Guest Post by Willis Eschenbach
Short Post. You can skip this if you understand the tidal force. Some folks seem confused about the nature of tidal forces. Today I saw this gem: “The tide raising force acts in both directions (bulge on each side in the simplistic model)” … the author of that statement may understand the tides, but that isn’t how tidal forces work, they don’t act in different directions.
To clarify the situation, let me give an example. Imagine three one-kilogram masses M1, M2, and M3, connected by a blue rope, that are in free-fall directly into the sun. Figure 1 shows the situation.
Figure 1. Forces on three one-kilogram masses M1—M3, which are in free-fall straight towards the sun, and are connected by the blue rope. The equations at the bottom show the force of gravity on each of the three masses. The “tidal force” is the difference between two gravitational forces on two objects. Solar tidal forces on a 1-kg mass at the earth’s surface average about half a micronewton (µN, or 10-6 newtons), and the lunar tidal forces are about twice that, about a micronewton. By comparison, Earth’s gravity exerts a force of about 10 newtons on a 1-kg mass at the surface …
Clearly, the gravitational force on M1 is greater than the force on M2, which in turn is greater than the force on M3. The difference between those two gravitational forces is the tidal force. There are two different tidal forces of interest in the diagram, which are GF1 – GF2, and GF2 – GF3. As a result of these tidal forces, as the three masses fall towards the sun. the near mass M1 moves the furthest, the far mass M3 moves the least, and the blue rope is always under tension.
For tidal forces on a theoretical planet, just imagine that the surface of the planet is the gray dotted line, and that “r” is the radius of the planet. The tidal force is the difference between the force of gravity on the unit mass M2 at the center of the planet, and the forces of gravity on the unit masses M1 and M3 at the points nearest and furthest from the sun.
Calculating the Tidal Forces
As shown in Figure 1, to calculate the tidal force you can just subtract the gravitational force on the unit mass at the center of the planet from the force on the other unit mass, for example GF1 – GF2. So we can use the equations in Figure 1 to calculate the force.
However, when the distance D is much, much larger than the radius r, we can make some simplifying assumptions. First, we can assume that the force between M1 and M2 is the same as the force between M2 and M3. It’s not, as you can see from the equations … but it is not a large error. For tides from the moon, the individual tidal forces are about ±5% different from their average value.
The next simplifying assumption we can make if D is much larger than r is that the average tidal force can be calculated as
Tidal Force = 2 * G * sunmass * r / D^3
This leads to tiny errors with respect to the sun’s average tidal force on the earth, and slightly larger errors with respect to the moon. For the sun we have
Near side true tidal force on earth (GF1 – GF2) = 0.50590 µN
Far side true force (GF2 – GF3) = 0.50583
True average = 0.50587
Approximation = 0.50587
So the approximation of the average tidal force is good to five significant digits …
And for the moon we have:
Near side true tidal force on earth (GF1 – GF2) = 1.1350 µN
Far side true force (GF2 – GF3) = 1.0796
True average = 1.1073
Approximation = 1.1067
For the moon, the approximation of the average is good to four significant digits. For most results, of course, those differences are far too small to be meaningful, and we can use the approximation without concern.
Why Are There Two Tidal Bulges?
Moving on, let’s look at why there are tidal bulges on both the near and far sides. It’s not because the tidal force acts in different directions, because tidal force is always directed towards the sun. So lets look at Figure 1 again, and this time we’ll include a planet and an ocean. As in Figure 1, the planet is simply free-falling into the sun. That result is shown in Figure 2:
Figure 2. Tidal forces elongating the planet and the ocean. Note that the planet is elongated as well, but this is not shown in the diagram because obviously, tides in the solid planet are much smaller than tides in the ocean. NOTE THAT THIS PLANET IS NOT THE EARTH.
