Re: It feels like Groundhog Day

[John Clark]

On Sun, May 8, 2022 at 8:09 AM Lawrence Crowell <
goldenfieldquaterni...@gmail.com> wrote:

*> I have not been following this. However, your formula for the
> gravitational force should be F = -GMm/r^2, with the acceleration a =
> -GM/r^2. It appears you wrote the equation for the velocity of an orbit.*



You're right of course, thanks for the correction.



John K ClarkSee what's on my new list at  Extropolis


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Re: It feels like Groundhog Day

[Lawrence Crowell]

I have not been following this. However, your formula for the gravitational 
force should be F = -GMm/r^2, with the acceleration a = -GM/r^2. It appears 
you wrote the equation for the velocity of an orbit.

LC

On Friday, May 6, 2022 at 6:36:38 AM UTC-5 johnk...@gmail.com wrote:

> I'm changing the title because I think it's bad form for the title of a 
> thread to contain the name of a list member, and because I really do feel 
> like Bill Murray, we've been over this exact same ground over and over 
> again almost verbatim. I'm (probably foolishly) going to do it one more 
> time:
>
> Suppose you wanted to measure the gravitational constant G, how would you 
> do it? You'd do it the same way Henry Cavendish did it 200 years ago, you'd 
> use Newton's formula F= √(GM/r) where F is the gravitational force of 
> attraction between 2 lead balls of equal mass if you chain one of the balls 
> to the earth's surface and let the other one swing freely. Now you'll need 
> to determine the mass of the balls, and you can do that by noting how fast 
> it accelerates when exposed to a known calibration force, for example a 
> force provided by a precisely made coiled clockspring, but if the energy in 
> the clockspring is half of what it was in Cavendish's day and the inertia 
> of the lead ball is also half of what it was in Cavendish's day then the 
> value of M you will write in your lab notebook will be the same as the 
> value Cavendish wrote in his lab notebook. And the amount of time it takes 
> for the freely swinging ball to hit the stationary ball that you write in 
> your lab notebook will be the same as the time Cavendish found.  So the 
> value of G that you write in your lab notebook will be the same value 
> Cavendish wrote in his.
>
> Newton says the orbital velocity of a planet a distance r from the sun is  v= 
> √ (GM/r) , so if G is the same and M is the same (because the inertial 
> and gravitational mass are always the same) then the orbital speed of a 
> planet is the same, and the solar system would look the same. And because 
> the gravitational acceleration on the surface of the earth g= GM/R^2 
> where M is the mass of the earth and R is the radius of the earth, g would 
> still be 9.8 m/s, and force would still equal mass times acceleration. 
>
> The one thing both Newton and Einstein agreed-upon is that gravitational 
> mass and inertial mass are always exactly the same, and that's why 
> Aristotle was wrong, something twice as heavy does not fall to the ground 
> twice as fast; that's also why even physicist who are adamantly opposed to 
> many worlds and love to badmouth it don't use the argument presented around 
> here that the solar system would become unstable. Some around here are 
> arguing in effect that Aristotle was right after all, an object twice as 
> heavy would fall to the ground twice as fast and we should just ignore 2000 
> years worth of progress in physics. Unless somebody says something new that 
> I haven't already responded to at least twice before (and at this point 
> that seems unlikely) I'm done with this and Groundhog Day is finally over.
>
> John K ClarkSee what's on my new list at  Extropolis 
> 
> dgd
>
>
>

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Re: It feels like Groundhog Day

[Alan Grayson]

On Friday, May 6, 2022 at 12:46:42 PM UTC-6 Alan Grayson wrote:

