Re: Mass was Re: Questions (Space)

On Sep 12, 3:51 pm, Ric Locke <warlo...@xxxxxxxxxxxx> wrote:
On Tue, 11 Sep 2007 23:16:36 -0000, goodbyeblues...@xxxxxxxx wrote:
On Sep 12, 12:48 am, Ric Locke <warlo...@xxxxxxxxxxxx> wrote:
On Tue, 11 Sep 2007 21:38:13 -0000, goodbyeblues...@xxxxxxxx wrote:
On Sep 11, 11:13 pm, Ric Locke <warlo...@xxxxxxxxxxxx> wrote:
On Tue, 11 Sep 2007 19:44:50 -0000, goodbyeblues...@xxxxxxxx wrote:
On Sep 11, 8:04 pm, Ric Locke <warlo...@xxxxxxxxxxxx> wrote:
On Tue, 11 Sep 2007 09:40:39 -0700, David Harmon wrote:
Is it constantly having kinetic energy added to it?

Yep. Of course it's always giving up kinetic energy at the same time, so
the total KE (which is a scalar, with no direction to it) remains
constant.

Ha ha!

Hey Ric, right now I'm putting a hat on your head. It's true! Of
course, I'm taking it off again at the same time, so you'll never be
the wiser. If you want me to stop, you'll have to pay me.

::shrug:: All you just said is that you didn't understand.

Regards,
Ric

Really? Well please enlighten me, because I can't make sense of what
you are saying.

When you write:
In the case of an orbit, at time T the orbiting body has a velocity
vector relative to the planet of X=V, Y=0, Z=0. At time T+epsilon the
velocity in the Z direction (which was toward/away from the center of
the planet at time T, and is now parallel to that line, i.e. it no
longer passes through the center) is no longer zero. At the same
T+epsilon the velocity in the X direction is equal to V minus some small
amount, such that SQRT((new X velocity)^2 plus (new Z velocity)^2) = V,
the original velocity. The object was "decelerated" in X and
"accelerated" in Y.

This "acceleration" in Y, is that your source of added kinetic energy?
And your "deceleration" in X, is that where the orbiter is giving up
kinetic energy? Because those two don't add up to zero... so that
can't be what you are talking about.

So what are you talking about?

How much kinetic energy is an orbiter constantly gaining and losing
according to you?

I think you need to explain yourself.

David Harmon said "acceleration means change in velocity, not change
in KE" and you said that was wrong. Well every textbook I've seen says
acceleration is rate of change of velocity.

David Friedman writes "acceleration at right angles to its velocity
does not increase its speed and so does not increase its kinetic
energy." That is also something which is featured in every textbook I
have seen.

I think you are confused.

This is one of those things that's hard without cocktail napkins.

Draw a dot. That's the center of mass of the planet. Draw a circle
around the dot. That's the orbit of the satellite. Now draw a line from
the dot, straight up, crossing the orbit, and at the point where that
line crosses the circle draw a tangent going left and right. You end up
with a right-angled cross, with the center of the orbit at the bottom
end and the orbit crossing the vertical at the same point the crossarm
does. The X axis is parallel to the crossarm; the Z axis is parallel to
the vertical; and the Y axis is perpendicular to the paper and doesn't
figure in this discussion. Now pick a direction for the satellite. (I
chose counterclockwise, and the below assumes that.)

At time T the satellite is at the point where orbit and crossarm cross
the vertical.

At time T+epsilon the satellite is at some small distance to the left of
the vertical. Draw another vertical line, parallel to the first one,
passing through the orbit where the satellite now is.

If there were no gravity, the satellite would travel along the crossarm
in a straight line. From the point where the crossarm crosses the new
vertical, draw a line diagonally back to the original point, the center
of mass of the planet. That line represents the gravitational pull of
the planet at the new position of the satellite. (Note that this is a
calculus problem, not an arithmetic one; the distances involved are so
small that the current and theoretical positions of the satellite are
not distinct radially; the curvature of the orbit is negligible compared
to the distance traveled along it.)

The gravitational pull is a force vector, which can be resolved into two
components. One, very small, points back toward the "start" position of
the satellite, that is, opposite to the direction it was going. That
force causes the satellite to decelerate /in that direction/, that is,
parallel to the X axis.

The second component of the force vector, larger, accelerates the
satellite in the Z /direction/, which is parallel to the Z /axis/ but
does not pass through the center of the planet.

So the satellite loses velocity in the X direction and gains velocity in
the Z direction. The accelerations are such that the total kinetic
energy (NOT a vector quantity) is constant, and the velocity vector is
the same size but rotated. The new vector will be tangent to the orbit
at the point where the satellite now is.

