Tides are caused by "differential gravity." That is, tides are caused by the slight difference in the gravitational attraction (for example) of the moon on the near side vs. the far side of the Earth.
In our previous discussion of Newton's law of gravity, F = G m_{1} m_{2} / R^{2} , we effectively assumed that the two objects in question have no size. So, let's be more realistic (but not completely realistic). Let's assume the moon has no size (all the mass of the moon is located exactly at a single point at the center of the moon). And let's assume the Earth has its normal size.
Now, the moon attracts every small piece of Earth, every tiny chunk, independently of all the other tiny chunks. Although all of the imagined Earthly chunks are identical in mass, they are not identical in distance from the moon. In fact, a piece of Earth on the near-side to the moon is 13,000 km closer to the moon than a piece of Earth on the far-side.
Assume the moon's orbit is perfectly circular with a semi-major axis of a = 384,000 km. Since a is taken from the center of the Earth, the force of gravity on the near-side is proportional to 1/(377,000 km)^{2} while the force of gravity on the far-side is proportional to 1/(391,000)^{2} , where we have rounded the value of the radius of the Earth to be 7,000 km, to make the math a bit cleaner.
I'm not interested in the exact numerical value of the gravitational force; but I am interested in the relative strengths of gravity of the near vs. the far side. So, let's compare the two forces. If we take a ratio, everything cancels out except the distances:
F_{near}/F_{far} = (GMm/r_{near}^{2}) / (GMm/r_{far}^{2}) = (1/r_{near}^{2}) / (1/r_{far}^{2}) = (r_{far}^{2}) / (r_{near}^{2}) = (r_{far} / r_{near})^{2}
F_{near}/F_{far}= (391,000 / 377,000)^{2} = (391/377)^{2} = 1.08
Or, the difference is 8%. That's quite a lot! If the Earth were perfectly elastic, and thus responded instantly, the Earth would become stretched along the line connecting the center of the Earth with the center of the moon. Of course, the oceans respond more easily than the rocks. But keep in mind that the "solid" earth is flexible; it stretches, flows and breaks. The rocky Earth near the moon actually rises by a few centimeters in response to the tidal pull of the Moon. Meanwhile, the oceans respond more strongly, in some places by a few meters, in other places by tens of meters, the local response being largely the result of topography of the ocean bottom and the continent-ocean interfaces.
This sounds like the tidal pull of the moon should produce high tide on the near-side and low tide on the far side. But this is wrong.
We can think of the 8% stretch as the near side of the Earth being stretched 4% harder than the center -- and being pulled away from the center, while the center is stretched 4% harder than the far side -- and is pulled away from the far side. Thus, the whole Earth is stretched. This effectively lifts the near side `off' the Earth, generating a tidal bulge, and leaves behind the far side, generating a high tide on the far side as well.
Thus, we have two high tides: one "under" the moon and the other "opposite" the moon. We also then have two low tides, on the two sides of the earth perpendicular to the Earth-Moon line. In the ideal situation, you would experience high tide exactly when the moon is directly overhead (noon at new moon; 6 PM at 1st quarter moon, etc.) and again exactly twelve hours later. You would have low tide six hours before and after the moon was directly overhead.
But the tides are not spaced in 24 hour cycles; they are spaced much closer to 24 hour 50 minute cycles. WHY?
Because the Moon orbits the Earth! After 24 hours (one rotation of the Earth) the moon has moved about 1/27th of it's orbital distance around the Earth. And 1/27th of 24 hours is about 50 minutes. So the Earth has to spin an extra 50 minutes worth to get the moon back to directly overhead. Thus, the moment of high tide cycles through the day as the moon orbits the Earth.
If we had no moon, would we have tides? Yes, due to the Sun.
What about the Sun?
The Sun also produces tides. But since the Sun is 149.6 million km away (assuming a perfectly circular orbit for the Earth), the differential effect is smaller than that caused by the moon.
F_{near}/F_{far}= (149,607,000 / 149,593,000)^{2} = 1.0002
Or, the difference is about 2/100 of 1%. That's more than 400 times weaker than the tidal effect of the moon. But since the gravitational pull of the Sun overall is 150 times stronger than that of the moon (because the Sun is so massive, despite being so far away), the net effect is that the differential pull generated by the sun is 150/400 = 0.4 times as strong as that of the moon. In other words, the effect of the sun is not negligible, but rather is almost half as important as the moon in causing tides.
So, what are the combined effects of the Sun and moon on tides?
If the Sun and moon are in perpendicular directions ("noon", 1st or 3rd quarter moon), "neap" tides (0.6 times as strong as if there were no Sun involved).
If the Sun and moon are in the same direction ("noon", new moon): "spring" tides (1.4 times as strong as if there were no Sun involved, or more than twice as strong as neap tides).