> On Friday, May 6, 2022 at 8:37:28 AM UTC-6 Alan Grayson wrote:
>
>> On Friday, May 6, 2022 at 5:36:38 AM UTC-6 johnk...@gmail.com wrote:
>>
>>> I'm changing the title because I think it's bad form for the title of a 
>>> thread to contain the name of a list member, and because I really do feel 
>>> like Bill Murray, we've been over this exact same ground over and over 
>>> again almost verbatim. I'm (probably foolishly) going to do it one more 
>>> time:
>>>
>>> Suppose you wanted to measure the gravitational constant G, how would 
>>> you do it? You'd do it the same way Henry Cavendish did it 200 years ago, 
>>> you'd use Newton's formula F= √(GM/r) where F is the gravitational force of 
>>> attraction between 2 lead balls of equal mass if you chain one of the balls 
>>> to the earth's surface and let the other one swing freely. Now you'll need 
>>> to determine the mass of the balls, and you can do that by noting how fast 
>>> it accelerates when exposed to a known calibration force, for example a 
>>> force provided by a precisely made coiled clockspring, but if the energy in 
>>> the clockspring is half of what it was in Cavendish's day and the inertia 
>>> of the lead ball is also half of what it was in Cavendish's day then the 
>>> value of M you will write in your lab notebook will be the same as the 
>>> value Cavendish wrote in his lab notebook. And the amount of time it takes 
>>> for the freely swinging ball to hit the stationary ball that you write 
>>> in your lab notebook will be the same as the time Cavendish found.  So the 
>>> value of G that you write in your lab notebook will be the same value 
>>> Cavendish wrote in his.
>>>
>>> Newton says the orbital velocity of a planet a distance r from the sun 
>>> is  v= √ (GM/r) , so if G is the same and M is the same (because the 
>>> inertial and gravitational mass are always the same) then the orbital speed 
>>> of a planet is the same, and the solar system would look the same. And 
>>> because the gravitational acceleration on the surface of the earth g= 
>>> GM/R^2 where M is the mass of the earth and R is the radius of the earth, g 
>>> would still be 9.8 m/s, and force would still equal mass times 
>>> acceleration. 
>>>
>>> The one thing both Newton and Einstein agreed-upon is that gravitational 
>>> mass and inertial mass are always exactly the same, and that's why 
>>> Aristotle was wrong, something twice as heavy does not fall to the ground 
>>> twice as fast; that's also why even physicist who are adamantly opposed to 
>>> many worlds and love to badmouth it don't use the argument presented around 
>>> here that the solar system would become unstable. Some around here are 
>>> arguing in effect that Aristotle was right after all, an object twice as 
>>> heavy would fall to the ground twice as fast and we should just ignore 2000 
>>> years worth of progress in physics. Unless somebody says something new that 
>>> I haven't already responded to at least twice before (and at this point 
>>> that seems unlikely) I'm done with this and Groundhog Day is finally 
>>> over.
>>>
>>> John K ClarkSee what's on my new list at  Extropolis 
>>> 
>>>
>>
>> Your analysis is in error because you've changed the problem -- which 
>> assumes the energy of a split universe decreases by 50%, where the wf 
>> allows two outcomes of equal probability. THEREFORE, since E = mc^2, the 
>> masses of the Sun and Earth ALSO decrease by 50%. Further, g, the 
>> acceleration of gravity at the surface of the Earth is *irrelevant* to 
>> whether the orbit can be maintained. Clearly, the orbit will be maintained 
>> if the Earth shrinks to any radius, for a given mass. On this latter point, 
>> I think Bruce made the same error. AG
>>
>
> That makes three members who regard your analysis as faulty, but for you 
> it's still groundhog day. Like I wrote (in effect) earlier, you're not 
> someone I would trust when buying a used car. AG 
>

For you it will always be groundhog day. Because, like a Trumper, you are 
unable to admit you're wrong. AG 

>
>>> dgd
>>>
>>>
>>>

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Re: It feels like Groundhog Day

[Brent Meeker]

On 5/6/2022 4:35 AM, John Clark wrote:
I'm changing the title because I think it's bad form for the title of 
a thread to contain the name of a list member, and because I really do 
feel like Bill Murray, we've been over this exact same ground over and 
over again almost verbatim. I'm (probably foolishly) going to do it 
one more time:


Suppose you wanted to measure the gravitational constant G, how would 
you do it? You'd do it the same way Henry Cavendish did it 200 years 
ago, you'd use Newton's formula F= √(GM/r) where F is the 
gravitational force of attraction between 2 lead balls of equal mass 
if you chain one of the balls to the earth's surface and let the other 
one swing freely.


??   F=GMM/r^2

Now you'll need to determine the mass of the balls, and you can do 
that by noting how fast it accelerates when exposed to a known 
calibration force, for example a force provided by a precisely made 
coiled clockspring, but if the energy in the clockspring is half of 
what it was in Cavendish's day and the inertia of the lead ball is 
also half of what it was in Cavendish's day then the value of M you 
will write in your lab notebook will be the same as the value 
Cavendish wrote in his lab notebook. And the amount of time it takes 
for the freely swinging ball to hit the stationary ball that you write 
in your lab notebook will be the same as the time Cavendish found.  So 
the value of G that you write in your lab notebook will be the same 
value Cavendish wrote in his.


