My Kepler Paradox

[Capiert]

(Please) take a look
 at the math.

If I reduce (=decrease)
 the eccentricity
 to almost zero
 (then) Kepler's_orbit_Period (3^rd law)
 is (still)
 T_Kepler=r_a^1.5;

 but (however) at zero eccentricity e=0
 (then) both the semi_major_axis r_a
 & the semi_minor_axis r_b
 are equal
 to the same (=identical),
 radius
 r=r_a=r_b if e=0
 (but) having the circular orbit_Period

 T_circle=2*Pi*((r/g)^0.5).

 

Surely,
 for an Earth's satellite,
 both Orbit_Periods
 T_{Kepler_(e=0)}#T_circle
 r^1.5#2*Pi*((r/g)^0.5), ^2

 are NOT the same
 &/or
 can NOT be the same.

Squaring both sides

 r³#4*Pi²*r/g, /r
 r²#4*Pi²/g, ^0.5
 r#2*Pi/g, *g
 g*r#2*Pi, let the centrifugal_acceleration a_c=g equal the gravitational free_fall (for an orbit)
 a_c*r#2*Pi, because
 the centrifugal_acceleration
 a_c=v_c²/r
 multiplied
 by the common radius r
 as
 a_c*r=v_c²

 is the (circular) orbit_speed squared v_c²;
 & (that v_c²) is (definitely) NOT 2*Pi.

 

v_c²=

 a_c*r=2*Pi*(2*Pi*r²/T²)
 is a factor
 of *(2*Pi*r²/T²)
 larger than 2*Pi.

Thus,
 (as far as I can see)
 Kepler's math
 is NONSENSE!
 (=it makes NO SENSE to me
 in its existing form);
 it has NO consistency
 with regular (Newtonian) Physics;
 & must have a different meaning.

I hope that explains
 my problem
 (with Kepler('s laws))
 well enough.

2 different (circular Orbit_period) formulas,
 which 1 is correct? (Kepler's or Newton's?)

Please DON'T tell me Newton('s centrifugal_acceleration) is WRONG!
..because that is from where I derived
 my circular Orbit_Period from.

Edited January 20, 2023 by Capiert
Typo.

[swansont]

1 hour ago, Capiert said:

If I reduce (=decrease)
 the eccentricity
 to almost zero
 (then) Kepler's_orbit_Period (3^rd law)
 is (still)
 T_Kepler=r_a^1.5;

Kepler’s 3rd law is a proportionality, not an equality

If your equation for period doesn’t have the form T^2/R^3 = constant, you’ve done it wrong. (g should not be in your equation. Put it in terms of M, R and G)

 

 

[Janus]

Newtonian Physics says the period of an orbiting object is T = 2pi R^(3/2)/(GM)^(1/2)

Thus 1/(2pi) = R^(3/2) / T(GM)^(1/2)

Square both sides:

1/(2pi)^2 = R^3 /  GM T^2

Move GM to the left side of the equation:

1/GM(2pi)^2 = R^3 / T^2

Invert both sides

GM(2pi)^2 = T^2/R^3

So what Kepler's law states is that for any central body, there is a specific relationship between R and T

 Newton keeps the relationship.   It just includes the mass of the central body, so if you know any two of T, R, or M. you can find the third.

 

 

[Capiert]

Thanks, both of you.

[Capiert]

On 1/21/2023 at 1:40 AM, swansont said:

Kepler’s 3rd law is a proportionality, not an equality

Thanks, that'( i)s exactly what I needed.
T_kepler~r_a^1.5
T_kepler=k*r_a^1.5

On 1/21/2023 at 1:40 AM, swansont said:

If your equation for period doesn’t have the form T^2/R^3 = constant, you’ve done it wrong.

Ok. Rooted gives
k=T/(r^{^1.5}).

On 1/21/2023 at 1:40 AM, swansont said:

(g should not be in your equation.

But, I do NOT see why (NOT).
Surely you mean ONLY the end result;
NOT the starting basis. ?

