Tides and the Roche limit

Builds on: Sections 3.7, S3 and S4
Reading time: 40 min
Exercises: 4
Olympiad problems: 3

On most coasts the sea rises and falls twice a day, with high tides about 12 hours 25 minutes apart, half of the 24 hours 50 minutes that the Moon takes to come back to the same place in the sky. The highest tides, called spring tides, come a day or two after new and full Moon, and the smallest, the neap tides, near the quarter Moons. The Moon pulls the sea towards itself, but that alone would give one high tide a day, under the Moon. In this section we see why there are two, and how the same stretching tears apart moons and comets that come too close to a planet.

3.8.1The tidal stretch

Before you readMake a guess first

The Sun pulls on Earth about 180 times as strongly as the Moon does. Which raises the bigger tides?

The Moon pulls every part of Earth towards itself, but not equally. The side facing the Moon is pulled a little more than Earth's centre, and the far side a little less. Earth as a whole falls towards the Moon with the acceleration of its centre, as it goes round the centre of mass of the two (Section 3.7). What is left over at each point, the pull there minus the pull at the centre, is the tidal acceleration. On the near side it points towards the Moon, and on the far side away from it, so the oceans are stretched into two bulges on opposite sides of Earth. At right angles to the line to the Moon the pulls converge slightly, which squeezes the oceans inwards.

ResultTidal acceleration(3.18)

\begin{aligned} a_{∥} & = \frac{2 G M R}{d^{3}} \\ a_{⟂} & = - \frac{G M R}{d^{3}} \end{aligned}

$a_{∥}$	the tidal acceleration at the points nearest to and furthest from the body that raises the tide, outwards
$a_{⟂}$	the tidal acceleration at points at right angles to the line to it, inwards
$M, d$	the mass of the body that raises the tide and its distance
$R$	the radius of the body on which the tide is raised

The rule holds when

R

is much smaller than

d

. At a point at angle

θ

from the line to the body, the part of the tidal acceleration along the vertical is

(G M R / d^{3}) (3 \cos^{2} θ - 1)

, and the part along the ground is

(3 G M R / 2 d^{3}) \sin 2 θ

, towards the point under the body.

Show the derivation

The pull of the Moon falls as $1 / d^{2}$ . By (S.12), a change $δ d = - R$ in the distance changes $G M / d^{2}$ by $(- 2 G M / d^{3}) (- R) = 2 G M R / d^{3}$ , which is the extra pull on the near side. The far side is pulled less by the same amount. At a point at right angles to the line, the distance hardly changes, but the pull points towards the Moon, at an angle $R / d$ to the line through Earth's centre. Its part across that line, towards Earth's centre, is $(G M / d^{2}) (R / d) = G M R / d^{3}$ , and the pull at Earth's centre has no such part.

For the Moon at Earth, $G M R / d^{3} = 5.50 \times 10^{- 7}$ m/s², so the stretch along the line to the Moon is $1.10 \times 10^{- 6}$ m/s², about $1.12 \times 10^{- 7}$ of Earth's gravity. That is tiny, but it acts on whole oceans. Earth turns under the two bulges, so most places pass through both each day and see two high tides. The Sun's tide is 0.46 times the Moon's. When the Sun and Moon are in line, at new and full Moon, their tides add to give spring tides, and when they are at right angles, at the quarter Moons, they partly cancel to give neap tides.

Figure 3.7The tidal stretch

Pulled by

Angle of A from the line to the Moon

Vertical part at A
Horizontal part at A

The figure shows Earth seen from above its North Pole, with arrows round the rim giving the tidal acceleration at each point. The dashed outline is the shape the oceans would take, with the bulges much exaggerated. The heavy arrow is the acceleration at the point A, and the table splits it into its parts along the vertical and along the ground. The buttons choose the Moon alone, the Sun alone at 0.46 of the Moon's strength, both in line as at new and full Moon, or both at right angles as at the quarter Moons. The figure can be set to , , where the pull along the ground is largest, or .

