Undeniable wrote in message
m...
Uncle Al wrote in message
...
Grok wrote:

Some of us computer scientists were sitting around, bored, and decided
to nudge the Earth towards the Sun. It'd be convenient if it would
hit within our lifetime, so we could finish at least one project on
time and under budget in our lifetimes!

I'm not a physicist, nor did I take enough math or physics to figure
this out, so would like your help.

My guessing says we have to slow the Earth's rotation so that its
gravitational acceleration to the Sun overtakes it's angular momentum,
allowing us to smack into the big one.

How much force is required to slow us down enough so that the big
splashdown is within 50 years?

Boy, are you ever muddled. The Earth masses 5.9742x10^27 grams. Its
average orbital acceleration around the sun (at 1 AU or 499.004782
light-seconds) is 0.593008 cm/sec^2. Stop it cold in its orbit, F=ma,
and it falls into the sun nice as you please. Was that so hard? Go
figure out how long a non-orbiting Earth requires to fall from here to
the sun and get back to us with your calculations,

[snip]

Let's see. It's been a long time I played with planet but...

I think Uncle Al that if you just stop the motion of the earth it's
total mechanical energy E will still be negative, E0 , due to just
the potential of the sun and it will again attain an elliptical orbit
around the sun but with some other radius and we will fry anyway.

All E 0 orbits are elliptical, including that for L = 0. By what
means will the Earth "again attain an elliptical orbit around the Sun"?
[Old Man]

Uncle Al wrote in message ...
DarkMatter wrote:

On Fri, 12 Sep 2003 11:18:59 -0400, "Greg Neill"
Gave us:

"Undeniable" wrote in message
om...

Let's see. It's been a long time I played with planet but...

I think Uncle Al that if you just stop the motion of the earth it's
total mechanical energy E will still be negative, E0 , due to just
the potential of the sun and it will again attain an elliptical orbit
around the sun but with some other radius and we will fry anyway.

Nope. It will be in a degenerate straight-line orbit
intersecting the Sun.

Hey, when you drop something on the Earth, does it go into
an elliptical orbit instead of hitting the ground?

Depends on how close you are to the spheroid, dip****.

No it does not. Stuff with zero orbital velocity falls straight down
vs. the fixed stars from any arbitrary height (ignoring wind) If you
look locally you subtract out ground motion and allow for ~1 cm/sec
horizontal Coriolus acceleration varying with latitude.

That's correct and I guess unless one is specific and uses the right
terms, like Uncle AL does, misunderstandings can arise.

The key word here is "fixed stars". If the answer is posed in this
context it's correct. However, in my answer I assumed an instantenuous
relative fix of earth's velocity with respect to a moving sun. This
case is a bit more complicated. Actually, everything is in a constant
relative motion and fixing anything could be hard. This is also one of
the reasons that in practice there is no such a thing as a "straight
line path" and what one perceives as a straight line in a free fall is
actually an arc of some curve.

This is to say, for those with a little hearing problem, that is there
is accelative motion the free falling planet towards the sun will also
attain a non-zero orbital velocity. This automatically activates conic
sections as the solutions and it's not a degenerate line orbit any
longer.

Have an idea: let's try it.

"Undeniable" wrote in message
...
Uncle Al wrote in message
...


No it does not. Stuff with zero orbital velocity falls straight down
vs. the fixed stars from any arbitrary height (ignoring wind) If you
look locally you subtract out ground motion and allow for ~1 cm/sec
horizontal Coriolus acceleration varying with latitude.

That's correct and I guess unless one is specific and uses the right
terms, like Uncle AL does, misunderstandings can arise.

The key word here is "fixed stars". If the answer is posed in this
context it's correct. However, in my answer I assumed an instantenuous
relative fix of earth's velocity with respect to a moving sun. This
case is a bit more complicated. Actually, everything is in a constant
relative motion and fixing anything could be hard.

Nah. Velocity may be relative but acceleration is absolute.
One need only adjust the circular velocity to result in a zero
centrifugal acceleration component. This is identical to
making the angular momentum zero and the degenerate straight-
line orbit then necessarily obtains.

This is also one of
the reasons that in practice there is no such a thing as a "straight
line path" and what one perceives as a straight line in a free fall is
actually an arc of some curve.

This depends entirely upon the local gravitational field and
motion of the bodies involved. One could easily propose to
do the experiment on a non-rotating planet and obtain straight
line results.

Now you'd better be careful and define what you mean by
a straight line in curved space.


This is to say, for those with a little hearing problem, that is there
is accelative motion the free falling planet towards the sun will also
attain a non-zero orbital velocity. This automatically activates conic
sections as the solutions and it's not a degenerate line orbit any
longer.