What if the Sun and moon are in opposite directions (sun @ "noon", moon @ full moon)?
1. tidal heating
Let's assume, for simplicity, that
b. 6 hours later, our high tide has relaxed to low tide.
c. 6 hours later, we are at high tide.
d. 6 hours later we are at low tide again.
This continues endlessly. Now, imagine that instead of the Earth, we are dealing with a perfectly spherical piece of modeling clay. First we squeeze it so that it has a long and short axis. Then we squeeze it so that the short axis becomes long and the long short. And then we repeat this, over and over again. What will happen to our chunk of modeling clay? It warms up, of course. This is due to friction inside: as we squeeze, we put energy of motion into the clay. Inside the clay, this motion becomes collisions between different pieces of clay. Thus, the energy becomes friction becomes heating, and the heating occurs INSIDE. Thus, amazingly, tides are a mechanism for transmitting energy in the form of heat to the INSIDE of the Earth. But you can't get something from nothing. So this energy must come from somewhere. So, Where does the energy come from that goes into tidal heating of the Earth? [the Earth's rotation]
The Earth is slowing down by 0.0016 seconds per century (about 1.6 thousandths of a second per century), or 1.6 x 10^{-5} s/yr.
Thus, in 1000 years, this amounts to
1000 yr * 1.6 x 10^{-5} s/yr = 0.016 sIn 1 million years,
10^{6} yr * 1.6 x 10^{-5} s/yr = 16 s.In 400 million years,
400 x 16 s= 6,400 s * 1 hour/3,600 s = 1.8 hoursso one day was about 22 hours, 400 million years ago.
In 4 billion years,
10 x 1.8 hours = 18 hours!So, this incredibly tiny effect adds up to an enormous amount over long time periods.
Thus, this discussion suggests that the Earth must once have rotated faster. Can we test this?
Yes.
Certain corals experience daily growth cycles, laying down one layer per day. They also experience annual cycles, like tree rings. Thus, by counting the number of small cycles per annual cycle, we can determine the number of days per year. The growth rings in 400 million years old, ancient corals, indicate that there were about 400 days per year, rather than 365.
The length of the year itself hasn't changed, we know that one year is equivalent to
1 year = 365 days x 24 hrs/day = 87,600 hours.In order to have 400 days in a year, the length of the day was very close to
1 (ancient) day = 87,600 hours/yr / 400 days/yr = 22 hours/day,which is almost exactly what we get in the above calculation!
A 1996 study of sedimentary rocks suggests the day was only 18 hours long 900 million years ago.
Now let's add another wrinkle of imperfection.
The Earth does not react perfectly, instantly. Thus, the tidal bulge does not fall directly along the Earth-Moon line of centers. Rather, it falls slighly ahead (in the direction of Earth's rotation) of this line.
In addition, the moon is not a perfectly small point. It has size,
just like the Earth. And, it rotates also.
And just as the moon has been slowing down Earth's rotation, the Earth
has been slowing down the Moon's rotation.
So, let's go back in time to a point when the Earth and Moon were both
spinning faster than today.
Both of them are stretched by the tides. Both of them respond
slowly so that the tidal bulges fall "ahead" of the line of centers.
Now, these tidal bulges pull back on each other, with the nearer bulges having the greater effect simply because they are closer.
The Earth's bulge pulls back on the moon, against it's rotation; the Moon's bulge does the same on the Earth. Thus, one can think of the tugs on the bulges as a mechanism for slowing down the rotation of both objects.
Since the Earth is bigger, it will have a greater effect on the Moon. And, in fact, it has slowed down the Moon so that the bulge is locked along the line of centers. If the Moon were spinning faster, the Earth would yank it back into place. If the Moon tried to spin more slowly, the Earth would again yank it back into place.
So, one side of the Moon is "locked" to face the Earth.
Given the effects of the tides, does the moon rotate? Yes. Once per orbit.
So, the tides have slowed down the Earth's rotation. And also, they have slowed down the moon's rotation and "locked" the rotational phase of the moon to the orbital phase.
Is this it? No. The tidal bulges of the Earth still pull on the Moon; but they are unable, any longer, to affect the rotation of the Moon. Now, instead, they end up pulling the entire Moon forward.
Effectively, they are transferring energy from the Earth's rotation to motion of the Moon. So, the tides yank the moon forward and make it move faster.
Now, let's go back to Kepler's laws and think about orbits. Kepler's 2nd law says that orbiting bodies "sweep out equal areas in equal times." So, a planet in an elliptical orbit is sometimes close to the Sun and sometimes far from the Sun. And when it is close, it moves fast and when it is far it moves more slowly. If it were in a perfectly circular orbit, it's speed would never change. Now, Kepler's 3rd law, P^{2 }= a^{3}, says that the period of the orbit depends on the size of the orbit. And the period says "how long" so if we know "how long" and we know the size of the orbit, we know the speed of the orbit. In other words, the size of the orbit determines the speed the object moves in orbit. So, for any given orbit the speeds are predetermined.