On 1/21/2023 at 1:40 AM, swansont said:

Put it in terms of M, R and G)

From Janus's post

On 1/21/2023 at 2:40 AM, Janus said:

Newtonian Physics says the period of an orbiting object is T = 2pi R^(3/2)/(GM)^(1/2)

Independent
 of this thread's presentation syntax:
 (as side_track)

I quickly (also) see**
 (in her 1st line)
 (that)
 Newton's gravitational_force
 F_g=G*(M/R)*(m/r)
 is 2 mass_per_distance (linear_)ratios (M/R)*(m/r)
 multiplied together,
 & then multiplied by a proportionality constant G,
 where their (common) center_to_center distances R=r
 are (intended as) identical(ly the same).

(I.e. Caution (poor) syntax, (clash):
Distance, NOT just a radius.
It's a ruff approximation.
 NOT to be confused with each (masses') radius separately
 (with their own (different) radius size);
 but instead the sum of both radii (distances + 1 height), (all) together.
 E.g. r=r_m+r_M+h=R, h=separation_height, surface to surface.
Disclaimer: That'( i)s how I intuitively interpreted F_g, in a flash.
That example of (symbol) r has NOTHING to do with my orbit radius r, later.)

**
The Key (move=mano[e]uver): is
 shift (=move)
 the rooted_G*M (denominator)
 1/((G*M)^{^0.5})
 (to) under the 2*π (numerator)
 (to obtain the proportionality constant k).
T=(2*Pi/((G*M)^(1/2)))*(r^(3/2)).

On 1/21/2023 at 2:40 AM, Janus said:

Thus 1/(2pi) = R^(3/2) / T(GM)^(1/2)

Square both sides:

1/(2pi)^2 = R^3 /  GM T^2

Move GM to the left side of the equation:

GM/(2pi)^2 = R^3 / T^2
 (is (surely) enough info
 to follow thru correctly,
 & ignore the typo)

On 1/21/2023 at 2:40 AM, Janus said:

1/GM(2pi)^2 = R^3 / T^2

Invert both sides

((2pi)^2)/GM = T^2 / R^3

On 1/21/2023 at 2:40 AM, Janus said:

GM(2pi)^2 = T^2/R^3

On 1/21/2023 at 2:40 AM, Janus said:

So what Kepler's law states is that for any central body, there is a specific relationship between R and T.

Newton keeps the relationship.  It just includes the mass of the central body, so if you know any two of T, R, or M, you can find the third.

Great inspiration.
Thanks.

---

So giving it another go, again.

A circular orbit_Period, is
 T_circle=2*Pi*((R/g)^{^0.5}).

Equating, the Weight (Force)
 Wt=m*g
 of a mass m,
 with Newton's gravity_Force
 F_g=G*M*m/(R^{^2})
 for the Earth's_mass M,
 separated
 by their total (radial) distance R
 (center to center),
 (& due to Newton's 3rd law
 of opposite & equal reaction)
 we have (equal & opposite forces, balancing (out))
 Wt=F_g
 m*g=G*M*m/(R^{^2}), /m
 & dividing both sides
 by the (small(er)) mass m
 we get
 the free_fall (gravitational) acceleration
 g=G*M/(R^{^2}).

(Please Not(ic)e:
 that g
 is typically measured
 near the Earth's surface;
 but (g) gets smaller
 as the separation ((e.g. orbit_)radius R)
 gets large(r),
 e.g. to a GSO
 (geo_stationary orbit's) radius
 R_GSO~g*(T^{^2})/(2*(Pi^{^2}))
 where the weightless(ness)
 g_GSO=0
 is zero.)

 Inserting that (g, as inverse factor
 1/g=(R^{^2})/(G*M))
 into the circular orbit_Period
 T_circle=2*Pi*((R*(1/g))^{^0.5}), gives
 T_circle=2*Pi*((R*(R^{^2})/(G*M))^{^0.5})
 & we get R^{^3} under the root(_sign)
 T_circle=2*Pi*((R^{^3})/(G*M))^{^0.5}).

The Key (move=mano[e]uver): is
 shift (=move) the rooted_G*M (denominator)
 1/((G*M)^{^0.5})
 (to) under the 2*π (numerator)
 (to obtain the proportionality constant k).