As the planet free-falls into the sun, the near side of the ocean is pulled the most. So it moves the most, pulling away from the planet. And the planet, in turn is pulled more than the far side, so it pulls away from the ocean. These relative motions create the “bulges”
In other words, the “bulges” on the two sides of the planet are simply a result of the tidal forces stretching the entire system. Indeed, the atmosphere is subject to the same phenomenon, so there are atmospheric tides as well. In prior centuries, these atmospheric tides were said to affect the weather, although in a cyclical rather than a secular manner. However, such connections, while certainly possible, have proven to be elusive and difficult to establish. One complicating factor is that the atmospheric tides that are caused by the sun occur in combinations with solar heating, which makes the atmosphere swell. Also, the variations in pressure caused by the atmospheric tides are small, on the order of a tenth of a millibar (or of a hectopascal), or a variation of only 0.01% …
Finally, I’ve said that the situation in Figure 2 shows that planet free-falling into the sun. As it gets closer and closer, the tidal forces stretching the system get greater and greater, stretching the system more and more. Eventually, at a distance called the “Roche limit”, the tidal forces can reach such strength that they rip the planet into pieces … dang, to have a ring-side seat right on the Roche limit line for that spectacle would be awesome, even if it were perhaps a bit warm …
My best to all,
w.
PS—As with gravity, every mass in the universe exerts a tidal force on every other mass, so you can personally take credit for raising a tide on the surface of the sun … yeah, it’s a small tide, but it’s real. Just sayin’, I had someone claim the opposite. Lots of tidal misinformation out there.
THE USUAL: If you disagree with me or anyone, please quote exactly what it is you disagree with, and tell us precisely why you disagree. That way, we can understand exactly what it is you object to.
stevefitzpatrick says:
February 15, 2014 at 2:26 pm
“OK you have convinced me; you will never understand this relatively simple problem.”
It is too simple. Don’t you realize that? Do you really think I would not understand your very simple setup?
But, it bears no relationship to reality. This is a more subtle problem than you yet apprehend. A lot of newbies make the same mistake as you do here.
Mike M says:
February 15, 2014 at 1:52 pm
This is trickier to answer than you may think
Imagine that you could have a big rocket engine mounted in the sky with the exhaust upwards from the Earth. Imagine that you started this engine when the exhaust pointed in the orbital “forward” direction. You should then expect the Earth was slowing down, right?
Well, it will not slow down, it will fall closer too the Sun and the orbital speed will increase. So if you try to brake the orbital velocity of the Earth the speed will increase.
The only way to slow down the orbital velocity of the Earth is to push it forward. The Earth will then move further away from the Sun and the velocity will counter-intuitively slow down.
This is not only a theoretical possibility, it actually happens each time we send a rocket away from the Earth. If for instance a rocket is sent away with destination Mars, it will be sent in the orbital “forward” direction, and the recoil will then make the Earth fall a little bit closer to the Sun and the speed will increase a little bit
To sum up: Closer orbits means higher velocity, but we have to brake the orbital objects to speed them up.
/ Jan
So why is the orbital speed of Earth faster at perihelion? Or is that something different due to an elliptical orbit?
RichardLH says: Indeed you do as I explained above. Less energy = LOWER orbit = higher speed. Strange isn’t it?
It is NOT a higher speed AT the same orbital radius – that’s the WHOLE point!
Tom in Florida says:
February 15, 2014 at 2:57 pm
So why is the orbital speed of Earth faster at perihelion? Or is that something different due to an elliptical orbit?
Because the closer you are to the Sun, the faster you have to move [‘sideways’] to stay in orbit.
stevefitzpatrick says:
February 15, 2014 at 2:26 pm
“OK you have convinced me; you will never understand this relatively simple problem.”
Try these for an explanation which is very similar to Wills’s work.
http://co-ops.nos.noaa.gov/restles3.html
http://www.lhup.edu/~dsimanek/scenario/tides.htm
“It is important to note that the centrifugal force produced by the daily rotation of the earth on it axis must be completely disregarded in tidal theory. This element plays no part in the establishment of the differential tide-producing forces.”
