A different take on gravity

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15 years 3 months ago #23804 by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
Aaron: 27 Jul 2009 : 05:05:35<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I loved the McGuyver reference, though. He was great, wasn't he?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">What made McGuyver's solutions memorable to me was not their novelty or ingenuity, but their total disregard of scale. Did you see the episode where he rescued a CIA operative from a dictators prison? He grabbed all the fire extinguishers on his way out of the prison. Hotly pursued by a squadron of heavily armed soldiers, he stopped at an overhanging boulder as big as a house with a crack in it, emptied his canteen into the crack, covered the boulder with CO2 from the fire extinguishers and continued on his way. Seconds later, the boulder breaks in two and falls directly on top of the lead vehicle of his pursuers. We're supposed to believe that the huge boulder froze solid in a matter of seconds. The water expanded as it became ice, completing the crack across the boulder, and then thawed in a matter of seconds. And McGuyver somehow knew exactly how long this would take and exactly when his pursuers would arrive. So, when I call your model a McGuyverism, you should not take that as a complement.

You are guessing that stellar wind from distant stars is pushing Earth toward the sun, and you don't bother to investigate how weak that stellar wind is. You don' t have to be a mathematician to see that the momentum of the stellar wind falls many orders of magnitude short of what is need to cause gravity. I merely looked up "solar wind" in Wikipedia to get a rough idea how hard it pushes the Earth, and it's obvious to me that our own solar wind has more influence on us that the stellar wind, which by the way is pushed aside by our solar wind at the boundary known as the heliopause.


Aaron: 27 Jul 2009 : 05:05:35<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I am not clear which force you are referring to in this section of your post of 23 July 2009 (that I have underlined) in this quoted portion of your post below:

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The pushing force of the stars is minuscule compared to gravity, as evidenced by the fact that planets orbit stars. If you divide the total power of sunlight Earth absorbs by the speed of light, you get the force exerted by sunlight on Earth. According to my calculations, the suns gravity pulls Earth about 14 billion times harder than sunlight pushes Earth. (I may have missed something because I think the ratio might be much greater than that. I also calculate that the <u>tangential component of that force</u> should be sufficient to make us fall into the sun in less than a year; that is obviously wrong.) At our place in the solar system, the suns magnitude is -26.74; thats 13 billion times brighter than Sirius (magnitude -1.46). Even if all the stars, other than the sun, were pushing us from one direction, their combined push would be less than the push of our own suns light. For the push of stars to be significant, they would have to have unseen and undetected energy output billions of times greater than the output that we can detect.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote"><hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Here's what I was trying to express: Sunlight impacts us radially in the reference frame centered on the sun and stationary relative to the stars; but in the reference frame centered on Earth, the sunlight hits us from our port bow, so to speak (looking down from the north). So the force vector has a radial component away from the sun and a tangential component directed opposite our orbital motion. Obviously I miscalculated because, multiplying that tangential force component by the number of seconds in a year, I came up with a number several times greater than our mean orbital speed. If you want to have a go at it, you can find all the necessary values in Wikipedia. I started with the total solar power absorbed by Earth and divided by the speed of light to get the force. The angle at which the light hits us (in our reference frame) is 2 pi radians times 8 minutes/1 year. The tangential component is the force times the sine of that angle. Divide the force by mass of Earth to get the acceleration, and multiply by the number of seconds in a year to get the change in our mean orbital speed per year. I hope you dont get the same result I did, because it means we're about to fall into the sun.

Aaron: 27 Jul 2009 : 05:05:35<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">It sounds like you are saying that my model couldn't be responsible for the amount of gravity that the Earth is experiencing at present using a Newtonian model for calculating the Solar Gravitational Force between the planets and the Sun, but that it could be strong enough to cause the Earth and other planets to "fall" into the sun very rapidly.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The fact that I got the tangential acceleration way too high means that the force is probably way less than what I calculated. Since the force I calculated is way to low to be responsible for gravity, the real force must be waaaay, waaaay to week to be responsible for gravity. And I'm talking about the force exerted by our sun, while you seem to be talking about the force of starlight and distant galaxy light, which is way less than that of sunlight. If you cant demonstrate that starlight is a credible force, then you needn't bother explaining how that force gets turned into gravity.

Aaron: 27 Jul 2009 : 05:05:35<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Personally I would like to see the Gravity effect formulated using just the particles and energies that are familiar and somewhat known to us, without having to invent exotic, far out, imaginative, unsupportable solutions, but I am very open minded if someone actually has a reasonable addition.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">If you understood the math that makes LeSage type gravitons a plausible carrier of gravity, you would see that the familiar particles are totally implausible as carriers of gravitymainly because they are way too slow.

None of us are immune to error. Invite all your buddies to come over and poke fun at me at " Might all forces propagate at speed of gravity? " I was off by 36 orders of magnitude!


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15 years 3 months ago #22985 by Larry Burford
<b>[Philj] "You are guessing that stellar wind from distant stars is pushing Earth toward the sun, and you don't bother to investigate how weak that stellar wind is."</b>

It is a good point, but in Aaron's defense he does say photons are the major contributor to the gravity effect in his model. And the solar photon flux <u>is</u> several orders of magnitude stronger than the solar wind.

Hmmm. I'm beginning to wonder why I don't start floating when I go into my closet and turn off the light. (No windows, so no photons when the lights go out.)

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