 T_circle=((2*Pi/(G*M))^{^0.5})*((R^{^3})^{^0.5}), rooting the (R^{^3})
 to (R^{^3})^{^0.5}=R^{^(3/2)}=R^{^1.5}
 we get
 T_circle=(2*Pi/((G*M)^{^0.5})*(R^{^3/2}), ^3/2=1.5
 T_circle=(2*Pi/((G*M)^{^0.5})*(R^{^1.5})

 or ingesting(=incorporating)
 the 2*π
 into under the root_sign,
 as rooted 4*(Pi^{^2});
 we have

 T_circle=((4*(Pi^{^2})/(G*M))^{^0.5})*(R^{^1.5}).

Janus's (wonderful method) tells me:
Newtonian Physics says the period of an orbiting object is

T=2*Pi*(R^{^(3/2)})/((G*M)^{^0.5}), *1/(T*2*Pi)

Thus

 1/(2*Pi)=(R^{^(3/2)})/(T*((G*M)^{^0.5})), ^2=Square both sides

 1/((2*Pi)^{^2})=(R^{^3})/((T^{^2})*G*M), *G*M

Move G*M to the left side of the equation
 (by multiplying both sides by G*M,
 gives):

(G*M)/((2*Pi)^{^2})=(R^{^3})/(T^{^2}), invert both sides

((2*Pi)^{^2})/(G*M)=(T^{^2})/(R^{^3}), swap sides

(T^{^2})/(R^{^3})=((2*Pi)^{^2})/(G*M), multiply by  R^{^3}

T^{^2}=(((2*Pi)^{^2})/(G*M))*(R^{^3}), ^0.5=root both sides

T=(2*Pi/((G*M)^{^0.5})*(R^{^1.5})

So Swansont's (searched (for (proportionality)))
constant, is

 k=2*Pi/((G*M)^{^0.5})

 for the circular orbit_period

 T=k*(R^{^1.5})

 with orbit_radius R.

Edited December 7, 2023 by Capiert

[swansont]

14 hours ago, Capiert said:

But, I do NOT see why (NOT).

g is the acceleration due to gravity on the surface of the earth. It depends on the mass and radius of the earth. Those are not orbital parameters, and are specific to the earth, not other celestial bodies. There’s no reason for it to show up in an orbital equation; objects generally do not orbit at the surface of the earth.

 

14 hours ago, Capiert said:

Please Not(ic)e:
 that g
 is typically measured
 near the Earth's surface;
 but (g) gets smaller
 as the separation ((e.g. orbit_)radius R)
 gets large(r),
 e.g. to a GSO
 (geo_stationary orbit's) radius
 R_GSO~g*(T^{^2})/(2*(Pi^{^2}))
 where the weightless(ness)
 g_GSO=0
 is zero.)

g is small at GSO, but it’s not zero.

Objects in orbit are in freefall. That’s why they are weightless. It’s not because the local g is zero.

(if there were no acceleration, it could not be in orbit. By Newton’s first law, if there is no acceleration, its motion would be in a straight line)

[Capiert]

On 12/7/2023 at 5:20 PM, swansont said:

g is the acceleration due to gravity

wrt

On 12/7/2023 at 5:20 PM, swansont said:

on the surface of the earth.

Motion is always relative to some reference.

In that (acceleration motion) case
 (which is the (experimental) observable)
 it sure looks
 to me like
 wrt the Earth's surface.

The rest (=explanation) is theory.

On 12/7/2023 at 5:20 PM, swansont said:

It depends on the mass and radius of the earth.

As the formula says.

On 12/7/2023 at 5:20 PM, swansont said:

Those are not orbital parameters,

Why NOT?
They are variables
 that can be mannipulated
 for other examples.

On 12/7/2023 at 5:20 PM, swansont said:

and are specific to the earth,

Their values are specific
 e.g. to Earth;
 but (as) the variables
 they do NOT need
 to be specific
 to (ONLY) the Earth.

Nature does NOT prefer
 for her (natural) laws.

Her laws are universal.

Thus other (different) examples
 (must) exist.

I'm NOT telling you anything new.

You know that already.

On 12/7/2023 at 5:20 PM, swansont said:

not other celestial bodies.

