Sensei

F=m*a That's correct formula for force, which any kid in the first class of physics in primary school is learning. How do we know mass of object? Because we can put it on weighting scale, and read value. Old-school weighting scale was comparison of two bodies. One with known weight. With one with unknown (analog weighting scale). How do we know acceleration I showed you post #17 http://www.scienceforums.net/topic/85792-gravity/?p=828512 Acceleration at the ground of Earth is [latex]a=\frac{M*G}{r^2}=9.81 \frac{m}{s^2}[/latex] M - mass of Earth G - gravitational constant r - radius of Earth How do we know mass of Earth? We can f.e. estimate average density [latex]p = \frac{M}{V}[/latex] density p, is mass divided by volume volume of Earth is [latex]V = \frac{4}{3}*\pi*r^3[/latex] For r = 6371000 meters radius [latex]V = \frac{4}{3}*\pi*6371000^3 = 1.083*10^{21} m^3[/latex] Density of Earth is approximately [latex]5514 \frac{kg}{m^3}[/latex] so [latex]M = p*V = 5514 \frac{kg}{m^3} * 1.083*10^{21} m^3 = 5.972*10^{24} kg[/latex] Calculate acceleration [latex]a=\frac{M*G}{r^2}=\frac{5.972*10^{24} kg * 6.67*10^{-11}\frac{N*m^2}{kg^2}}{6371000^2} = 9.81 \frac{m}{s^2}[/latex]

Relative, you have no basic knowledge about units. See couple the first posts of this thread and learn from it. ps. Where you have been whole primary school, when kids were learning "what is Force"? etc. kind of things.. If you multiply m*a*G, you certainly won't get Newtons unit..

Mars doesn't have magnetosphere. Does it also not have gravity? What with rock meteors? What with ice comets?

Remote desert? It's just 8 km in straight line from Nile.. Currently the first green trees are 100 meters from Pyramids.

String.substring() returns middle part of string. If length is missing it's probably going to the end of string. lastChar is of course last character, last letter, of the string. That code was checking whether it's dot. psn is variable name. Shortcut from position-index inside of string.

In that case you should learn how to measure 9.81 m/s^2 acceleration at home doing experiment. I have showed whole experiment in this thread: http://www.scienceforums.net/topic/84336-if-pi-ratio-was-squared-and-98-mss-how-would-this-change-the-whole-of-science/?p=815752 Camera and cable to computer and some video editing application will make easier calculations.

Entropy?! s is second, unit of time...

Force by definition has unit Newton, which is [latex]kg\frac{m}{s^2}[/latex]

I bet you have no idea how to calculate (measure) from scratch [latex]g=9.81 \frac{m}{s^2}[/latex].... Show me that I am wrong..

I am just interpreting what you wrote in image, and showing you that there is basic unit mismatch.

M has unit kg P has unit N/m^2 kg/(N/m^2) doesn't match N..

Imagine such device: Each block of rock used to build has average 2.5 tons. We can lift it up on little "boat", place rock on it, surround it by wood walls, fill container with water, and block is 1 level up. It's moved to it's place. Water is released. Boat goes down. New block of rock is placed on boat, and it's filled by water again, and cycle is repeated. We can imagine row of such water-lifters, around whole pyramid. Obviously when one level is finished, new lifters have to be build on above level. We would have to pump water to the top most lifter. Egyptians were using pumps since ever to irrigate grain fields, so technology is present. Water released from lifter in above level, can flow to below lifter, so it's reused, and less water have to be pumped.

I am telling this to you for months... Here: Basically you select some text, and press that button, and you have quoted text... Everybody on this forum is using it hundred times per day.. Except you.

Have you bothered reading Rosetta Stone article, I provided? There are 3 languages used on it, and more or less, the same content.

What a nonsense you are writing.. How we would know names of kings and pharaohs without being able to identify symbols.. ? Have you heard about Rosetta stone? http://en.wikipedia.org/wiki/Rosetta_Stone

You could alternatively take broken/not used old electronics, like video player, amplifier etc. and find needed parts in it.

From going to electronic shop for breadboard and electronic elements?

Hawking radiation is hypothetical.

