# How can we slow down g?

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53 minutes ago, Strange said:

I didn't use ANY coordinates.

If you don't know (=recognize, exactly) what you are doing

until you recognize it yourself.

That might take you some time. Good luck.

I can only hint,

& hope you get it someday.

If the radius stays the same,

what is falling? Answer: only x & y's change (but that's cartesian).

r alone demonstrates no change, thus no acceleration (of the height).

Got it?

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Are you deliberately lying

No. Not that I know.

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or are you just incapable of understanding what you read?

Maybe. If it makes nonsense.

radius r=((x^2)+(y^2))^0.5, is (a constant or variable) legal for cartesian x,y coordinates;

but x & y are not necessarily legal (as in reverse) for radial coordinates.

Either you're discussing (only) radial (which does NOT include x,y because they are named Cartesian)

or else Cartesian (which can include the radius).

I guess your professors weren't very fussy.

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All I can see is you making a hash of ... well, everything.

I see your education has made you incapable of distinguishing some things.

Let's drop (the radial theme), you don't get.

The significant (speculative) part has been said

(but you're still against it).

Edited by Capiert
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1 minute ago, Capiert said:

If you don't know (=recognize, exactly) what you are doing

I know exactly what I am doing: Trying to explain schoolboy physics to someone who is too arrogant to learn anything.

2 minutes ago, Capiert said:

what is falling?

Any object in orbit.

It doesn't matter what coordinate system you use. It is still falling.

The fact that you think changing the coordinates makes a difference is another illustration of how profoundly ignorant your are.

4 minutes ago, Capiert said:

Its not my choice. I haven't even reported it to the mods (your immaturity and ignorance is mildly amusing).

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10 minutes ago, Strange said:

I know exactly what I am doing: Trying to explain schoolboy physics to someone who is too arrogant to learn anything.

(What is falling?)

Any object in orbit.

Why doesn't its height change then

if it is suppose to be falling?

I don't see any fall involved.

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It doesn't matter what coordinate system you use. It is still falling.

I'm sorry that makes no sense.

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The fact that you think changing the coordinates makes a difference is another illustration of how profoundly ignorant your are.

Thanks, profoundly.

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Its not my choice. I haven't even reported it to the mods (your immaturity and ignorance is mildly amusing).

Thanks, I guess we have to leave the (radial) theme.

Edited by Capiert
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1 minute ago, Capiert said:

Why doesn't its height change then

if it is suppose to be falling?

I don't see any fall involved.

So, you didn’t read the explanations in the links I provided. Lazy as well as arrogant.

9 minutes ago, Capiert said:

I'm sorry that makes no sense.

Of course not. If you understood it, you wouldn’t be making such ludicrous assertions.

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47 minutes ago, Strange said:

So, you didn’t read the explanations in the links I provided. Lazy as well as arrogant.

Sorry, I had to reinstall the adobe flash player. But you're right.

Am I so arrogant: to try new things, not known to you?

Is that being arrogant?

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Of course not. If you understood it, you wouldn’t be making such ludicrous assertions.

I can't say I liked the explanation.

It's a (little) deceptive (=misleading) picture, making use of falling on 1 (e.g. left) side,

& (but) ascending on the other (e.g. right side).

So the spacecraft just falls all the way around the Earth, never hitting the surface. The curve of the spacecraft's path is about the same as the curve of Earth's surface.

So astronauts orbiting Earth aren't really weightless, they are just falling . . . and falling, and falling!

That doesn't make (complete) sense.

It only exploits the earthly falling mentality,

such as falling (=starting to fall),

& a roller coaster looping.

E.g. with gravity always working at the bottom of the picture,

even below the earth.

Why should there be (even more) gravity below (under) the earth in that picture?

I'm content with the (circular) orbital period

T=2*Pi*(R/g)

is (like) the pendulum period

T=2*Pi*(L/g),

the pendulum's length L=R the orbit's radius;

& the circumferential_speed vc=cir/T=2*Pi*R/T.

That doesn't work for the earth to moon distance d,

in which I must conclude

the barycenter radii would have to be in line,

as to create maybe

Re+d=Rm

or

d=Rm-Re.

& the only way to decern

would be to notice a wooble (magnitude or amount) wrt the stars (positions, after a sidereal day 23h 56m 4s).

Edited by Capiert
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3 hours ago, Capiert said:

Radially there is no fall acceleration (neither) up (n)or down

So you've failed basic physics. No surprise.

Objects with no acceleration mover in a straight line (Newton's first law). Are you familiar with Newton's laws of motion? They're kind of important.

