On 4/29/2022 10:55 PM, Alan Grayson wrote:
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 <[email protected]> wrote:
On Fri, Apr 29, 2022 at 6:30 PM Alan Grayson
<[email protected]> 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
A body entering the solar system has enough energy to leave the solar
system, unless it collides with something. So it's orbit will be an
hyperbola. It won't be captured.
Brent
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
--
You received this message because you are subscribed to the Google
Groups "Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send
an email to [email protected].
To view this discussion on the web visit
https://groups.google.com/d/msgid/everything-list/9b1f8f9b-3379-4c3e-9614-26f957331977n%40googlegroups.com
<https://groups.google.com/d/msgid/everything-list/9b1f8f9b-3379-4c3e-9614-26f957331977n%40googlegroups.com?utm_medium=email&utm_source=footer>.
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To view this discussion on the web visit
https://groups.google.com/d/msgid/everything-list/36cfca64-fc94-7673-711e-759e97dca08a%40gmail.com.