Other celestial bodies have their own values
 for the parameters.

On 12/7/2023 at 5:20 PM, swansont said:

There’s no reason for it to show up in an orbital equation;

The same variables
 can be used
 more or less universally.
You just have NOT seen the connection yet.
Or am I wrong?
You will naturally say yes
 if I am NOT mistaken.

On 12/7/2023 at 5:20 PM, swansont said:

objects generally do not orbit at the surface of the earth.

If you are on the Earth,
 then you are (also) moving with it
 ((as) circular motion),
 without a (visible) change in height.

An orbit can be equated
 (at least by me)
 to circular motion
 (which is)
 without height change.

At least I can attempt
 to (try &) do that
 (if you can NOT).

On 12/7/2023 at 5:20 PM, swansont said:

g is small at GSO, but it’s not zero.

If there is NO_fall vertically
 then I see NO acceleration (vertically, either).

On 12/7/2023 at 5:20 PM, swansont said:

Objects in orbit are in freefall.

But a GSO has NO vertical_fall.
It (vertical_"fall") is NOT observable.

I think you are confusing
 that
 objects are "free" (NOT bound)
 to fall
 if they could;
 but they do NOT fall ("down", vertically)
 (perhaps because they are moving?).

Each orbit radius (value) r=(v_c^{^2})/a_c
 has its own circumferential_speed vc=cir/T=2*Pi*r/T;
 but that centrifugal_acceleration ac=(v_c^{^2})/r
 is only a math_construct, anyway.

It stems from squaring
 the circumferential_speed
 v_c^{^2}=a_c*r
 & then splitting
 that into an acceleration a_c
 & (r radius_)distance product.

It'( i)s otherwise total NONSENSE
 to express an orbit
 in "area" (units)
 per time_squared;
 when (circumferential_)speed vc=cir/T
 will do (already).

The a_c*r product was only created for convenience.

On 12/7/2023 at 5:20 PM, swansont said:

That’s why they are weightless.

Weightless per word definition
 is "less" weight;
 NOT NO weight.

But here I use it
 as otherwise intended,
 meaning:
Weightless (as like floating)
 indicates
 zero (vertical_)acceleration.

E.g. Einstein's Equivalence.

I DON'T care what you "believe"
 (to explain),
 I am interested in the (experimental) observables
 (in order to formulate).

If the weight
 Wt=m*g
 but the mass m
 is NOT falling
 (e.g. NOT changing its vertical_position)
 then its weight is (also) zero.

Its (=The mass's)
g=0 wrt the Earth's surface.

It (=The mass) does NOT change vertical_height h=constant.
Its (=The mass's) vertical motion is zero wrt the Earth's surface.
That means (both): NO speed, & NO acceleration wrt the Earth's surface.
(I can NOT understand why you think so rigidly.
I suspect you forget that you are using only (math_)constructs.)

On 12/7/2023 at 5:20 PM, swansont said:

It’s not because the local g is zero.

What do you mean by local (there)?
Up in the sky at the mass? (y?);
 or down on the Earth's surface (n?).
The reference
 is the Earth's surface;
 but it'( i)s observing the mass
 (from there=Earth's_surface).

On 12/7/2023 at 5:20 PM, swansont said:

(if there were no acceleration, it could not be in orbit.

Why NOT?
Orbit is ZERO vertical_acceleration (observed).

On 12/7/2023 at 5:20 PM, swansont said:

By Newton’s first law, if there is no acceleration, its motion would be in a straight line)

You can still have (1D) "linear" acceleration (or deceleration, completely) without (circular) orbits.
(Thus) That is NOT an exclusive decisive example to rely on.
It (=Your(=That (particular)) argument)
does NOT decide ANYTHING.
(It'( i)s NOT a double blind proof.)

I would need a (much) better example than that to convince me otherwise.
Sorry.

Edited December 8, 2023 by Capiert

[swansont]

27 minutes ago, Capiert said:

wrt

Motion is always relative to some reference.

This is about acceleration. Acceleration is not motion. You can have an acceleration when v = 0

Acceleration is not relative. If an object is accelerating, you can tell.

