The inverse square law

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21 years 3 weeks ago #6876 by EBTX
Replied by EBTX on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">How does your theory deal with these allegedly fatal objections ...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I shall give them ... but first ... I lost some sleep over the following (actually several hours over two nights ... and I've always said to myself that no idea is worth losing sleep over because it will still be there tomorrow ;o).

From Newton ...

<center><font size="4">(2M x 2M) / r^2 = (M x M) / (r/2)^2</font id="size4"></center>
where M is two different bodies of the same mass. (You don't have subscripts so I didn't use 1 and 2 there)

Let the Ms be two spherical asteroids some distance apart with a pole between to keep them from gravitating together. One end of the pole pins a bathroom scale to one asteroid and reads "100 pounds".
_____________________

Now, we double the mass of the asteroids making sure to keep their centers the same distance apart. The force registered on the bathroom scale should now be "400 pounds".
The total inverse square radiative flux on each mass is doubled. We might then conjecture that 2 x 2 on each mass accounts for the quadrupling of the force registered on the bathroom scale.
_______________________

Now we return to the original setup and this time leave the masses alone and put those bodies at half the original distance taking care to keep their centers at that half distance. Then the bathroom scale once again reads "400 pounds".
The total inverse square radiative flux on each mass is quadrupled. If we apply the same logic as above we must conjecture that 4 x 4 on each mass accounts for a 16 fold increase in force (1600 pounds) in contradiction with the 400 which should be registered on the bathroom scale as per Newton.
_____________________

So, all inverse square radiative flux theories of gravitation must then be false or, at best, rendered problematic.

Q.E.D.





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21 years 3 weeks ago #6877 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by EBTX</i>
<br />Let the Ms be two spherical asteroids some distance apart with a pole between to keep them from gravitating together. One end of the pole pins a bathroom scale to one asteroid and reads "100 pounds".<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

As I tried to explain, force and weight are arbitrary and require definition. You personally will weigh a different amount on every planet or moon you visit. In Earth orbit, you are weightless yet falling in a strong gravitational field. You are using "weight" as if it meant "mass" when it does not. If you stick to acceleration and mass, which are always well-defined, no paradoxes will arise.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If we apply the same logic as above we must conjecture that 4 x 4 on each mass accounts for a 16 fold increase in force (1600 pounds) in contradiction with the 400 which should be registered on the bathroom scale as per Newton.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

The "same logic as above" was that "400 pounds" meant the same as 2 x 2 on each mass <i>at the original distance</i>. If you don't mix up mass and weight, you can't get confused in this step. Weight can be anything without violating Newton's law. It even depends on horizontal speed, as when orbiting. Mass is invariant. -|Tom|-

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21 years 3 weeks ago #6984 by 1234567890
Replied by 1234567890 on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by EBTX</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">How does your theory deal with these allegedly fatal objections ...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I shall give them ... but first ... I lost some sleep over the following (actually several hours over two nights ... and I've always said to myself that no idea is worth losing sleep over because it will still be there tomorrow ;o).

From Newton ...

<center><font size="4">(2M x 2M) / r^2 = (M x M) / (r/2)^2</font id="size4"></center>
where M is two different bodies of the same mass. (You don't have subscripts so I didn't use 1 and 2 there)

Let the Ms be two spherical asteroids some distance apart with a pole between to keep them from gravitating together. One end of the pole pins a bathroom scale to one asteroid and reads "100 pounds".
_____________________

Now, we double the mass of the asteroids making sure to keep their centers the same distance apart. The force registered on the bathroom scale should now be "400 pounds".
The total inverse square radiative flux on each mass is doubled. We might then conjecture that 2 x 2 on each mass accounts for the quadrupling of the force registered on the bathroom scale.
_______________________

Now we return to the original setup and this time leave the masses alone and put those bodies at half the original distance taking care to keep their centers at that half distance. Then the bathroom scale once again reads "400 pounds".
The total inverse square radiative flux on each mass is quadrupled. If we apply the same logic as above we must conjecture that 4 x 4 on each mass accounts for a 16 fold increase in force (1600 pounds) in contradiction with the 400 which should be registered on the bathroom scale as per Newton.
_____________________

So, all inverse square radiative flux theories of gravitation must then be false or, at best, rendered problematic.

Q.E.D.






<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Yes, and an equivalent way to look at it would be to
consider M1 and M2 different from one another in Newton's
gravitational force equation. Say the total mass
of the system is 10 (i.e. M1 + M2 = 10). Now, this mass
can be distributed as 4 and 6 ; 2 and 8; 3 and 7; etc. So
the total mass of the system adds up to 10 but when you multiply
the masses, you get a different total each time for the square area of the force.

The flux of gravitational force (or energy) is not conserved in Newton's Force equation because multiplication of masses is used
instead of addition.

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21 years 3 weeks ago #7443 by EBTX
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<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If you don't mix up mass and weight, you can't get confused in this step. <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
So ... you agree that the scale would read 400 pounds in each case?

