Gravity weakens with distance, so the near side of a body is pulled harder than its far side. That difference, the tidal force, stretches the body along the line to the other mass. It raises ocean tides, it tears apart moons that stray too close, and near a small black hole it would pull you apart. The tool is linear approximation applied to gravity outside a sphere.
💡 Why it works
The acceleration toward a mass $M$ is $a(r) = GM/r^2$. Two points separated by a small $\Delta r$ along the line to $M$ feel accelerations that differ by about $$\Delta a \approx \left|\frac{da}{dr}\right|\Delta r = \frac{2GM}{r^3}\,\Delta r.$$ The derivative of $1/r^2$ is $-2/r^3$, so the tidal effect falls off one power faster than gravity itself. In the frame of the body's center, the near side is pulled toward $M$ and the far side is left behind: two bulges, not one.
🔑 Key idea
Tidal acceleration across a length $\Delta r$ at distance $r$: $\Delta a \approx 2GM\,\Delta r/r^3$. It goes as $M/r^3$, so a near, small body (the Moon) can out-tide a far, huge one (the Sun).
The difference between the Moon's pull on each part of Earth and its pull on Earth's center. The arrows point away from Earth on both the near and far sides: two tidal bulges.
Worked examples
✏️ Worked example: The Moon and the Sun on Earth's oceans
Compare the tidal accelerations from the Moon ($7.35 \times 10^{22}$ kg at $3.84 \times 10^{8}$ m) and the Sun ($1.99 \times 10^{30}$ kg at $1.50 \times 10^{11}$ m) across Earth's radius ($6.37 \times 10^{6}$ m).
Sun: $5.0 \times 10^{-7}$ m/s², so the Sun's tidal effect is about $0.45$ of the Moon's.
Answer: The Moon raises tides about twice as strong as the Sun's, even though the Sun pulls Earth about 180 times harder.
✏️ Worked example: The Roche limit
A rock sits on the near side of a moon of radius $r$ and density $\rho_m$ orbiting a planet of radius $R_P$ and density $\rho_P$ at distance $d$. Where does the planet's tidal pull beat the moon's own gravity?
Self-gravity: $Gm/r^2$ with $m = \tfrac43\pi r^3\rho_m$. Tidal pull: $2GM\,r/d^3$ with $M = \tfrac43\pi R_P^3\rho_P$.
Equal when $d^3 = 2Mr^3/m$, so $d = R_P(2\rho_P/\rho_m)^{1/3}$.
Answer: For icy bodies ($900$ kg/m³) near Saturn ($R_P = 60{,}300$ km, $687$ kg/m³) that is about 69,000 km; a fluid moon breaks up farther out, near $2.44R_P(\rho_P/\rho_m)^{1/3} \approx 135{,}000$ km. Saturn's bright main rings lie roughly inside that distance.
✏️ Worked example: Falling into a black hole
Find the tidal difference across a 2 m body at the horizon of a $10\,M_\odot$ black hole ($r_s = 2.95 \times 10^{4}$ m).
Answer: About 20 million $g$: fatal "spaghettification" well outside the horizon. At the horizon of Sagittarius A* ($4.3 \times 10^{6}\,M_\odot$) the same difference is only $10^{-3}$ m/s², since $\Delta a$ at $r_s$ goes as $1/M^2$.
⚠️ Common mistake
Thinking tidal force falls off as $1/r^2$, like gravity itself. It is a difference of gravity across a body, and the difference falls off as $1/r^3$: double the distance and gravity drops to a quarter, but tides drop to an eighth. Check: if tides went as $M/r^2$, the Sun's tides would be 180 times the Moon's instead of about half.
In 1992 comet Shoemaker–Levy 9 passed inside Jupiter's Roche limit and was torn into more than 20 pieces, which struck Jupiter in July 1994.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.