omega^2 (R^2 + 2Rh)
Solving for h: h = \frac{\omega^2 R^2}{2(g - \omega^2 R)}
Since \omega^2 R is a small quantity, we can ignore it in the denominator, yielding the oblateness as: \frac{h}{R} = \frac{\omega^2 R}{2g}
This is actually half of the result we obtained in a previous article. Some readers might wonder: didn’t the previous result already fit the data quite well? If we take half of it now, won’t the deviation be larger? Indeed, the deviation is larger, but this result is derived based on physical laws and is theoretically persuasive. As for the large deviation, it is due to our idealized model; many other factors also affect the Earth’s oblateness.
II. Tides
The same model can be used to estimate the wave height on Earth caused by the tidal forces of the Moon and the Sun. The Moon is the primary source, with its effect being approximately twice that of the Sun. The magnitude of the tidal force is \frac{2GMm}{d^3}x, where M is the mass of the Moon, d is the Earth-Moon distance, and x can be simply understood as the radius of the latitude circle where a point on Earth is located. Then, the tidal potential energy brought by the tidal force is -\frac{GMm}{d^3}x^2, and the Earth’s gravitational potential energy is mgh. It should be noted that the interaction between seawater and tidal forces does not reach hydrostatic equilibrium. However, because seawater is easily deformed, it does not stay fixed in one position when subjected to tidal forces but oscillates around the equilibrium position. Therefore, it is reasonable to use the hydrostatic equilibrium position to estimate the wave height.
Thus, hydrostatic equilibrium tells us: gh - \frac{GM}{d^3}x^2 = C
Similarly, at the poles, we should have h=0 and x=0, so C=0. At the equator: gh = \frac{GM}{d^3}x^2 \approx \frac{GM}{d^3}R^2 Which gives: h = \frac{GMR^2}{gd^3} \approx 0.37\text{m}
This yields a wave height of 37 cm. From this perspective, the estimate is reasonable. Of course, there are much higher waves on Earth, which are formed by the combined action of various factors. For the Moon’s gravity, this is essentially the contribution it provides. During the New Moon (the first day of the lunar month), the tidal forces of the Moon and the Sun add up, and the above calculation would yield a wave height of about 50 cm, which was called the “spring tide” in ancient times. During the last quarter (around the 22nd day of the lunar month), the tidal forces of the two partially cancel each other out, and the tide-generating effect reaches its minimum, known as the “neap tide.” Of course, this is only a theoretical estimate and is meaningful only as a long-term average effect.
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