microgravity - I stand corrected

Maybe it would help to be explicit that these gravity wells have two
spatial dimensions that correspond to the two dimensional orbit, and
the third dimension of the gravity well is not spatial at all. The
height of the well is ENERGY. The zero point of energy is the
"stretched rubber sheet" prior to "deformation". This totally flat
sheet is a representation of ZERO GRAVITY. That is actual zero
gravity, as in no gravity at all to warp the spatial dimensions.

Then when a massive body is introduced into this space, the sheet
deforms and produces a gravity well. Notice that nowhere on the sheet
will you find zero gravity any more for any finite distance from this
mass.

So the energy well has stretched below the zero point. Any satellite
with negative total energy will be captured by this body. But a
thrusting maneuver can increase kinetic energy, raising its energy,
depicted as height above the rubber sheet. If that kinetic energy
increases to the height of the original undeformed sheet (the point of
zero total energy) then you have a parabolic orbit. If the kinetic
energy increases to the point of excess total energy (positive) then
you have a hyperbolic trajectory.


~ CT


Stuf4 wrote:
From sal:

Just to be nit-picky, I thought I'd point out that, if the Earth looks
like a point mass and gravity is Newtonian, then the path _is_ a true
parabola _if_ you launch the projectile with just enough energy to escape.

So, the issue isn't really that it's only a parabola for _small_ powder
loads. It's that it's only a parabola for one particular whopping _big_
powder load.

The proof's a bit tedious, and is part of the reason Newton got so
famous :-) . FWIW, here's my version, which I just put up, mostly 'cause I
was so pleased at actually getting through all the details:

http://www.physicsinsights.org/orbit_shapes_1.html

Those are some very messy equations you have there on that page. An
extremely clean way to analyze escape vs capture is graphically with an
energy potential well diagram. It's just like those coin wells, except
that it visualizes the energy of the orbiting body in terms of kinetic
energy being the height above the potential well surface. Energy is
conserved so the total (sum of potential + kinetic) stays constant. So
the height of the trajectory doesn't change.

As the spacecraft ventures farther from the primary body, kinetic
energy decreases while potential increases. So the three classes of
orbits become:

- Spacecraft kinetic energy is insufficient to reach the top of the
well (ELLIPTICAL CLASS),
- Spacecraft kinetic energy *exactly* reaches the top of the well
(PARABOLIC CLASS),
- Spacecraft kinetic energy exceeds the top of the well (HYPERBOLIC
CLASS).

The concept can be clearly described with pretty pictures and no messy
equations.


~ CT

On Sun, 19 Nov 2006 21:19:09 -0800, Stuf4 wrote:

From sal:

Just to be nit-picky, I thought I'd point out that, if the Earth looks
like a point mass and gravity is Newtonian, then the path _is_ a true
parabola _if_ you launch the projectile with just enough energy to
escape.

So, the issue isn't really that it's only a parabola for _small_ powder
loads. It's that it's only a parabola for one particular whopping _big_
powder load.

The proof's a bit tedious, and is part of the reason Newton got so
famous :-) . FWIW, here's my version, which I just put up, mostly 'cause
I was so pleased at actually getting through all the details:

http://www.physicsinsights.org/orbit_shapes_1.html

Those are some very messy equations you have there on that page. An
extremely clean way to analyze escape vs capture is graphically with an
energy potential well diagram.

Oh, absolutely. That's the way one normally does it, for sure; you are
correct.

However, the point of the page was to actually solve the equations to find
the orbit shapes, not just the escape velocity.

In particular, you can easily find the escape velocity with an energy
argument -- but can you prove the trajectory at escape velocity is a
parabola that way, and an ellipse if you miss escaping but have too much
energy for a circle? AFAIK that proof is a mess no matter how you tackle
it -- but I'd be glad to learn otherwise.


It's just like those coin wells, except
that it visualizes the energy of the orbiting body in terms of kinetic
energy being the height above the potential well surface. Energy is
conserved so the total (sum of potential + kinetic) stays constant. So
the height of the trajectory doesn't change.

As the spacecraft ventures farther from the primary body, kinetic energy
decreases while potential increases. So the three classes of orbits
become:

- Spacecraft kinetic energy is insufficient to reach the top of the well
(ELLIPTICAL CLASS),

But how do you prove it's really an ellipse, without the mess?

- Spacecraft kinetic energy *exactly* reaches the top of the well
(PARABOLIC CLASS),

How do you prove it's a parabola, without the mess of solving the
equations and verifying the form of the result?

- Spacecraft kinetic energy exceeds the top of the well (HYPERBOLIC
CLASS).

And again, how do you prove it's really a hyperbola, short of solving the
equations of motion?


The concept can be clearly described with pretty pictures and no messy
equations.


~ CT

--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org

It's also interesting to analyze what happens with the velocity-energy
relationships as the spacecraft ventures out to infinite distance. For
the hyperbolic case, you can easily determine what the final velocity
will be. This is the velocity corresponding to its excess energy
(height above the zero point of the rubber sheet).

