Cap'n Refsmmat

If you're in orbit, you're moving sideways, around the Earth. Gravity is pulling straight down, perpendicular to your velocity. Hence its force does not slow you down. Consider a weight on the end of a string. You can hold the string over your head and get the weight spinning, so that it sweeps out horizontal circles in the air over your head. The string is pulling on the weight constantly, but the weight maintains its speed.

Tell me, if you're in an elevator and the cables are cut, do you stay pinned to the floor of the elevator as it falls, or do you float inside it?

That's because you edited your post to fix it, after I quoted it.

Indeed. (Chanel is only valid when capitalized as the name of the perfume, so it's interesting that you capitalized it)

I think it's silly to correct someone's errors while making errors yourself. Try this: http://theoatmeal.com/comics/misspelling The first one is applicable.

It looks like the WYSIWYG editor will be completely revamped for the next version, so let's hope that improves things. In the mean time, you can go to My Settings and uncheck "Enable visual (RTE) editor?" to prevent it from attempting to outsmart you. I've had to fight with the editor a good bit -- you should see what it took to get LaTeX working.

Then you can discuss the new ideas in Speculations. In the Physics forum, we answer questions with already-accepted physics, not speculative new ideas.

Well, they go together. The mass is converted into energy, so a decrease in mass corresponds to a creation in energy. But to be simple, the energy in two hydrogen nuclei is greater than the energy in one helium nucleus, so the spare energy goes somewhere. You might try to skim this: http://en.wikipedia.org/wiki/Nuclear_fusion http://en.wikipedia.org/wiki/Muphry%27s_law

Could you be slightly less condescending when debating? I'd prefer this thread not degenerate while on its first page.

Yes, imath is the inline math tag, for LaTeX inside blocks of text. The difference: imath: [imath]\frac{1}{a}[/imath] math: [math]\frac{1}{a}[/math]

Hong Kong Central Library is the legal deposit library, and publishers submit copies of every book they publish in Hong Kong to the library. This is a common practice in many countries; for example, the Library of Congress in the United States receives two copies of every book published in the country. Now, I'm not sure if every book submitted will appear in the catalog immediately.

Then please do not bring it up as an answer to mainstream science questions. New hypotheses should be discussed in the Speculations forum.

imath and ce trip different environments in LaTeX; imath does the inline environment, and ce uses mhchem.

[imath]\frac{1}{a} \neq a[/imath], unless [math]a = 1[/math]. No, [imath]\frac{1}{a} = 1 \div a[/imath], which is different. As for the second bit: Perhaps this can be best explained another way. This makes sense, right: [math]\frac{a^5}{a^2} = a^3[/math] because: [math]\frac{a^5}{a^2} = \frac{a\times a\times a\times a\times a}{a \times a}[/math] and you cancel stuff out. So it works something like this: [math]\frac{a^m}{a^n} = a^{m - n}[/math] When you divide exponents, you subtract. Now, we also know that [imath]a^0 = 1[/imath], so you can do this: [math]a^{-m} = a^{0 - m} = \frac{a^0}{a^m} = \frac{1}{a^m}[/math] since that's just the same thing I did before, but in reverse. Is that making sense, or is it too complicated?

"Cancelling" is really division. I mean, here's a concrete example: [math]\frac{4 \times (2 + 1)}{4 \times (7 - 2)} = \frac{2 + 1}{7 - 2}[/math] You can do that because [imath]\frac{4}{4} = 1[/imath], and they divide out. On the other hand: [math]\frac{4 + 2}{1 + 4} \neq \frac{2}{1}[/math] You can see that just be doing the addition. Because division is the opposite of multiplication, you can "cancel" (divide) things that are multiplied. But you can't "cancel" things that are added, because you're not undoing any multiplication.

What swansont is explaining goes something like this: Gravity has a [imath]\frac{1}{r^2}[/imath] relation with distance. That is, the gravitational force experienced by something goes down proportional to the square of the distance. Go twice as far away and the gravity will be one-fourth as strong. Magnetic dipoles, on the other hand, have a [imath]\frac{1}{r^3}[/imath] relation with distance. Go twice as far away and the magnetism will be one-eighth as strong. The strength of the gravitational force has been measured for centuries with the torsion balance: http://en.wikipedia.org/wiki/Torsion_spring#Torsion_balance There actually was a several-story-tall torsion balance in one of the buildings on campus here years ago.

Yeah, they're aliases to each other.

Indeed. You can't do this: [math]\frac{a^2 + 1}{a + 1} \neq \frac{a^2}{a}[/math] Or this: [math]\frac{a^2 + 1}{a + 1} \neq \frac{a + 1}{1}[/math] You can only do something like this: [math]\frac{a(a^2 + 1)}{a(a+1)} = \frac{a^2 + 1}{a+1}[/math] You can only cross out factors that are multiplied. Division is the opposite of multiplication, so [math]a \times a \div a = a[/math]. It's not the case that [math]a \times a \div (a+1) = a[/math].

I don't know of any magnetic fields in this solar system that'd be strong enough to cause significant influence between planets. The Earth's magnetic field is fairly wimpy. This difference in behavior makes it unlikely that gravity is just magnetism. (Also, magnetism can cause electric currents; gravity does not. There's a significant number of behavioral differences.) Now, I haven't watched the entire video you posted, so I can't go through everything he mentions. He mentions a lot of puzzles about the Sun, though, so you might be interested in NASA's STEREO probe: http://www.nasa.gov/mission_pages/stereo/main/index.html

How? Temperature is an average; any individual atom can have a range of energies inside the material. Changing the average doesn't affect that. The only way to constrain an individual atom's energy is to observe it with greater precision than it was observed before.

Your problem is here: 1 - 1 = 0.0000....1 There are an infinite number of 0s that go first, so you can never reach that 1 at the end. It's only 0.999 infinitely repeating that is equal to 1.

Uncertainty isn't required for the first part of my statement. Temperature is average, and some particles will be hotter than others. That's just how temperature works; no uncertainty principle required. Now, I don't think that at low temperatures you know the energy with higher certainty. You'd only know that if you tried to measure an individual atom's kinetic energy.

Hmm, checking the PM, it looks like that's not included. It definitely should be. I'll remember that when I revise it.

No.

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