So you're equating forces Kx=GMM/r^2 where K is the spring constant in 
Cavendish's experiment and x is the deflection.  Then you halve K and 
the M's.  Then G must double??




Newton says the orbital velocity of a planet a distance r from the sun 
is  v= √ (GM/r) ,



So the orbital velocity at the same radius would stay the same...which 
contrary to my idea that stellar aberration would change.  But now the 
number for G in books has changed.  But wait, there's more.  Distances 
are measured by time*SoL, and the speed of light, time is measured by 
cesium atoms energy levels.  So halving energy will double periods and 
wavelengths and the meter will double.  So if G->2G, M->M/2, and r->r/2  
then v does increase by sqrt(2).


Brent

so if G is the same and M is the same (because the inertial and 
gravitational mass are always the same) then the orbital speed of a 
planet is the same, and the solar system would look the same. And 
because the gravitational acceleration on the surface of the earth g= 
GM/R^2 where M is the mass of the earth and R is the radius of the 
earth, g would still be 9.8 m/s, and force would still equal mass 
times acceleration.


The one thing both Newton and Einstein agreed-upon is that 
gravitational mass and inertial mass are always exactly the same, and 
that's why Aristotle was wrong, something twice as heavy does not fall 
to the ground twice as fast; that's also why even physicist who are 
adamantly opposed to many worlds and love to badmouth it don't use the 
argument presented around here that the solar system would become 
unstable. Some around here are arguing in effect that Aristotle was 
right after all, an object twice as heavy would fall to the ground 
twice as fast and we should just ignore 2000 years worth of progress 
in physics. Unless somebody says something new that I haven't already 
responded to at least twice before (and at this point that seems 
unlikely) I'm done with this andGroundhog Day is finally over.


John K Clark    See what's on my new list at Extropolis 


dgd


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Re: It feels like Groundhog Day

[Alan Grayson]

On Friday, May 6, 2022 at 8:37:28 AM UTC-6 Alan Grayson wrote:

> On Friday, May 6, 2022 at 5:36:38 AM UTC-6 johnk...@gmail.com wrote:
>
>> I'm changing the title because I think it's bad form for the title of a 
>> thread to contain the name of a list member, and because I really do feel 
>> like Bill Murray, we've been over this exact same ground over and over 
>> again almost verbatim. I'm (probably foolishly) going to do it one more 
>> time:
>>
>> Suppose you wanted to measure the gravitational constant G, how would you 
>> do it? You'd do it the same way Henry Cavendish did it 200 years ago, you'd 
>> use Newton's formula F= √(GM/r) where F is the gravitational force of 
>> attraction between 2 lead balls of equal mass if you chain one of the balls 
>> to the earth's surface and let the other one swing freely. Now you'll need 
>> to determine the mass of the balls, and you can do that by noting how fast 
>> it accelerates when exposed to a known calibration force, for example a 
>> force provided by a precisely made coiled clockspring, but if the energy in 
>> the clockspring is half of what it was in Cavendish's day and the inertia 
>> of the lead ball is also half of what it was in Cavendish's day then the 
>> value of M you will write in your lab notebook will be the same as the 
>> value Cavendish wrote in his lab notebook. And the amount of time it takes 
>> for the freely swinging ball to hit the stationary ball that you write 
>> in your lab notebook will be the same as the time Cavendish found.  So the 
>> value of G that you write in your lab notebook will be the same value 
>> Cavendish wrote in his.
>>
>> Newton says the orbital velocity of a planet a distance r from the sun is  
>> v= 
>> √ (GM/r) , so if G is the same and M is the same (because the inertial 
>> and gravitational mass are always the same) then the orbital speed of a 
>> planet is the same, and the solar system would look the same. And because 
>> the gravitational acceleration on the surface of the earth g= GM/R^2 
>> where M is the mass of the earth and R is the radius of the earth, g would 
>> still be 9.8 m/s, and force would still equal mass times acceleration. 
>>
>> The one thing both Newton and Einstein agreed-upon is that gravitational 
>> mass and inertial mass are always exactly the same, and that's why 
>> Aristotle was wrong, something twice as heavy does not fall to the ground 
>> twice as fast; that's also why even physicist who are adamantly opposed to 
>> many worlds and love to badmouth it don't use the argument presented around 
>> here that the solar system would become unstable. Some around here are 
>> arguing in effect that Aristotle was right after all, an object twice as 
>> heavy would fall to the ground twice as fast and we should just ignore 2000 
>> years worth of progress in physics. Unless somebody says something new that 
>> I haven't already responded to at least twice before (and at this point 
>> that seems unlikely) I'm done with this and Groundhog Day is finally 
>> over.
>>
>> John K ClarkSee what's on my new list at  Extropolis 
>> 
>>
>
> Your analysis is in error because you've changed the problem -- which 
> assumes the energy of a split universe decreases by 50%, where the wf 
> allows two outcomes of equal probability. THEREFORE, since E = mc^2, the 
> masses of the Sun and Earth ALSO decrease by 50%. Further, g, the 
> acceleration of gravity at the surface of the Earth is *irrelevant* to 
> whether the orbit can be maintained. Clearly, the orbit will be maintained 
> if the Earth shrinks to any radius, for a given mass. On this latter point, 
> I think Bruce made the same error. AG
>