For the parabolic case, the potential well diagram shows that the
farther the spacecraft gets from the primary body, the closer its
kinetic energy gets to zero. So the farther it gets, the slower it
gets. The ramification of this trend toward zero velocity as it
approaches infinite distance is that the trajectory will also take
infinite time.


....of course, this analysis is only two body theory. An extremely
clinical version of the universe. There are many other factors as to
what actually happens to spacecraft like the Voyagers and such.


~ CT


Stuf4 wrote:
Maybe it would help to be explicit that these gravity wells have two
spatial dimensions that correspond to the two dimensional orbit, and
the third dimension of the gravity well is not spatial at all. The
height of the well is ENERGY. The zero point of energy is the
"stretched rubber sheet" prior to "deformation". This totally flat
sheet is a representation of ZERO GRAVITY. That is actual zero
gravity, as in no gravity at all to warp the spatial dimensions.

Then when a massive body is introduced into this space, the sheet
deforms and produces a gravity well. Notice that nowhere on the sheet
will you find zero gravity any more for any finite distance from this
mass.

So the energy well has stretched below the zero point. Any satellite
with negative total energy will be captured by this body. But a
thrusting maneuver can increase kinetic energy, raising its energy,
depicted as height above the rubber sheet. If that kinetic energy
increases to the height of the original undeformed sheet (the point of
zero total energy) then you have a parabolic orbit. If the kinetic
energy increases to the point of excess total energy (positive) then
you have a hyperbolic trajectory.


~ CT


Stuf4 wrote:
From sal:

Just to be nit-picky, I thought I'd point out that, if the Earth looks
like a point mass and gravity is Newtonian, then the path _is_ a true
parabola _if_ you launch the projectile with just enough energy to escape.

So, the issue isn't really that it's only a parabola for _small_ powder
loads. It's that it's only a parabola for one particular whopping _big_
powder load.

The proof's a bit tedious, and is part of the reason Newton got so
famous :-) . FWIW, here's my version, which I just put up, mostly 'cause I
was so pleased at actually getting through all the details:

http://www.physicsinsights.org/orbit_shapes_1.html

Those are some very messy equations you have there on that page. An
extremely clean way to analyze escape vs capture is graphically with an
energy potential well diagram. It's just like those coin wells, except
that it visualizes the energy of the orbiting body in terms of kinetic
energy being the height above the potential well surface. Energy is
conserved so the total (sum of potential + kinetic) stays constant. So
the height of the trajectory doesn't change.

As the spacecraft ventures farther from the primary body, kinetic
energy decreases while potential increases. So the three classes of
orbits become:

- Spacecraft kinetic energy is insufficient to reach the top of the
well (ELLIPTICAL CLASS),
- Spacecraft kinetic energy *exactly* reaches the top of the well
(PARABOLIC CLASS),
- Spacecraft kinetic energy exceeds the top of the well (HYPERBOLIC
CLASS).

The concept can be clearly described with pretty pictures and no messy
equations.


~ CT

By the way, yesterday I was out at Walmart with Joe Engle, in case
anyone was interested.


~ CT

From sal:
On Sun, 19 Nov 2006 21:19:09 -0800, Stuf4 wrote:
snip
Those are some very messy equations you have there on that page. An
extremely clean way to analyze escape vs capture is graphically with an
energy potential well diagram.

Oh, absolutely. That's the way one normally does it, for sure; you are
correct.

However, the point of the page was to actually solve the equations to find
the orbit shapes, not just the escape velocity.

In particular, you can easily find the escape velocity with an energy
argument -- but can you prove the trajectory at escape velocity is a
parabola that way, and an ellipse if you miss escaping but have too much
energy for a circle? AFAIK that proof is a mess no matter how you tackle
it -- but I'd be glad to learn otherwise.


Agreed. I'm not aware of any way to derive the trajectory geometry
from the energy analysis. Thanks for clearing up for me what you're
accomplishing with those detailed equations.


~ CT








It's just like those coin wells, except
that it visualizes the energy of the orbiting body in terms of kinetic
energy being the height above the potential well surface. Energy is
conserved so the total (sum of potential + kinetic) stays constant. So
the height of the trajectory doesn't change.

As the spacecraft ventures farther from the primary body, kinetic energy
decreases while potential increases. So the three classes of orbits
become:

- Spacecraft kinetic energy is insufficient to reach the top of the well
(ELLIPTICAL CLASS),

But how do you prove it's really an ellipse, without the mess?

- Spacecraft kinetic energy *exactly* reaches the top of the well
(PARABOLIC CLASS),

How do you prove it's a parabola, without the mess of solving the
equations and verifying the form of the result?

- Spacecraft kinetic energy exceeds the top of the well (HYPERBOLIC
CLASS).

And again, how do you prove it's really a hyperbola, short of solving the
equations of motion?


The concept can be clearly described with pretty pictures and no messy
equations.

From sal:
On Sun, 19 Nov 2006 21:19:09 -0800, Stuf4 wrote:

From sal:

Just to be nit-picky,
snip

I can be also contacted through http://www.physicsinsights.org

Hey, sal. Out of curiosity I poked around your website a bit. Looks
like some neat stuff there. If you're open to constructive feedback, I
saw that on this page:
http://www.physicsinsights.org/lagrange_1.html

....you've got Newton's Second Law as F=ma.