Therefore the satellite lost kinetic energy (velocity) in the X
direction, but gained kinetic energy (velocity) in the Z direction, the
magnitudes of gain and loss being such that the total kinetic energy
remained the same. The transfer was mediated by the gravitational field
it is immersed in.

Clearer?

No, not at all. I have drawn your drawing, and it doesn't help.

You have not told me what the magnitude of the gain and loss of
kinetic energy is like I asked. You have not shown that the two cancel
each other. You have just postulated that they do. Instead of showing
it, you have jumped ahead to the answer: The speed remains the same,
so the kinetic energy doesn't change.

You write yourself that it is a calculus problem, so the positions of
the satelite are not distinct radially. And yet you go on right after
that to write that the gravitational vector in the second position has
a component which points back towards the first position. When we
finish our calculations by letting epsilon go to zero, this
deceleration and associated loss of kinetic energy you are talking
about also goes to zero. Which is in conflict with what you started
off postulating.

Which is why I said it was a "calculus problem". The whole point of
calculus is adding up zeroes (or, rather, things that /look/ like
zeroes, infinitesimals) to get a nonzero result. 1/(1-X) is zero if X is
1. The integral of (1/(1-X))dx is not.

The theoretical position of the satellite if there were no gravity and
its actual position aren't distinct /radially/. Along the satellite's
path, which for this infinitesimal we are taking as parallel to the X
axis, the positions /are/ distinct. The difference is delta-X.

The force slowing the satellite down is equal to the gravitational force
acting on the satellite times the sine of arctan(X'-X)/R, where R is the
radius of the orbit. The force pushing it in the Z direction ("down" at
this particular instant) is F times the cosine, and F is the gravitation
equation. If you set it up in those terms and integrate, you will get
the equation for a satellite's orbit or the exact (rather than an
approximation) equation for the path of any projectile in Newtonian
space. I no longer have an orbital mechanics reference, and my skills
with that sort of math are not so much rusted as corroded to nasty dust,
so I can't do it for you, but I do /remember/ doing it, and it works.

The reason it's important is that it gives a foundation for realizing
that the difference between "straight line" motion (i.e. along a
geodesic of space) is fundamentally different from motion along a curve,
the difference being that the latter can only result from acceleration.
The reason /that/ is important is that the difference between Special
and General Relativity is that in Special Relativity we consider the
case of entities that /have been/ accelerated; in General Relativity we
consider the process of acceleration itself. General Relativity is
/much/ hairier than Special.

And I owe David Friedman something of an apology, which I hereby render.
Sorry, David. I failed to realize that I was indulging in one of my
personal crotchets, which in their own way are just as bizarre as
Tina's. Specifically, I find it easier to visualize situations like this
in terms of "momentum", that is, kinetic energy with a vector attached.
Yes, the kinetic energy of the satellite remains constant if the orbit
is circular. The "momentum" in the X and Y directions of Cartesian
coordinates changes constantly. I'm well aware that there ain't no sech
animal in that particular zoo, but it's a useful crutch for me in
setting up the problem.

Regards,
Ric

Oh, so all this time you've said kinetic energy but meant momentum?
That's the end of the discussion then. You should be more careful with
your words in the future.

.

Relevant Pages

  • Re: Mass was Re: Questions (Space)
    ... That's the orbit of the satellite. ... The X axis is parallel to the crossarm; the Z axis is parallel to ... Now pick a direction for the satellite. ... direction, but gained kinetic energy in the Z direction, the ...
    (rec.arts.sf.composition)
  • Re: Mass was Re: Questions (Space)
    ... T+epsilon the velocity in the X direction is equal to V minus some small ... That's the orbit of the satellite. ... Now pick a direction for the satellite. ... Therefore the satellite lost kinetic energy in the X ...
    (rec.arts.sf.composition)
  • Re: Mass was Re: Questions (Space)
    ... T+epsilon the velocity in the X direction is equal to V minus some small ... That's the orbit of the satellite. ... Now pick a direction for the satellite. ... Therefore the satellite lost kinetic energy in the X ...
    (rec.arts.sf.composition)
  • Re: Mass was Re: Questions (Space)
    ... T+epsilon the velocity in the X direction is equal to V minus some small ... That's the orbit of the satellite. ... Now pick a direction for the satellite. ... Therefore the satellite lost kinetic energy in the X ...
    (rec.arts.sf.composition)
  • Re: Mass was Re: Questions (Space)
    ... T+epsilon the velocity in the X direction is equal to V minus some small ... That's the orbit of the satellite. ... Now pick a direction for the satellite. ... direction, but gained kinetic energy in the Z direction, the ...
    (rec.arts.sf.composition)