Do you realize it would have to violate either Baryon number conservation and Lepton number conservation?

Neutrinos are taken into account. See below "Sum of energies". I am adding just once 0.42 MeV, assuming 50% of energy is taken by neutrino. 73.46% Hydrogen and 24.85% Helium from photosphere are pretty accurate measurements (similar content has Jupiter), as they're obtained from spectral lines of light from the Sun. What is inside is prediction from models, and will remain this way. After all we won't send there any device to check it experimentally.

That's true it's unlikely to collide. But as I showed at the beginning of post, it must happen [latex]9.17*10^{37}[/latex] times per second anyway. Otherwise Earth would be able to get 1367 J of energy per meter^2 per second. Helium-4 fusion is even more unlikely to happen, because it requires 3 atoms at the same time (Be-8 has more mass-energy than 2 He-4 alone, and additionally decays to 2 He-4 + 92 keV, very rapidly): [latex]_2^4He + _2^4He + _2^4He \rightarrow _6^{12}C + \gamma + 7.275 MeV[/latex] More likely is fusion between He-3 and He-4 (if your suggestion about larger amount of He-3 in core is true): [latex]_2^3He + _2^4He \rightarrow _4^{7}Be + \gamma + 1.5861 MeV[/latex] Beryllium-7 is unstable and decays to Lithium-7 via electron capture: [latex]_4^7Be + e^- \rightarrow _3^7Li + V_e + 0.861893 MeV[/latex] This is exactly what is triggering Chlorine based neutrino detector (neutrinos from proton-proton have up to 0.42 MeV too low to trigger this kind of detector). True. We're ashes remaining from older generation of stars. He-3 can probably only two (or three) ways to be created: by decay from Tritium, or by fusion between proton and Deuterium.

I didn't calculate from whole Sun. But multiplied by 0.34 (34%) first (calculated 60% from just core). 0.34 * 0.6 = 0.204 Yes, I am aware of no convection (according to theory/model). But I don't care about convection in this equation at all.. If you start reaction with x particles (no matter if it's star or chemical reaction at home/lab), and then it's growing to x+y after t seconds. x remain constant. y/t is rate of production. x is initial amount of Helium x+y is current amount of Helium Meaningless amount of He-4 fused further (according to wiki CNO cycle is 0.8% of energy source at the moment). I have no idea. But 2 He-3 would fuse quickly to He-4 and two protons at the beginning of life of star, so it might be meaningless.. That's why it's so rare isotope (1.34(3)×10^−6)