The forces that arise from moving around the barycentre are tiny compared to those. In there as described above but swamped by all the rest.
Jan Kjetil Andersen says February 15, 2014 at 2:56 pm ” This is trickier to answer than you may think”
—
Mike M says:
February 15, 2014 at 1:52 pm
My ORIGINAL question: If earth’s orbital velocity around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Answer the question, A or B?
—
Not ‘tricky’ at all …in the context of my explanation of a mass M on the equator going faster at midnight or slower at noon with respect to earth’s orbital velocity – the answer is A…. PERIOD.
The mass experiences more net force toward the sun at noon and less at midnight by virtue of the difference in its orbital velocity as imparted by the rotation of the earth. (The difference in gravity itself by difference of one earth’s diameter versus 93 million miles is miniscule as WIllis already calculated.)
Mike M says:
February 15, 2014 at 2:58 pm
“It is NOT a higher speed AT the same orbital radius – that’s the WHOLE point!”
But it is impossible to stay at the same radius with lower radial velocity. In this universe anyway.
As you move closer and closer to the Earth’s surface you will get faster and faster as you do. Like a skater bringing in their arms in a spin.
As the Moon gains energy from the Earth by the friction on the tides, it is moving further and further away as Lief says. More potential energy = less radial velocity = larger orbital distances.
Mike M says:
February 15, 2014 at 3:08 pm
“My ORIGINAL question: If earth’s orbital velocity around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Answer the question, A or B?”
If it has a lower orbital radial velocity (and therefore higher potential energy to be stable) it will be B.
If it has a higher orbital radial velocity (and therefore lower potential energy to be stable) it will be A.
As noted in the link above, you fire a forward thruster to loose height but gain orbital speed and vice versa.
We can think of the of the Sun and earth or the earth and moon being stationary at small point in time would the earth not have a bulge nearest the moon and furthest from the moon at this small instant of time.I think it would.
I just dropped in to see what condition my condition was in.
I considered the gravitational force between two bodies and the centrifugal force of one rotating around another with a fluid surface of the smaller. This is where the discussion seemed to move to even though Willis’ original post was the attraction of a non orbiting body to another.
The previous discussion has me concerned that I may not know sh*t from apple butter. I will not cook breakfast at home tomorrow but will instead eat @ur momisugly a fast food restaurant w/ no apple butter.
Therefore my condition is unknown by me at this moment in time.
LOL
————-
Sorry mods, I post this comment on another thread by accident. My condition seems to be deteriorating.
RichardLH says: As noted in the link above, you fire a forward thruster to loose height but gain orbital speed and vice versa.
No.. not INITIALLY! Initially your velocity HAS TO slow down because you are DEcelerating! Yes, it speeds back up as you get lower and gravity increases BUT per my tidal assertion, you DO NOT get lower – you are staying within an earth radius of earth’s orbit around the sun.
The question about what will happen if a satellite “slows down” or “fires retro-rockets” is ill-posed. There are two different interpretations with two very different answers. Let’s assume the orbit is circular of radius “r1” to start with.
1) If on-board rockets are fired backwards briefly, the rocket will slow at that point in the orbit. It will stop orbiting in the circular orbit and orbit in an elliptical orbit with a semimajor axis r2 < r1. The time it takes to orbit will become less. It will have less overall energy, less average gravitational potential energy, and more average kinetic energy.
2) If on-board rockets are fired forwards briefly, the rocket will speed at that point in the orbit. If on-board rockets are fired forwards AGAIN briefly when the satellite reaches apogee so that it takes on a new circular orbit with r2 > r1, then the new orbit will have more average gravitational potential energy, and less average kinetic energy. The average speed over the whole orbit will decrease. (Rather counter-intuitive that your can fire the rockets forward TWICE and still end up with less KE!)
Basically, you have to know if you are forcing the satellite back into a circular orbit or not.
Interesting bit of trivia — for circular orbits, the potential energy is always twice the magnitude and opposite in sign to the kinetic energy.