Quick checkThe Sun's tide against the Moon's

The Moon has a mass of $7.342 \times 10^{22}$ kg and is 384 400 km away. The Sun has a mass of $1.988 \times 10^{30}$ kg and is $1.496 \times 10^{8}$ km away. What is the Sun's tidal acceleration on Earth as a fraction of the Moon's?

3.8.2Tidal friction and locking

Earth turns once a day, faster than the Moon goes round it, and friction in the oceans drags the bulges a little ahead of the line to the Moon. The Moon pulls back on the nearer bulge, which slows Earth's spin, and the bulge pulls the Moon forwards, which raises its orbit. Today the day lengthens by about 2 milliseconds a century, and the Moon moves away by 3.8 cm a year. Earth raises much larger tides on the Moon, and long ago they slowed the Moon's spin until it turned once in each orbit. This is called tidal locking, and it is why the Moon always shows us the same face.

3.8.3The Roche limit

A moon held together only by its own gravity is torn apart if it comes too close to its planet, where the tidal stretch across it is stronger than its own pull. The distance at which this happens is the Roche limit, after Édouard Roche, who worked it out in 1848.

ResultThe Roche limit(3.19)

d_{R} = k R_{P} {(\frac{ρ_{P}}{ρ_{m}})}^{1 / 3}

$d_{R}$	the closest distance from the planet's centre at which the moon can hold together
$R_{P}, ρ_{P}$	the radius and the mean density of the planet
$ρ_{m}$	the mean density of the moon
$k$	a number that depends on how the moon deforms, $2^{1 / 3} = 1.26$ for a rigid sphere and 2.44 for a fluid one

The limit does not depend on the size of the moon, only on the densities. Small bodies held together by their own strength, such as spacecraft and rocks, can go much closer. Saturn's bright rings lie close to the planet, mostly inside its Roche limit for bodies of ice.

Show the derivation

Take a small piece of mass $μ$ on the surface of a rigid moon of radius $r$ and mass $m$ , on the side facing the planet of mass $M$ . The moon pulls it inwards with $G m μ / r^{2}$ , and by (3.18) the tidal stretch pulls it outwards with $2 G M μ r / d^{3}$ . The piece comes away when

\frac{2 G M r}{d^{3}} = \frac{G m}{r^{2}}, d^{3} = 2 \frac{M}{m} r^{3} .

With $M = \frac{4}{3} π R_{P}^{3} ρ_{P}$ and $m = \frac{4}{3} π r^{3} ρ_{m}$ , this gives $d = 2^{1 / 3} R_{P} (ρ_{P} / ρ_{m})^{1 / 3}$ . A fluid moon is stretched out of shape by the tide, which makes it easier to pull apart, and Roche's own calculation for it gives 2.44 in place of 1.26.

Example 3.13The comet that broke up at Jupiter

In July 1992 the comet Shoemaker-Levy 9 passed within about 40 000 km of Jupiter's cloud tops, and was found afterwards as a string of more than twenty pieces, which struck Jupiter in 1994. Take the comet's density as 500 kg/m³. Jupiter's radius is 71 492 km and its mean density 1326 kg/m³. Was the comet inside Jupiter's Roche limit?

Solution

The density ratio gives $(1326 / 500)^{1 / 3} = 1.384$ , so by (3.19) the Roche limit is $1.26 \times 1.384 = 1.74$ Jupiter radii for a rigid body and $2.44 \times 1.384 = 3.38$ Jupiter radii for a fluid one. The comet passed $(71 492 + 40 000) / 71 492 = 1.56$ Jupiter radii from Jupiter's centre.

AnswerThe comet passed inside both limits, so its own weak gravity could not hold it together, and it was pulled apart into a chain of fragments.

Example 3.14How close the Moon could come to Earthfinish it

Earth has a mean density of 5514 kg/m³ and a radius of 6371 km, and the Moon a mean density of 3344 kg/m³. Treating the Moon as a fluid, what is its Roche limit round Earth?