Hello! Each [latex]m^2[/latex] of surface of Earth is receiving 1367 W (ignoring atmosphere influence). We know inverse-square law: [latex]P = \frac{P_0}{4*\pi*r^2}[/latex] After reversing it, we're receiving: [latex]P_0 = P*4*\pi*r^2[/latex] Distance to the Sun is approximately [latex]r = 150*10^9[/latex] meters. After plugging numbers we get: [latex]P_0 = 1367*4*\pi*(150*10^9)^2=3.8651*10^{26} W[/latex] As you can see this pretty much agree with wikipedia Sun page http://en.wikipedia.org/wiki/Sun (Luminosity on the right table) It's energy Sun is emitting every single second. One of possible fusion path is: Fusion of 4 protons to 2 Deuterium: [latex]p^+ + p^+ \rightarrow D^+ + e^+ + V_e + 0.42 MeV[/latex] (neutrino takes part of energy) [latex]p^+ + p^+ \rightarrow D^+ + e^+ + V_e + 0.42 MeV[/latex] [latex]e^+ + e^- \rightarrow \gamma + \gamma + 1.022 MeV[/latex] [latex]e^+ + e^- \rightarrow \gamma + \gamma + 1.022 MeV[/latex] Fusion of 2 Deuterium to 2 Helium-3: [latex]p^+ + D^+ \rightarrow _2^3He + \gamma + 5.49 MeV[/latex] [latex]p^+ + D^+ \rightarrow _2^3He + \gamma + 5.49 MeV[/latex] Fusion of 2 Helium-3 to 1 Helium-4: [latex]_2^3He + _2^3He \rightarrow _2^4He + p^+ + p^+ + 12.86 MeV[/latex] Sum of energies: E = 0.42 MeV + 1.022 MeV + 1.022 MeV + 5.49 MeV + 5.49 MeV + 12.86 MeV = 26.304 MeV Conversion to Joules: [latex]26.304 MeV * 10^6 * 1.602*10^{-19} = 4.2139008*10^{-12} J[/latex] Divide energy Sun has to emit by single fusion cycle to get quantity of reactions per second: [latex]\frac{3.8651*10^{26} W}{4.2139008*10^{-12} J} = 9.17*10^{37}s^{-1}[/latex] Additionally we can calculate quantity of neutrinos: [latex]9.17*10^{37}*2 = 1.8344*10^{38}[/latex] [latex]\frac{1.8344*10^{38}}{4*\pi*(150*10^9)^2}=6.488*10^{14} m^{-2}=64.88 \frac{bln}{cm^2}[/latex] Which also nicely fits with wiki page neutrino flux 65 bln. http://en.wikipedia.org/wiki/Neutrino Mass of the Sun is [latex]1.98855*10^{30} kg[/latex] Single atom of Hydrogen has mass = [latex]\frac{\frac{1.007825}{1000}}{6.022141*10^{23}}=1.6735*10^{-27} kg[/latex] Single atom of Helium has mass = [latex]\frac{\frac{4.0026}{1000}}{6.022141*10^{23}}=6.6465*10^{-27} kg[/latex] If we know mass of object, and what it's made of, we can calculate quantity of atoms. According to this website: http://en.wikipedia.org/wiki/Solar_core Sun's core has 34% of mass of the Sun. And it's made of Helium in ~60%, and ~40% Hydrogen. Phrase "so the innermost portion of the Sun is now roughly 60% helium" can be found on net and Sun wiki page http://en.wikipedia.org/wiki/Sun [latex]1.98855*10^{30} kg * 0.34 * 0.4 = 2.7*10^{29} kg / mass_H = 1.616*10^{56}[/latex] Hydrogen atoms [latex]1.98855*10^{30} kg * 0.34 * 0.6 = 4.0566*10^{29} kg / mass_{He} = 6.1*10^{55}[/latex] Helium atoms If core has 34% of mass then in outer regions of the Sun there will be: [latex]1.98855*10^{30} kg * 0.66 * 0.7346 = 9.6412*10^{29} kg / mass_H = 5.761*10^{56}[/latex] Hydrogen atoms [latex]1.98855*10^{30} kg * 0.66 * 0.2485 = 3.2614*10^{29} kg / mass_{He} = 4.907*10^{55}[/latex] Helium atoms In total we can find now [latex]6.1*10^{55}+4.907*10^{55}=1.101*10^{56}[/latex] Helium atoms Initial quantity of Helium atoms according to theories was 27.4%, so: [latex]1.98855*10^{30} kg * 0.274 = 8.1978*10^{55}[/latex] Helium atoms Time needed to fuse can be calculated by f.e.: [latex] t = \frac{Q_{current}-Q_{init}}{rate}[/latex] (Q - quantity) [latex] t = \frac{1.101*10^{56}-8.1978*10^{55}}{9.17*10^{37}}=3.07*10^{17} seconds = 9.7[/latex] bln years Why so large discrepancy from 4.56 bln years? Where is error? Abundance of Helium-4 in the core is lower than 60% as all materials are suggesting? And this graph would make Sun even more older than that: The smaller luminosity (power), the smaller amount of Helium-4 produced per second, and star must be even older.. Faster rate of production = more energy emitted to cosmos = vaporization of planets.. ps. In older version of calc, I was taking into account also lost of mass due to flares etc. but don't want to be boring you too much..

You misunderstood my objections. Photon, or kinetic energy of particles is causing gravitation also. Gravitation doesn't disappear, just because unstable isotope decayed, as long as energy is still there. Gamma photon will be absorbed by other particles near it.

You're treating word "son" quite literally. These statements are perfectly fine (according to science): "We're sons and daughters of the Sun" "We're children of the Milky Way galaxy" Our bodies are ashes of dead supernova star. "We're children of the God" (if we treat God = Universe = Energy of the all particles)