With as exacting as you have shown to be, I would expect that you'd
cringe as much as I do whenever that kind of oversimplification to the
point of inaccuracy is done. It is cited about as commonly as the term
"zero gravity" is used, but hardly as egregious. Heh.


~ CT

On Mon, 20 Nov 2006 00:07:36 -0800, Stuf4 wrote:

From sal:
On Sun, 19 Nov 2006 21:19:09 -0800, Stuf4 wrote:

From sal:

Just to be nit-picky,
snip

I can be also contacted through http://www.physicsinsights.org

Hey, sal. Out of curiosity I poked around your website a bit. Looks like
some neat stuff there. If you're open to constructive feedback, I saw
that on this page:
http://www.physicsinsights.org/lagrange_1.html

...you've got Newton's Second Law as F=ma.

With as exacting as you have shown to be, I would expect that you'd cringe
as much as I do whenever that kind of oversimplification to the point of
inaccuracy is done. It is cited about as commonly as the term "zero
gravity" is used, but hardly as egregious. Heh.

Errr .... I wrote down Newton's laws on that page from memory, didn't
double check 'em. Time for a big "Ooops", maybe?

Gotta run off right now but I'll check that out later today. Tx.



~ CT

--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org

On Mon, 20 Nov 2006 00:07:36 -0800, Stuf4 wrote:

From sal:
On Sun, 19 Nov 2006 21:19:09 -0800, Stuf4 wrote:

From sal:

Just to be nit-picky,
snip

I can be also contacted through http://www.physicsinsights.org

Hey, sal. Out of curiosity I poked around your website a bit. Looks like
some neat stuff there. If you're open to constructive feedback, I saw
that on this page:
http://www.physicsinsights.org/lagrange_1.html

...you've got Newton's Second Law as F=ma.

Ah, hmm... yeah, I looked that over again. I should rewrite that section.

The problem is one of definition, of course. Newton's original, though
written in Latin rather than calculus, translates pretty directly into
f=ma, but, of course, the more useful form is f=dP/dt since it
generalizes to nonconstant mass. The problem with the latter is how we
define P.

The definition which is probably closest to correct is "That vector
quantity which is conserved and which behaves like mv in simple cases" but
that's hard to work with.

OTOH if it's defined simply as "mv", and we take m to be constant (for
Newtonian mechanics), then we get back to f=ma in one step, and that's
probably what I should say on that page.

If we _define_ it as "@T/@v" then we've skipped a lot of steps -- that
doesn't seem quite right for the initial definition of P.

Anyway, upon looking back at that page I think it could use a little more
explanation between equations (4) and (5) and I also think
you're right, I should restate the second law.



With as exacting as you have shown to be, I would expect that you'd
cringe as much as I do whenever that kind of oversimplification to the
point of inaccuracy is done. It is cited about as commonly as the term
"zero gravity" is used, but hardly as egregious. Heh.


~ CT

--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org

From sal:
On Mon, 20 Nov 2006 00:07:36 -0800, Stuf4 wrote:

Hey, sal. Out of curiosity I poked around your website a bit. Looks like
some neat stuff there. If you're open to constructive feedback, I saw
that on this page:
http://www.physicsinsights.org/lagrange_1.html

...you've got Newton's Second Law as F=ma.

Ah, hmm... yeah, I looked that over again. I should rewrite that section.

The problem is one of definition, of course. Newton's original, though
written in Latin rather than calculus, translates pretty directly into
f=ma, but, of course, the more useful form is f=dP/dt since it
generalizes to nonconstant mass. The problem with the latter is how we
define P.

The definition which is probably closest to correct is "That vector
quantity which is conserved and which behaves like mv in simple cases" but
that's hard to work with.

OTOH if it's defined simply as "mv", and we take m to be constant (for
Newtonian mechanics), then we get back to f=ma in one step, and that's
probably what I should say on that page.

If we _define_ it as "@T/@v" then we've skipped a lot of steps -- that
doesn't seem quite right for the initial definition of P.

Anyway, upon looking back at that page I think it could use a little more
explanation between equations (4) and (5) and I also think
you're right, I should restate the second law.

The irony I saw there is that you went from F=ma and propogated it into
the 2nd Law! I imagined your reason for doing so is that the majority
of your audience would not have recognized it as the 2nd Law if you had
simply stated it as:
_
F=dp/dt

Heh!

(And to reiterate what I've responded to OM, this distinction is
especially important on this forum because you can't get to the rocket
equation from F=ma.)


~ CT

sal wrote:

...you've got Newton's Second Law as F=ma.

Errr .... I wrote down Newton's laws on that page from memory, didn't
double check 'em. Time for a big "Ooops", maybe?

Gotta run off right now but I'll check that out later today. Tx.

F = d(mv)/dt (rate of change of momentum)

If m = constant, F = ma.

- - -

At this point, I would respectfully request that you cease encouraging
CT to begin spinning his net once again.

Thanks.

--
Dave Michelson