The start

The Moon is treated as a fluid, so $k = 2.44$ in (3.19).

Your stepNow find

d_{R}

Common errorsSection 3.8

Tides fall as the cube of the distance, while the pull itself falls as the square. Comparing the pulls of the Sun and the Moon gives the wrong answer for their tides.
The far side of Earth has a high tide too. It is pulled less than Earth's centre, so it is left behind as Earth's centre falls towards the Moon.
The tidal acceleration is measured relative to the centre of the body on which the tide is raised. The pull at that centre must be taken away, or there is no second bulge.
The Roche limit compares densities, not sizes, and holds only for bodies held together by their own gravity. A moon of rock and a grain of sand on the same orbit are affected very differently.

ExercisesSection 3.84 questions

3.8.1
If the Moon were twice as far away, by what factor would its tides on Earth change?
3.8.2
At which phases of the Moon are the tides largest?
3.8.3
What is $G M R / d^{3}$ for the Sun's tide on Earth? Take the Sun's mass as $1.988 \times 10^{30}$ kg, its distance as $1.496 \times 10^{11}$ m and Earth's radius as 6371 km.
× 10⁻⁷ m/s²
3.8.4
Why do we always see the same face of the Moon?

Olympiad problemsPast IOAA problems, restated3 problems

2008The Moon's pull along the sea20 points

Adapted from IOAA 2008 (Bandung), theory problem 8. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

The pulls of the Sun and the Moon raise and lower the surface of the sea. Two points A and B lie on the equator, with a difference of longitude $φ$ between them, and A is on the sea. The Moon is overhead at B. Find the horizontal acceleration of the sea water at A caused by the Moon, as measured by observers on Earth, in terms of $φ$ , Earth's radius $R$ and the distance $r$ from Earth's centre to the Moon.

I solved it without the solution

AnswerThe horizontal acceleration towards B is $(G M \sin φ / r^{2}) [(1 + R^{2} / r^{2} - 2 (R / r) \cos φ)^{- 3 / 2} - 1]$ , where $M$ is the Moon's mass. It is close to $(3 G M R / 2 r^{3}) \sin 2 φ$ , largest 45° from the point under the Moon and zero under the Moon and 90° from it.

2011Days in a year long ago10 points

Adapted from IOAA 2011 (Katowice), theory problem 6. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

Tides exert a torque on Earth. Assume that for the last several hundred million years this torque and the length of the sidereal year have stayed the same, at $6.0 \times 10^{16}$ N m and $3.15 \times 10^{7}$ s. How many days were there in a year 600 million years ago? The moment of inertia of a uniform solid sphere of mass $m$ and radius $R$ is $I = \frac{2}{5} m R^{2}$ . Take Earth's mass as $5.972 \times 10^{24}$ kg and its radius as 6371 km.

I solved it without the solution

AnswerA year held about 424 sidereal days, or 423 solar days, 600 million years ago. The official solution writes the earlier spin as the present spin minus the change, which would make the day longer in the past, though its final figure of about 420 days uses the right sign. Growth bands in fossil corals of about 400 million years ago show about 400 days in a year.

2021Gravity on the two faces of Charon15 points

Adapted from IOAA 2021 (Bogotá, held online), theory problem 9. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

Charon, the largest moon of Pluto, has one eighth of Pluto's mass. Both move on circular orbits round their common centre of mass, and both are tidally locked. The distance between their centres is $R = 19 640$ km, and Charon's radius is $r = 593$ km. Let $g_{0}$ be the acceleration due to Charon's own gravity at its surface. Let A be the point of Charon's surface facing Pluto, and B the point opposite it. Find the difference between the accelerations at A and B as a percentage of $g_{0}$ .

I solved it without the solution

AnswerThe accelerations at A and B differ by about 0.004% of $g_{0}$ . Each is about 0.07% less than $g_{0}$ , and the near side is reduced slightly more than the far side.