You are supposing that the experimenter has incompetently left
a residual angular momentum. As shown above, this is not a
necessary feature.


Have an idea: let's try it.

We await your detailed proposal with the greatest anticipation.
Particularly the bit about how to go about generating the
required thrust.

"Greg Neill" wrote in message ...
"Undeniable" wrote in message
...
Uncle Al wrote in message
...


No it does not. Stuff with zero orbital velocity falls straight down
vs. the fixed stars from any arbitrary height (ignoring wind) If you
look locally you subtract out ground motion and allow for ~1 cm/sec
horizontal Coriolus acceleration varying with latitude.

That's correct and I guess unless one is specific and uses the right
terms, like Uncle AL does, misunderstandings can arise.

The key word here is "fixed stars". If the answer is posed in this
context it's correct. However, in my answer I assumed an instantenuous
relative fix of earth's velocity with respect to a moving sun. This
case is a bit more complicated. Actually, everything is in a constant
relative motion and fixing anything could be hard.

Nah. Velocity may be relative but acceleration is absolute.
One need only adjust the circular velocity to result in a zero
centrifugal acceleration component. This is identical to
making the angular momentum zero and the degenerate straight-
line orbit then necessarily obtains.

Wait a second. Where did you find out that all acceleration is
absolute? In my view this is only due to an idealized situation where
in a local experiment the effect of other matter accelerating in the
universe is null and there is no difference in accelerating a particel
with +a or -a. This problem is still unsolved I think. Newtonian
mechanics and Newton himself adhere to absolute rotation and
acceleretion. But if you ever heard of Mach's principle, there are
other opinions that only relative acceleration is only significant.
This is also the perspective Einstein took.


This is also one of
the reasons that in practice there is no such a thing as a "straight
line path" and what one perceives as a straight line in a free fall is
actually an arc of some curve.

This depends entirely upon the local gravitational field and
motion of the bodies involved. One could easily propose to
do the experiment on a non-rotating planet and obtain straight
line results.

Just go ahead and propose one. I'm all ears. You simply talking about
a no susceptibility to experimental investigation, a common fallacious
appeal.


Now you'd better be careful and define what you mean by
a straight line in curved space.


This is to say, for those with a little hearing problem, that is there
is accelative motion the free falling planet towards the sun will also
attain a non-zero orbital velocity. This automatically activates conic
sections as the solutions and it's not a degenerate line orbit any
longer.

You are supposing that the experimenter has incompetently left
a residual angular momentum. As shown above, this is not a
necessary feature.


Above,nothing was shown but only a postulation was made that
acceleration is absolute , which in my opinion is total nonsense. You
just performed a thought experiment with no bearing to reality.


[snip]

On 13 Sep 2003 02:51:28 -0700, (Undeniable)
wrote:

Uncle Al wrote in message ...
DarkMatter wrote:

On Fri, 12 Sep 2003 11:18:59 -0400, "Greg Neill"
Gave us:

"Undeniable" wrote in message
om...

Let's see. It's been a long time I played with planet but...

I think Uncle Al that if you just stop the motion of the earth it's
total mechanical energy E will still be negative, E0 , due to just
the potential of the sun and it will again attain an elliptical orbit
around the sun but with some other radius and we will fry anyway.

Nope. It will be in a degenerate straight-line orbit
intersecting the Sun.

Hey, when you drop something on the Earth, does it go into
an elliptical orbit instead of hitting the ground?

Depends on how close you are to the spheroid, dip****.

No it does not. Stuff with zero orbital velocity falls straight down
vs. the fixed stars from any arbitrary height (ignoring wind) If you
look locally you subtract out ground motion and allow for ~1 cm/sec
horizontal Coriolus acceleration varying with latitude.

That's correct and I guess unless one is specific and uses the right
terms, like Uncle AL does, misunderstandings can arise.

The key word here is "fixed stars". If the answer is posed in this
context it's correct. However, in my answer I assumed an instantenuous
relative fix of earth's velocity with respect to a moving sun. This
case is a bit more complicated. Actually, everything is in a constant
relative motion and fixing anything could be hard. This is also one of
the reasons that in practice there is no such a thing as a "straight
line path" and what one perceives as a straight line in a free fall is
actually an arc of some curve.

This is to say, for those with a little hearing problem, that is there
is accelative motion the free falling planet towards the sun will also
attain a non-zero orbital velocity. This automatically activates conic
sections as the solutions and it's not a degenerate line orbit any
longer.

Have an idea: let's try it.

But where's the new angular momentum coming from?

"DarkMatter" wrote in message
...
On Fri, 12 Sep 2003 23:29:38 -0400, "Greg Neill"
Gave us:


DarkMatter earns a place in the hall of shame for failing
to think before spewing, and then being rude about it.