If so, we could replace the "force" of gravity with a little rocket at 400 pounds thrust on each of our masses. The acceleration produced should then be twice as great on the small mass as on the doubled mass.

But ... the flux on the smaller mass is 4 times greater on the small mass than on the original mass. And ... it's the flux which is the "rocket". This is the bone of contention. Do you see what I mean?

The problem is the flux at four fold increase. If flux just caused a genuine acceleration, it would be four times greater in the small mass than the doubled mass with a doubled flux propelling it ... not just twice the acceleration.

Or, look at it this way. You have four, 100 pound thrusters on the small mass in the halved distance scenario ... and ... two, 100 pound thrusters on the doubled mass in its scenario ... so the smaller mass will accelerate at four times the rate of the doubled mass.

Thus, an inverse square flux cannot be the "effective cause" of gravity.
_______________

Or ... are you saying that the scales in each case would read something different? If so, would this not constitute a clear violation of Newton's laws?
___________________

And I agree, it is confusing. The closer you look at something, the more it tends to drip through your fingers. I am open to enlightenment ;o)

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21 years 3 weeks ago #6880 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by EBTX</i>
<br />So ... you agree that the scale would read 400 pounds in each case?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">As I understand your setup, which is not well, I agree.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If so, we could replace the "force" of gravity with a little rocket at 400 pounds thrust on each of our masses. The acceleration produced should then be twice as great on the small mass as on the doubled mass.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">No, 200 pounds on each mass for a total of 400 pounds. The total weight measured by the scale should logocally depend on both masses, not just one of them; but it depends on how you choose to define "weight", which is ambiguous until specified. "Weight" normally has meaning only with respect to some one particular body. But if either one of the asteroids were hollow and nearly massless, the reading for the combined forces would drop from 400 to 200 pounds total.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">But ... the flux on the smaller mass is 4 times greater on the small mass than on the original mass. And ... it's the flux which is the "rocket". This is the bone of contention. Do you see what I mean?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Not really. But with the preceding correction, maybe it will make sense to you. If not, please try to explain your setup again, with special attention to definitions of force and weight, which are ambiguous here. I've never thought about gravity in terms of weight before because it requires a combination of a gravitational force and a non-gravitational force (e.g., a spring in the scales), and requires one to define "weight" for each application. That can get tricky because different laws apply to the non-gravitational force, and because there is no universal menaing to "weight".

If you use Newton's gravitational acceleration law, you can't go wrong because only one mass at a time is involved.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The problem is the flux at four fold increase. If flux just caused a genuine acceleration, it would be four times greater in the small mass than the doubled mass with a doubled flux propelling it ... not just twice the acceleration.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I don't see why, but feel I may now have completely lost contact with your setup. One double mass means double acceleration for the other mass with no change in its own. And vice versa. So if both masses double, that gives four times the acceleration. What can be ambiguous about that?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Or, look at it this way. You have four, 100 pound thrusters on the small mass in the halved distance scenario ... and ... two, 100 pound thrusters on the doubled mass in its scenario ... so the smaller mass will accelerate at four times the rate of the doubled mass.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I'm not following you at all now. I can't even guess what you might mean. Is there a typo in this last paragraph? If you meant "double" where you said "four times" in the last line, I might agree, but wonder what the question was.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Thus, an inverse square flux cannot be the "effective cause" of gravity.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">My failure to understand this is surely derivative from my failure to understand the preceding. But don't you have an inate sense of discomfort when you are challenging something that millions of minds before you have thought hard about over centuries and found no fault with? You are still entitled to have your questions answered. It's the presumption in your writing that you have just discovered soemthing that millions have thought hard about and not noticed before that I find ... oblivious.

Clear, unambiguous procedure:

* acceleration of body m is a1 = GM/r^2
* acceleration of body M is a2 = Gm/r^2
* total relative acceleration is a1 + a2
* total relative force depends on definition
* weight depends on definition

These last two depend on whether you want the one-body force or weight (relative to which body?), or the two-body force or weight. Define your terms carefully and you can go there. Otrherwise, be safe and stick with masses and accelerations as everyone else does and no ambiguities will ever arise. -|Tom|-

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21 years 3 weeks ago #6995 by EBTX
Replied by EBTX on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">But don't you have an inate sense of discomfort when you are challenging something that millions of minds before you have thought hard about over centuries and found no fault with?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
No, not at all. I don't have to worry about my reputation or credibility since I have not got any to start with ;o)

What I am proposing is simple logical raplacement of gravity with rocket propulsion which the scale cannot identify by any means whatsoever.

If you say that a mass of 1 unit produces shadowing effect of 1 unit ... then won't two masses produce 2 units of shadowing effect? Is this not correct by your theory? If it produces 4 units of shadowing, I would indeed be wrong ... but how so? The shadowing effect seems fairly straightforward to me and appears to be additive.

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