Tidal Forces

A tidal force is not a separate fundamental force, but rather the differential gravitational force experienced across an extended body caused by a non-uniform gravitational field.

Because gravity drops off with distance according to an inverse-square relationship, the side of an object closest to a gravitational source experiences a stronger pull than its center, which in turn experiences a stronger pull than the far side.

Deriving the Tidal Force Gradient

Consider an extended body of radius $R$ at a center-to-center distance $r$ from a mass $M$ . The gravitational force on a unit mass at the near side versus the far side is:

F_{\text{near}} = \frac{GM}{(r - R)^2}, \quad F_{\text{far}} = \frac{GM}{(r + R)^2}

The stretching force $\Delta F$ across the object is the difference between these forces:

\Delta F = GM \left( \frac{1}{(r - R)^2} - \frac{1}{(r + R)^2} \right)

Combining terms over a common denominator:

\Delta F = GM \left( \frac{(r + R)^2 - (r - R)^2}{(r^2 - R^2)^2} \right) = GM \left( \frac{4rR}{(r^2 - R^2)^2} \right)

Because the distance to the external body is much greater than the size of the object ( $r \gg R$ ), we can approximate $r^2 - R^2 \approx r^2$ :

\Delta F \approx \frac{4GM R r}{r^4}

\Delta F \approx \frac{4GMR}{r^3}

(1)

This demonstrates that tidal forces scale with the inverse-cube of distance ( $1/r^3$ ), making them far more sensitive to distance than standard gravitational force ( $1/r^2$ ).

This inverse-cube gradient explains why the Moon exerts roughly twice the tidal force on Earth as the Sun does. While the Sun's overall gravitational pull on Earth is much stronger, the Moon's proximity creates a much steeper gradient across Earth's diameter.

Solar and Lunar Alignment

The vector sum of the Moon's and Sun's tidal forces varies as the Moon orbits Earth:

Spring Tides: When the Sun, Earth, and Moon align during New and Full Moon phases, their tidal forces constructively combine to produce the highest high tides and lowest low tides.
Neap Tides: When the Moon is at a right angle to the Sun during First and Third Quarter phases, the tidal forces destructively interfere, resulting in the smallest tidal range.

Astrophysical Applications

Differential gravitational forces cause several important phenomena throughout the universe:

Tidal Friction and Locking: Differential forces continuously deform an orbiting body. Over time, energy lost to internal friction slows the body's rotation until it becomes tidally locked, facing the same side toward its primary.
Tidal Heating: Rapid periodic stretching generates massive internal frictional heat, such as Jupiter's tidal flexing of its moon Io.
The Roche Limit: The minimum distance a celestial body can approach a larger mass before the tidal force pulling it apart exceeds its own internal self-gravity.

Example 1

Find the ratio of the Moon's tidal force on Earth to the Sun's tidal force on Earth.

Given $M_{\text{Sun}} = 1.99 \cdot 10^{30} \text{ kg}$ , $M_{\text{Moon}} = 7.35 \cdot 10^{22} \text{ kg}$ , Earth-Sun distance $r_{\text{Sun}} = 1.50 \cdot 10^{11} \text{ m}$ , and Earth-Moon distance $r_{\text{Moon}} = 3.84 \cdot 10^8 \text{ m}$ .

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Use the inverse-cube relationship from Equation 1 to form a direct ratio of tidal forces.

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Using Equation 1, the ratio of tidal forces is:

\frac{\Delta F_{\text{Moon}}}{\Delta F_{\text{Sun}}} = \frac{M_{\text{Moon}}}{M_{\text{Sun}}} \left( \frac{r_{\text{Sun}}}{r_{\text{Moon}}} \right)^3

Plugging in the known values:

\frac{\Delta F_{\text{Moon}}}{\Delta F_{\text{Sun}}} = \left( \frac{7.35 \cdot 10^{22} \text{ kg}}{1.99 \cdot 10^{30} \text{ kg}} \right) \left( \frac{1.50 \cdot 10^{11} \text{ m}}{3.84 \cdot 10^8 \text{ m}} \right)^3

\frac{\Delta F_{\text{Moon}}}{\Delta F_{\text{Sun}}} = (3.69 \cdot 10^{-8}) \cdot (5.96 \cdot 10^7) \approx 2.2

The Moon's tidal force is approximately 2.2 times stronger than the Sun's tidal force on Earth.