3 hours ago, Capiert said:

I haven't (really) noticed when the moon fell down onto the earth, & crashed.

It didn't crash — it missed us. And keeps missing us, because of the way it's moving. But it's accelerating toward us.

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1 hour ago, Capiert said:

Sorry I forgot the rooting

I'm content with the (circular) orbital period

T=2*Pi*((R/g)^0.5)

is (like) the pendulum period

T=2*Pi*((L/g)^0.5)

the pendulum's length L=R the orbit's radius;

& the circumferential_speed vc=cir/T=2*Pi*R/T.

21 minutes ago, swansont said:

So you've failed basic physics. No surprise.

I'm exclusively discussing vertically. i.e. radius r.

21 minutes ago, swansont said:

Objects with no acceleration mover in a straight line (Newton's first law). Are you familiar with Newton's laws of motion? They're kind of important.

No bout adout it in x & y.

21 minutes ago, swansont said:

It didn't crash — it missed us. And keeps missing us, because of the way it's moving. But it's accelerating toward us.

Quite right.

Again x,y,z.

To summarize,

I don't find a delta r wrt time;

I only find delta x

& delta y

wrt time.

You'll need a different strategy

to convince otherwise

I cannot lie to myself

& believe it.

Sorry it won't work until you guys start making sense

on that point,

so it's better to drop the radial theme

til then.

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I.e. I can not confirm what does not exist, namely a change in r, wrt time,

either as speed or acceleration.

But instead, (x,y) cartesian coordinates are perfect (for the job),

I can confirm (fall of an orbit) with using all 3 (r, x, & y) parameters in 2D;

or else in x,y,z, all 4 (r^2=x^2 + y^2 + z^2) parameters in 3D.

Edited by Capiert
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7 hours ago, Capiert said:

Is that being arrogant?

Yes. Asking a question and then refusing to read the explanations is arrogant.

You prefer to think that your random meaningless ideas must be right and hundreds of years of science back by theory and experiment must be wrong.

7 hours ago, Capiert said:

It's a (little) deceptive (=misleading) picture, making use of falling on 1 (e.g. left) side,

& (but) ascending on the other (e.g. right side).

It is not "ascending" on the right side.d.

So you looked at the pictures but didn't understand them.

8 hours ago, Capiert said:

That doesn't make (complete) sense.

It does.

The problem is not with the explanation but with your inability to understand it.

8 hours ago, Capiert said:

That doesn't work for the earth to moon distance d,

6 hours ago, Capiert said:

Quite right.

Again x,y,z.

No one mentioned x,y,z apart from you. Therefore this is a straw man argument (ie a fallacy).

6 hours ago, Capiert said:

Sorry it won't work until you guys start making sense

The problem is not on our side.

You won't make any progress until you are able to learn.

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9 hours ago, Capiert said:

I'm exclusively discussing vertically. i.e. radius r.

You can only talk about radius in e.g. a spherical or cylindrical coordinate system, and constant r in such systems is a circle. The path is not straight. Therefore, there MUST be an acceleration.

This is introductory physics. Often one of the first physics concepts presented. If you don't get this, then there's no hope of understanding more advance concepts.

8 hours ago, Capiert said:

I.e. I can not confirm what does not exist, namely a change in r, wrt time,

either as speed or acceleration.

But instead, (x,y) cartesian coordinates are perfect (for the job),

I can confirm (fall of an orbit) with using all 3 (r, x, & y) parameters in 2D;

or else in x,y,z, all 4 (r^2=x^2 + y^2 + z^2) parameters in 3D.

The choice of coordinate system doesn't matter. If it's accelerating, it's accelerating.

If it's not accelerating, the force on it must be zero. There's gravity, and...?  What is counterbalancing gravity?

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14 hours ago, Capiert said:

That's usually determined by which 1 is different of 2 from the 3rd (or else 3 watches).

Really?  How do you know that the 1 watch isn't correct and the other 2 are out?  Or that all 3 are just wrong?   -  You can't.

An moving object in orbit NEEDS a force on it to stop it going off in a straight line at a tangent from the earth. This force provides an acceleration in the direction of the earth (Down). This force curves the path of the object (that would normally go in a straight line) so that it orbits the earth. If the force it too weak then the object flies out of orbit, if too strong it falls....  if balanced then the object stays in orbit. The force and acceleration are still there - they are keeping the motion of the object circular rather than straight.  Accept it - it is how it is and is shown in actuality by weights on ropes and things in orbit around the earth. Cut the rope (remove the force) and the weight flies off in a straight line.

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