That makes three members who regard your analysis as faulty, but for you 
it's still groundhog day. Like I wrote (in effect) earlier, you're not 
someone I would trust when buying a used car. AG 

>
>> dgd
>>
>>
>>

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Re: It feels like Groundhog Day

[Alan Grayson]

On Friday, May 6, 2022 at 5:36:38 AM UTC-6 johnk...@gmail.com wrote:

> I'm changing the title because I think it's bad form for the title of a 
> thread to contain the name of a list member, and because I really do feel 
> like Bill Murray, we've been over this exact same ground over and over 
> again almost verbatim. I'm (probably foolishly) going to do it one more 
> time:
>
> Suppose you wanted to measure the gravitational constant G, how would you 
> do it? You'd do it the same way Henry Cavendish did it 200 years ago, you'd 
> use Newton's formula F= √(GM/r) where F is the gravitational force of 
> attraction between 2 lead balls of equal mass if you chain one of the balls 
> to the earth's surface and let the other one swing freely. Now you'll need 
> to determine the mass of the balls, and you can do that by noting how fast 
> it accelerates when exposed to a known calibration force, for example a 
> force provided by a precisely made coiled clockspring, but if the energy in 
> the clockspring is half of what it was in Cavendish's day and the inertia 
> of the lead ball is also half of what it was in Cavendish's day then the 
> value of M you will write in your lab notebook will be the same as the 
> value Cavendish wrote in his lab notebook. And the amount of time it takes 
> for the freely swinging ball to hit the stationary ball that you write in 
> your lab notebook will be the same as the time Cavendish found.  So the 
> value of G that you write in your lab notebook will be the same value 
> Cavendish wrote in his.
>
> Newton says the orbital velocity of a planet a distance r from the sun is  v= 
> √ (GM/r) , so if G is the same and M is the same (because the inertial 
> and gravitational mass are always the same) then the orbital speed of a 
> planet is the same, and the solar system would look the same. And because 
> the gravitational acceleration on the surface of the earth g= GM/R^2 
> where M is the mass of the earth and R is the radius of the earth, g would 
> still be 9.8 m/s, and force would still equal mass times acceleration. 
>
> The one thing both Newton and Einstein agreed-upon is that gravitational 
> mass and inertial mass are always exactly the same, and that's why 
> Aristotle was wrong, something twice as heavy does not fall to the ground 
> twice as fast; that's also why even physicist who are adamantly opposed to 
> many worlds and love to badmouth it don't use the argument presented around 
> here that the solar system would become unstable. Some around here are 
> arguing in effect that Aristotle was right after all, an object twice as 
> heavy would fall to the ground twice as fast and we should just ignore 2000 
> years worth of progress in physics. Unless somebody says something new that 
> I haven't already responded to at least twice before (and at this point 
> that seems unlikely) I'm done with this and Groundhog Day is finally over.
>
> John K ClarkSee what's on my new list at  Extropolis 
> 
>

Your analysis is in error because you've changed the problem -- which 
assumes the energy of a split universe decreases by 50%, where the wf 
allows two outcomes of equal probability. THEREFORE, since E = mc^2, the 
masses of the Sun and Earth ALSO decrease by 50%. Further, g, the 
acceleration of gravity at the surface of the Earth is *irrelevant* to 
whether the orbit can be maintained. Clearly, the orbit will be maintained 
if the Earth shrinks to any radius, for a given mass. On this latter point, 
I think Bruce made the same error. AG

>
> dgd
>
>
>

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