On Friday, April 29, 2022 at 5:21:50 PM UTC-6 Bruce wrote:

> On Sat, Apr 30, 2022 at 9:00 AM John Clark <johnk...@gmail.com> wrote:
>> On Fri, Apr 29, 2022 at 6:30 PM Alan Grayson <agrays...@gmail.com> wrote:
>> * > If m in f = ma, decreases by 50% on every single split, I'm pretty 
>>> sure (but not certain) that planetary orbits would change, making life 
>>> impossible on Earth. g might not change, but G likely will.*
> If by G you mean Newton's universal gravitational constant, then that will 
> not change if all energies are reduced by 50% because it is not an energy. 
> Despite the claims made by some here, the gravitational acceleration at the 
> surface of the earth, g, will certainly change. From the text-books, g = 
> GM/r^2, where G is Newton's constant, M is the mass of the earth, and r its 
> radius. If you reduce M by 50%, then g will also reduce by 50%.

What is the relationship between g and the orbital elements of the Earth, 
or any body entering the solar system and being captured? I once had a 
course in Orbital Mechanics a LONG TIME AGO, and in trying to reconstruct 
what I learned, I think most, possibly all of the orbital elements, are 
purely *accidental*, say for a body entering the solar system and being 
captured; namely, the eccentricity, the ap and perihelions, the location of 
the foci (for a closed orbit) and their separation distance, the 
inclination of the orbit referencing the plane defined by the rotation of 
the Sun, and so forth. All the elements are purely accidental in defining 
the orbit except possibly its mass, and depend on the parameter of the 
incoming vector. But how does the gravitational acceleration at the surface 
of such a body fit into this picture, if indeed it does? AG

> As I've a explained before and will not explain again, If the sun's 
>> gravitational attraction to the earth, f, is reduced by 50% (because the 
>> sun's mass/energy is reduced by 50%) then the earth's inertia must also be 
>> reduced by 50% (because the earth's mass/energy is also reduced by 50%) so 
>> the two changes would cancel out and the earth's orbit would not change by 
>> one nanometer. And for the same reason if you performed an experiment, 
>> after the two changes were made to determine the value of G, the experiment 
>> would look exactly as it did before and so you'd get the same numerical 
>> value for G.
> As I said, Newton's constant. G, is not an energy so it will not change. 
> But gravitational forces depend on the product of the masses involved, not 
> their ratio. So changing the energy will certainly change planetary orbits 
> and most other things gravitational. The argument about inertia is spurious 
> because that will affect only one of the masses, say the mass of the moon, 
> m, in the gravitational attraction between earth and moon. Newton's 
> gravitational law says that the force between the earth and the moon is 
> GMm/r^2, where r is the earth-moon distance, and the other factors are as 
> before. The inertia of the moon will change with m, but the inertia of the 
> earth  will not change with its mass M (because in this simple 
> approximation, the earth is not moving). The moon's mass, m, then cancels 
> out, and the orbital attraction changes according to the change in the 
> Earth's mass, M. So the orbit will necessarily change if the mass M is 
> reduced by 50%. So just as the acceleration at the surface of the earth, g, 
> decreases by the same factor as the energy, the attraction between the 
> earth and the moon decreases by the same factor. You could certainly 
> measure the change in g directly by experiments on the surface of the 
> earth. Or you could watch the earth-moon distance increasing. In either 
> case, the changes induced by reducing all energies by 50% will be very 
> evident.
> Bruce

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