 

27 minutes ago, Capiert said:

In that (acceleration motion) case
 (which is the (experimental) observable)
 it sure looks
 to me like
 wrt the Earth's surface.

At the earth's surface.

g = GM/r^2 where r is the radius of the earth, i.e. it is determined at the earth's surface

 

 

27 minutes ago, Capiert said:

The rest (=explanation) is theory.

As the formula says.

Why NOT?
They are variables
 that can be mannipulated
 for other examples.

g does not depend on anything being in orbit.

You can use g in an equation, but then you have to do things like correct for the fact that the actual acceleration is not g if you are not on the surface of the earth, which is an unnecessary complication of the formula. Follow the KISS principle (Keep It Simple, Stupid)

27 minutes ago, Capiert said:

Their values are specific
 e.g. to Earth;
 but (as) the variables
 they do NOT need
 to be specific
 to (ONLY) the Earth.

But causes unnecessary complication.

Compare how many lines is took Janus to derive the equation and how many lines it took you to do it.

27 minutes ago, Capiert said:

Other celestial bodies have their own values
 for the parameters.

Indeed they do. 

27 minutes ago, Capiert said:

 

If you are on the Earth,
 then you are (also) moving with it
 ((as) circular motion),
 without a (visible) change in height.

If you are standing on earth you are not in orbit. You are not in freefall.

27 minutes ago, Capiert said:

An orbit can be equated
 (at least by me)
 to circular motion
 (which is)
 without height change.

At least I can attempt
 to (try &) do that
 (if you can NOT).

An orbit has more conditions than a circular motion. 

 

27 minutes ago, Capiert said:

If there is NO_fall vertically
 then I see NO acceleration (vertically, either).

Any object moving in circular motion has an acceleration toward the center of the circle

27 minutes ago, Capiert said:

But a GSO has NO vertical_fall.
It (vertical_"fall") is NOT observable.

The vertical fall is observable. But there is an equal amount of sideways motion as well, which is why the path is circular, and why vertical (y) and horizontal (x) aren't the most useful descriptions. In a circular coordinate system you use radial and tangential.

27 minutes ago, Capiert said:

I think you are confusing
 that
 objects are "free" (NOT bound)
 to fall
 if they could;
 but they do NOT fall ("down", vertically)
 (perhaps because they are moving?).

An object in orbit is not free; it must be in a bound state. You have to add energy to get it to an arbitrary distance 

Freefall just means you are acceleration at the local gravitational acceleration.

27 minutes ago, Capiert said:

Each orbit radius (value) r=(v_c^{^2})/a_c
 has its own circumferential_speed vc=cir/T=2*Pi*r/T;
 but that centrifugal_acceleration ac=(v_c^{^2})/r
 is only a math_construct, anyway.

The acceleration is centripetal (center-seeking)

 

27 minutes ago, Capiert said:

It stems from squaring
 the circumferential_speed
 v_c^{^2}=a_c*r
 & then splitting
 that into an acceleration a_c
 & (r radius_)distance product.

It'( i)s otherwise total NONSENSE
 to express an orbit
 in "area" (units)
 per time_squared;
 when (circumferential_)speed vc=cir/T
 will do (already).

It's not expressed as an area per time squared.

 

27 minutes ago, Capiert said:

The a_c*r product was only created for convenience.

Weightless per word definition
 is "less" weight;
 NOT NO weight.

Weightless (especially in this context) means no weight.

 

27 minutes ago, Capiert said:

But here I use it
 as otherwise intended,
 meaning:
Weightless (as like floating)
 indicates
 zero (vertical_)acceleration.

Again, we should speak of radial, and there is a radial acceleration.

I refer you again to Newton's first law. If there was no acceleration, the object would travel in a straight line.

Do you deny the correctness of Newton's laws?

 

27 minutes ago, Capiert said:

It (=The mass) does NOT change vertical_height h=constant.
Its (=The mass's) vertical motion is zero wrt the Earth's surface.
That means (both): NO speed, & NO acceleration wrt the Earth's surface.
(I can NOT understand why you think so rigidly.
I suspect you forget that you are using only (math_)constructs.)