For simplicity, pretend that the moon is in a perfectly circular orbit with a radius of 384,0000 km. The speed of the moon in orbit is determined by that size.
Now, suddenly, we yank the moon forward, thereby increasing it's speed.
What is the net effect of this yank on the moon?
Suddenly, the moon is moving too fast for it's orbit. It is as if it has been switched from a circular orbit to an elliptical orbit. Now, it is moving too fast for it's distance so it must rise away from the Earth and slow down. The "yank" point represents the close point of an elliptical orbit. So, the net effect is to increase the size of the moon's orbit, to push the moon away from the Earth.
In 1969, the astronauts placed a mirror on the moon and NASA built a special laser at McDonald Observatory in Texas. The laser was used to bounce a light signal off the lunar mirror. Since the speed of light is constant and well known, the time for the light to travel the round trip to the moon can be used to measure the distance to the moon. And, in fact, the moon is receding from the Earth at a rate of about 3 cm per year!
In 100 years, this amounts to 100 x 3 cm = 300 cm = 3 m.
In 1000 years, this is 30 m.
In 1 million years, this is 1000 x 30 m = 30,000 m = 30 km.
In 1 billion years, this is 1000 x 30 km = 30,000 km.
In 4.5 billion years, this is 4.5 x 30,000 = 135,000 km, which is more than one-third the current distance to the Moon!
So, again, small numbers add up to very large numbers over enormous
amounts of time! Of course, since as the moon gets closer to the
Earth (as we move backwards in time), the tidal forces increase and therefore
all the effects just discussed become larger. Thus, in the past, the rate
at which the moon moved away from the Earth was greater in the past and
our estimate above is an extreme lower limit to how far outwards the moon
has evolved.
The effects of TIDES in our solar system
1. Mercury: spin-orbit coupling
orbital (sidereal) period: 87.969 days
rotation (sidereal) period: 1407.6 hrs = 58.650 days
ratio orbital/rotation periods = 1.4999 = 1.5 = 3 : 2
2a. Moon: spin-orbit coupling
orbital (sidereal) period: 27.3217 days
rotation (sidereal) period: 27.3217 days
ratio orbital/rotation periods = 1.0000 = 1 : 1
2b. Earth: slowing of Earth's rotation
3. moons of Mars: spin-orbit coupling
Note that Phobos is tidally locked in an orbit that is less than one Martian day (which is very close to 24 hours). Thus, Phobos revolves faster than Mars rotates. Consequently, the effects of the tides on Phobos will be dramatically different from the tidal effects on most other moons in the solar system. Why? What will the tidal effects be?
Phobos:4. moons of Jupiter
orbital (sidereal) period: 0.31891 days
rotation (sidereal) period: 0.31891 days
ratio orbital/rotation periods = 1.0000 = 1 : 1Deimos:
orbital (sidereal) period: 1.26244 days
rotation (sidereal) period: 1.26244 days
ratio orbital/rotation periods = 1.0000 = 1 : 1
orbital period of Io: 1.769138 days5. moons of Saturn
rotation period of Io: 1.769138 days
ratio orbital/rotation periods = 1.0000 = 1 : 1tidal heating: the special case of Io and Europa
orbital period of Europa: 3.551181 days
rotation period of Europa: 3.551181 days
ratio orbital/rotation periods = 1.0000 = 1 : 1orbital period of Ganymede: 7.154553 days
rotation period of Ganymede: 7.154553 days
ratio orbital/rotation periods = 1.0000 = 1 : 1orbital period of Callisto: 16.689018 days
rotation period of Callisto: 16.689018 days
ratio orbital/rotation periods = 1.0000 = 1 : 1ratio of orbital periods of Europa : Io = 2.007
ratio of orbital periods of Ganymede: Europa = 2.015
ratio of orbital periods of Callisto: Ganymede = 2.333
ratio of orbital/rotation period of Mimas: 1.0006. moons of Uranus
ratio of orbital/rotation period of Enceladus: 1.000
ratio of orbital/rotation period of Tethys: 1.000
ratio of orbital/rotation period of Dione: 1.000
ratio of orbital/rotation period of Rhea: 1.000
ratio of orbital/rotation period of Iapetus: 1.000
ratio of orbital/rotation period of Epimetheus: 1.000
ratio of orbital/rotation period of Janus: 1.000
ratio of orbital/rotation period of Miranda: 1.0007. moons of Neptune
ratio of orbital/rotation period of Ariel: 1.000
ratio of orbital/rotation period of Umbriel: 1.000
ratio of orbital/rotation period of Titania: 1.000
ratio of orbital/rotation period of Oberon: 1.000
ratio of orbital/rotation period of Triton: 1.0008. the Pluto/Charon system
orbital period of Charon: 6.38725 days
rotation period of Charon: 6.38725 days
rotation period of Pluto: 6.38725 days
Phobos revolves around Mars in less than the time required for Mars to rotate. Thus, the tidal bulge raised on Mars by Phobos is behind the line connecting the centers of Mars and Phobos, whereas normally this bulge is ahead of the line of centers. Consequently, Phobos is pulled backwards - against the direction of its orbit - by the tides, thereby slowing down Phobos and removing energy from its orbit. As a result, Phobos spirals into rather than away from Mars. If we look into the future, we find that Phobos will spiral inwards ultimately crashing into the Martian surface in about 10 million years. On the other hand, if we run this "movie" backwards, Phobos would spiral outwards until it reached synchronous orbit (orbiting in the same amount of time as Mars' rotation period). If it were ever outside of synchronous orbit, Phobos would have evolved outwards; thus, it must have begun its life as a Martian moon no further from Mars than just inside synchronous orbit. We find that Phobos would spiral outwards (going backwards in time) in about 100 million years. This means that Phobos has been orbiting Mars for no more than 100 million years. Given that Mars is about 4500 million years old, this is barely 2% of Mars' lifetime.
