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Comment Re:200 Million Yahoo "Users" (Score 2) 169

But apparently the security questions and answers were stored in plain text. That's like locking your front door with a triple lock, a fingerprint reader and iron bars but then leaving the ground floor window wide open with a neon sign "enter here" pointing to it. And then claiming that you take security seriously. And when someone enters, you don't tell anyone for two years because you're afraid your parents will find out.

Comment Re:Screw you (Score 5, Insightful) 133

I "know someone who" wanted to rent a 5 year old movie on iTunes not long ago. He was ready to pay for it. The rights holders, however, had decided that this particular movie was only to be made available for purchase, not rental. More than twice the price of a rental. So guess what he did...

Other example, same guy, rented a movie on iTunes then decided he liked it so much he wanted to purchase it. Do you think they would let him convert the rental into a purchase? Nope, full price on top of rental. So guess what he did...

Bad service turns potential customers into pirates. In both examples above the rights holders missed out on the money someone was willing to spend because they were simply too greedy. It's easy to blame the pirates, though.

Comment Re: It's the Sun, actually (Score 1) 130

Thank you very much for providing an article that proves my point. Too bad you apparently didn't read it yourself, or you would have realised you were wrong and I was right all along.

From that very website:
"The distortion of water and earth that we call "tidal bulges" is the result of deformation of earth and water materials at different places on earth in response to the combined gravitational effects of moon and sun. It is not simply the size of the force of attraction of these bodies at a certain point on earth that determines this. It is the variation of force over the volumes of materials (water and earth) of which the earth is composed."

You know, that variation that you claimed was completely insignificant and could never be strong enough to cause tides? Yes, that very variation turns out to be the cause of the tides after all, just like I wrote it was.

The article goes into more detail than I did: it turns out these differential gravitational forces don't just mean more gravity one one side and less on the other, but also have sideways components due to the change in the direction of gravity. At the poles the resulting force is even towards the surface of the earth. That's something I didn't even know, I would have thought those would be insignificant.

Anyway, the part about "A closer look at centrifugal forces" goes on to debunk your very theory about tides being caused by centrifugal forces.

"So the bottom line is that centrifugal forces on the earth due to the presence of the moon are not tide-raising forces at all. They cannot be invoked as an "explanation" for any tide, on either side of the earth or anywhere else."

Seriously, do me a favor and read the article. All of it. I couldn't explain it better than it does.

Comment Re: It's the Sun, actually (Score 1) 130

Your calculation are wrong byl orders of magnitude. The gravitational pull decreases by the inverse square law. The difference in distance is 60 times. the inverse square law gives a difference of 1/3600 between the near side and far side

Come on, this is basic physics.

Gravity at the surface of the moon (1737.1 km from the center) is 1.62 m/s^2
Gravity at 384400 km from the center of the moon is 1.62 * 1737.1^2 / 384400^2 = 33.08 micron per second squared.
Gravity at 390771 km from the center of the moon is 1.62 * 1737.1^2 / 390771^2 = 32.01 micron per second squared.
Gravity at 378029 km from the center of the moon is 1.62 * 1737.1^2 / 378029^2 = 34.21 micron per second squared.

A difference of 1/60 in the distance does not result in a difference of only 1/3600. Derivatives don't work that way, you are squaring the wrong thing.

Comment Re: It's the Sun, actually (Score 1) 130

Look, I don't have time for this right now. I've got a book about gravity here ("Gravity from the ground up" by Bernard Schutz), which starts from the basics and works up to General Relativity and cosmology in a very scientific, not-dumbed-down way with actual math, and it explains tides exactly like I did. So does the Wikipedia article on tides.

I hate to use authoritative arguments, but I really do have better things to do, sorry. I'm not being paid for this.

Spinning around an axis creates a centrifugal (pseudo-)force, but orbiting by gravity (translation rather than rotation) does not. Otherwise the occupants of the ISS would not feel weightless inside the station but would stick to one side of it like I already explained. The motion of the earth around the center of gravity of the earth-moon system is caused by the attraction of the moon which acts on all of the earth, the only difference between different points being the gravitational gradient. I calculated that tidal gravitational gradient for you in my previous post (about 6% between the two sides, hardly negligible). There's not much more I can do, it really does work that way.