RichardLH says: But it is impossible to stay at the same radius with lower radial velocity. In this universe anyway.
No – not when you’re stuck to earth by 10 m/sec^2 gravity it isn’t impossible.
It is fun to try to wrap (or would it be warp) my mind around all these ideas.
But something keeps nagging me…
Why is the rope blue ?
Tim Folkerts says: February 15, 2014 at 3:35 pm
The question about what will happen if a satellite “slows down” or “fires retro-rockets” is ill-posed. There are two different interpretations with two very different answers. Let’s assume the orbit is circular of radius “r1″ to start with.
1) If on-board rockets are fired backwards briefly, the rocket will slow at that point in the orbit.
…………………
STOP! Right there! … that’s all I was ever talking about!
Mike M says:
February 15, 2014 at 3:38 pm
STOP! Right there! … that’s all I was ever talking about!
and what has that to do with the tides that are around all the time.
Mike M says:
February 15, 2014 at 3:32 pm
“No.. not INITIALLY! Initially your velocity HAS TO slow down because you are DEcelerating! Yes, it speeds back up as you get lower and gravity increases BUT per my tidal assertion, you DO NOT get lower – you are staying within an earth radius of earth’s orbit around the sun.”
You seem to have a very strange view of the universe. It takes a while to catch up with what you do.
Unfortunately in the real universe, if you fire the thruster for a millisecond or an hour. the rate of change will happen instantaneously and so will velocity/height/etc.
If you can’t manage going down. Think the other way instead. As your rocket engine – now pointing along the orbit – pushes you forwards, you will slow down in forward radial velocity as you rise in height. The energy has gone into potential energy for height above the surface and thus lost from radial forward velocity.
Below stable orbital radial velocity you will of course get faster as you rise to it. It is what happens from then we are talking about.
lsvalgaard says:
February 15, 2014 at 3:41 pm
“and what has that to do with the tides that are around all the time.”
Nothing. But people don’t get how to manoeuvre in space at all. Most of the games have it wrong as well.
RichardLH says: “…will happen instantaneously and so will velocity/height/etc.”
Nope. You fire the rocket, your velocity increases as the rocket fires and THEN you start going up. As you go up, OVER TIME, your velocity then decreases until your reach a higher orbit at a lower velocity than that at which you started. You do NOT “instantaneously” change orbit altitude!
Mike M says:
February 15, 2014 at 3:51 pm
“Nope. You fire the rocket, your velocity increases as the rocket fires and THEN you start going up. As you go up, OVER TIME, your velocity then decreases until your reach a higher orbit at a lower velocity than that at which you started. You do NOT “instantaneously” change orbit altitude!”
Fire it for a millisecond in an hour and you will still see a change in orbit. Or are all the guys who fly in orbit wrong as well?
Mr. Eschenbach–
” In prior centuries, these atmospheric tides were said to affect the weather, although in a cyclical rather than a secular manner.” I am confused (or, Kung-Fu-sed, as my son’s karate teacher likes to say): I have never used “secular” in such a context as you have here. What does it mean here? (If you respond to this, thank you.)
William Larson says:
February 15, 2014 at 3:56 pm
I have never used “secular” in such a context as you have here. What does it mean here?
From the prayers IN OMNIA SAECULA SAECULORUM meaning [roughly] ‘for ever and ever’, thus ‘long-term’
Mike M says:
February 15, 2014 at 3:36 pm
“No – not when you’re stuck to earth by 10 m/sec^2 gravity it isn’t impossible.”
But you’re not in free orbit at that point. If you were to try and stay in free orbit just above the surface (1cm) you will need to have an airless planet and a very flat planet to achieve it. The required orbital velocity is just too great.
u.k.(us) says:
February 15, 2014 at 3:37 pm
“It is fun to try to wrap (or would it be warp) my mind around all these ideas.
But something keeps nagging me…
Why is the rope blue ?”
Cheapest on offer at K-Mart?