This raises some interesting questions. Why, after 4400 million years, did Mars suddenly end up with a Moon in this unusual place. Clearly, Mars did not begin its life as a planet with this Phobos in place, else Phobos would have crashed into Mars 4400 million years ago. Therefore, Phobos was not in orbit around Mars when Mars formed or for the first 4400 million years of Martian history. So where did Phobos come from? Almost certainly, it is a captured asteroid, one that was kicked out of the asteroid belt and came close enough to Mars that Mars could capture it.
This scenario raises another question, one that is much more profound: why did Mars capture this moon in the recent past, just recently enough that it is still around for us to in effect watch it crash into Mars? This must be a very special time in the history of the solar system! But no! We don't believe this statement. It is a violation of the Copernican Principle.
Copernicus, of course, in 1543 asserted that the Earth orbits the Sun, not vica versa. This bold proposition effectively moved the center of the universe from the Earth to the Sun. As a consequence, the Earth becaues merely one of six planets orbiting the Sun. The Earth, as normal planet, is not special and does not occupy a special or privileged place in the universe. This is the Copernican Principle. An extension of the Copernican Principle is to assert that there is nothing special or privileged about this moment in time in the universe. Thus, for us to happen to witness the only captured Moon to spiral into it's demise on the surface of its host planet is a violation of the Copernican Principle. How do we get out of this? We must assert that what we see is commonplace, that asteroids are captured all the time into orbits around planets and that some of them are captured into sub-synchronous orbits, like Phobos.
In fact, we do believe that many of the small moons around the giant
planets are captured asteroids. Quite a few orbit in the wrong direction
(see "R" for retrograde motion on the satellite "Fact Sheets" you can find
on the Surfing the Web page), which is a very strong indication these objects
are captured asteroids (or comets).
The Moons of Jupiter tour
Satellite comparison factsheet
The Galilean moons, side by side
Volcanic eruptions, full moon image
Arizona-sized
eruption on Io
Conceptually, the Roche limit addresses the idea that 1) objects are bound into a whole by the force of gravity but 2) tidal stretching forces can be greater than the internal bonding strength. Consequently, the tides can rip an object apart.
This general concept was first developed by the Frenchman, Edouard Roche, in the 19th century. The distance from a planet at which these two forces are equal is known as the Roche limit. Inside of this distance, moons are unstable and will be ripped apart; beyond this distance, large moons can exist.
The Roche limit equation assumes that the moon has no internal strength other than gravity (no molecular bonds; zero material strength). Think of the moon as a spherical pile of rubble:
r_{roche} = 2.5 (p_{planet} / p_{m} )^{1/3} Ramazingly, the Roche limit depends on the ratio of the densities of the planet and moon, scaled by the radius of the planet. The size (diameter/radius) of the moon is irrelevant! Note that the Roche limit decreases as the density of the moon increases.where
r_{roche} = the Roche limit
p_{planet} = density of the planet
p_{m} = density of the moon
R = radius of the planet
Example calculation:
let the planet be made of rock: p_{planet} = 3.5 gm cm^{-3} = 3500 kg m^{-3}
let the moon be made of ice: p_{m} = 1.0 gm cm^{-3} = 1000 kg m^{-3}
let the planet be Earth. Thenr_{roche} = 2.5 (3500/1000)^{1/3} R_{earth }= 3.8 R_{earth}= 24,000 km
Given this limit, why don't the tides rip apart
the Hubble Space Telescope (orbital elevation 350 km; distance from
center of the Earth = 1.05 R_{Earth}= 6730 km)?
images of rings: Giant Planet tour
Jupiter's Ring