Comment Re: It's the Sun, actually (Score 1) 130

No, I'm sorry, but your theory with centrifugal forces is completely wrong.

First of all, you need to realise that centrifugal forces are not real forces but pseudo-forces that only show up when you are using an accelerating reference frame. If, for example, you take a turn with your car, the tires are making the car turn (providing centripetal force towards the center of the turn) while your body does not receive that force and wants to go straight ahead. This feels like a force that pushes you to the outside of the turn, but that's only because you are using the car as your reference frame. If you take the road as the reference, your body is merely continuing its straight trajectory until it hits the inside wall of the car which is taking a different, curving trajectory.

Now, if the turn is not caused by the friction of the tires but by some sort of gravitational force (like, for example, the ISS orbiting the earth with a centripetal force of almost 1g), you will not feel any centrifugal force. The bodies of the astronauts receive the same gravitational acceleration as the space station so they feel weightless relative to the ISS. If you would spin the ISS at the same distance with a rope instead of the force of gravity, the astronauts would be pinned to the outer side with a centrifugal acceleration of almost 1g.

The earth is indeed orbting the common center of gravity of the moon-earth system (which is inside the earth). If it were orbiting that point because it was attached to it by some fourth-dimensional pin that forcibly made it orbit that point, then yes, you would get centrifugal forces if you took the earth as your reference frame. But this is not the case: it is not orbiting that point because it is attached to it, but rather because the moon is pulling it towards it.

Now here's the contradiction in your theory: IF it were true that the moon was exerting "the same amount of gravity on every cubic mile of the earth with a difference that is completely insignificant", then that would result in no centrifugal force whatsoever. Just like the astronauts feel no centrifugal force in the ISS.

In reality, there IS a significant difference between the moon's attraction on both sides of the earth. A quick back-of-the-envelope calculation, using the moon's average distance of 384400 km, its radius of 1737.1 km, its surface gravity of 1.62 m/s^2 and the earth's radius of 6371 km, gives me about 33 micron/s^2 at the center of the earth, 32 micron/s^2 on the far side of the moon, and 34 micron/s^2 on the near side. That may not seem like much, but since it's a steady force continuously revolving around, it's enough to keep a mass of water moving around the planet.

Of course the water moves horizontally. The water in the tital bulges comes from the sides, where else would it come from? I agree with you on that bit.

And yes, there are local scenarios where particular coastlines and ocean depths change the tides so they happen a bit earlier or later. But that does not change the fact that the major force driving the tides is simply the gradient of the moon's gravitational field.

By the way, gravitational gradients are enough to tear apart asteroids passing too close by Jupiter. You would think that, on an asteroid only a few km across, the difference in attraction to Jupiter on both sides would be completely insignificant. But apparently it's enough to tear rocks apart.

Comment Re: It's the Sun, actually (Score 5, Informative) 130

It has nothing to do with the total amount of gravitational pull.

Tidal stress is the difference between the amounts of gravitational pull at different distances. If you are floating in space near a massive object (planet, star, moon), with your feet pointing towards it, your feet will be closer to it and will therefore experience more attraction, while your head will be further away from it and therefore experience less attraction. This will tend to stretch your body a little bit (or a whole lot if you happen to fall into a black hole). If the gravitational gradient is strong enough, it will make your hair stand up. Your entire body is actually accelerating towards the object, but your feet are being pulled harder while your hair gets left behind. Relative to the pull on your gravitational center, it feels like there are two forces pulling at you from opposite sides trying to tear you apart.

On earth, that means there's a high tide on the side of the moon (closer to the moon, therefore attracted more to it) but also on the other side (further from the moon, therefore attracted less and bulging the other way). The effect is the same on both sides because it's not the amount of attraction that matters, but only the difference with the attraction experienced by the earth's center.

The sun also creates a similar effect, though smaller.

When the sun and the moon are either on the same side (new moon) or on opposite sides (full moon), the two effects are added together and you get spring tide. And, apparently, possibly more or stronger earthquakes.

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