Clearly, if the planet has zero angular momentum about
the primary and no radial velocity, it must subsequently
fall straight in.

One wonders if Darkmatter (Dull Grey Matter?) has any clue
at all. What is this nonsense about proximity to some
spheroid?



Dumb****. Drop from ten feet on a 26,000 mile spheroid.

Straight fall.

Drop from 150 miles out, and observe.

Fire from 150 miles out, slightly off the line that points toward
the center of the spheroid, and observe.

Fire on a line that misses the spheroid completely, but very close
and observe.

My remark that it depends on how close one is to the spheroid is
exactly that.

Down here, ON THE surface, it would be very hard to observe an
influence for an off axis drop. Even a mile up

From the 150 mile out mark, however, it would be quite easy.

Does it make more sense now, the previously posted remark?

It should.

DarkMatter has made one thing clear; He's an idiot.

DarkMatter cannot tell the difference between dropping something
in a spherical gravitational potential, wherein the acceleration
vector is necessarily directed at precisely the center of mass,
and imparting an initial non-central velocity.

Conclusion:
DarkMatter = Doesn't Matter.

"Undeniable" wrote in message
...

This is to say, for those with a little hearing problem, that is there
is accelative motion the free falling planet towards the sun will also
attain a non-zero orbital velocity.

Why?

My hearing is fine. I think there is a problem with your writing.

"Undeniable" wrote in message
m...

Above,nothing was shown but only a postulation was made that
acceleration is absolute , which in my opinion is total nonsense. You
just performed a thought experiment with no bearing to reality.

So when do you expect your letter from the Nobel committee?

Undeniable wrote in message
m...
"Greg Neill" wrote in message
...
"Undeniable" wrote in message
...
Uncle Al wrote in message
...


No it does not. Stuff with zero orbital velocity falls straight
down
vs. the fixed stars from any arbitrary height (ignoring wind) If
you
look locally you subtract out ground motion and allow for ~1 cm/sec
horizontal Coriolus acceleration varying with latitude.

That's correct and I guess unless one is specific and uses the right
terms, like Uncle AL does, misunderstandings can arise.

The key word here is "fixed stars". If the answer is posed in this
context it's correct. However, in my answer I assumed an instantenuous
relative fix of earth's velocity with respect to a moving sun. This
case is a bit more complicated. Actually, everything is in a constant
relative motion and fixing anything could be hard.

Nah. Velocity may be relative but acceleration is absolute.
One need only adjust the circular velocity to result in a zero
centrifugal acceleration component. This is identical to
making the angular momentum zero and the degenerate straight-
line orbit then necessarily obtains.

Wait a second. Where did you find out that all acceleration is
absolute? In my view this is only due to an idealized situation where
in a local experiment the effect of other matter accelerating in the
universe is null and there is no difference in accelerating a particel
with +a or -a. This problem is still unsolved I think. Newtonian
mechanics and Newton himself adhere to absolute rotation and
acceleretion. But if you ever heard of Mach's principle, there are
other opinions that only relative acceleration is only significant.
This is also the perspective Einstein took.

Undeniable's opinion doesn't have a single empirical leg to stand-on.
Mach's "principle" is not empirically falsifiable. It doesn't even rate
as a theory. Rotation rate is self-referential and absolute; Foucault
pendulum; laser ring gyroscope. Reference to the "fixed stars" is not
necessary. Such references involve non-local measurements whereof
the laws of physics are not guaranteed to be invariant. According to
Einstein and GTR, measurement of deviation of an observer's reference
frame from a gravitational geodesic (free-fall) is self-referential and
absolute. The statement that acceleration is absolute is empirically
falsifiable and is consistent with all observations. [Old Man]

"DarkMatter" wrote in message
...
On Sat, 13 Sep 2003 14:08:05 -0400, "Greg Neill"
Gave us:

DarkMatter has made one thing clear; He's an idiot.

Ahhh...

DarkMatter cannot tell the difference between dropping something
in a spherical gravitational potential, wherein the acceleration
vector is necessarily directed at precisely the center of mass,
and imparting an initial non-central velocity.

A non-central velocity such as that which would occur if an orbiting
object were to have its path disrupted? What is being simulated here,
Chucko?

It is postulated that the planet's orbital velocity is changed
so as to result in zero angular momentum about the Sun. That
is to say, it's orbital velocity is eliminated. This situation
is indistinguishible from dropping an object straight down to
the Sun from the height of the Earth's orbit.

DarkMatter is utterly confused when he imagines that the mechanism
employed to bring the planet to a halt prior to releasing it for
its decent can have any bearing on the subsequent dynamics.


Conclusion:
DarkMatter = Doesn't Matter.

Try again.

The conclusion holds. DarkMatter has failed to impress.