No, I'm using physics definitions. If you're going to use physics terminology, you have to use the same definitions.

If you make up your own definitions you can't communicate ideas.

27 minutes ago, Capiert said:

What do you mean by local (there)?

At the location under discussion

27 minutes ago, Capiert said:

 

Why NOT?
Orbit is ZERO vertical_acceleration (observed).

Patently untrue

27 minutes ago, Capiert said:

You can still have (1D) "linear" acceleration (or deceleration, completely) without (circular) orbits.

Acceleration is a change in velocity. It works in more than one dimension. Velocity is a vector; it has a magnitude (the speed) and a direction. If you change direction, there is an acceleration, even if speed is constant.

If the discussion is about circular orbits, you can't be looking at this in one dimension. A circle has two dimensions

 

[Capiert]

On 12/8/2023 at 8:56 PM, swansont said:

This is about acceleration. Acceleration is not motion.

That'( i)s new to me.
What then is your definition
 of motion?
Mine is, a change of position
 with time.

On 12/8/2023 at 8:56 PM, swansont said:

You can have an acceleration when v = 0.

Please explain.

I can NOT imagine acceleration
 without a change
 of position
 (wrt time).

On 12/8/2023 at 8:56 PM, swansont said:

Acceleration is not relative.

Please fill me in, there.

(=That (statement) does NOT make sense
 to me.
E.g. I can NOT have a chicken
 without the (prerequisite=)egg, 1st.
I can NOT build(=continue)
 upon NO foundation.)

On 12/8/2023 at 8:56 PM, swansont said:

If an object is accelerating, you can tell.

As I said (=implied),
 I need help there.

On 12/8/2023 at 8:56 PM, swansont said:

At the earth's surface.

g = GM/r^2 where r is the radius of the earth, i.e. it is determined at the earth's surface

So it (=g) is wrt radius r. (y/n?)

On 12/8/2023 at 8:56 PM, swansont said:

g does not depend on anything being in orbit.

g is only an acceleration
(it's called free_fall acceleration),
& it is vertical.

(That means:)
It has magnitude, & direction.
(So it must (also) be a vector, too. ?)

On 12/8/2023 at 8:56 PM, swansont said:

You can use g in an equation, but then you have to do things like correct for the fact that the actual acceleration is not g

That sentence is a head_twister
 for me.
Would you mind explaining
 a bit better.
I assume your 1st "g" is a symbol,
 but your 2nd "g" makes NO sense
 to me.
My examples stated how (my) g varied,
 e.g. wrt height,
 because they were wrt to a reference.
At least I knew (exactly) what & where
 I was dealing with.
But with your definitions
 I go off
 into nirvana
 (because they seem not_founded
 e.g. NOT specified enough,
 or arbitrarily ambiguous).

I DON'T mind
 an extra complication
 if it helps me understand clearly,
 instead of get lost.

On 12/8/2023 at 8:56 PM, swansont said:

if you are not on the surface of the earth, which is an unnecessary complication of the formula.

Unnecessary? complicated? maybe for you;
 but NOT for me.

On 12/8/2023 at 8:56 PM, swansont said:

Follow the KISS principle (Keep It Simple, Stupid)

I DON'T want to make it stupid;
 I want to make it thorough.
It'( i)s too easy
 to get lost
 if a definition
 is lost (=missing).

On 12/8/2023 at 8:56 PM, swansont said:

But causes unnecessary complication.

I DON'T like the complexity either;
 but your (physics) definitions
 (e.g. calculus)
 which dominate the scene,
 force me
 into a more (extensive) complex syntax
 just to make the (algebraic) distinction.

Considering it (=my syntax)
 is only algebra
 I should NOT (even) need
 to state "average_" every time;
 NOR delta, etc.
 for simple differences.
But I do (have to)
 (just to make the (algebraic) distinction.)
It'( i)s that simple!
 but has become
 that complicated.

On 12/8/2023 at 8:56 PM, swansont said:

Compare how many lines is took Janus to derive the equation and how many lines it took you to do it.

Yes! 2..3 lines.

T_kepler=k*r_a^1.5

  On 1/21/2023 at 1:40 AM, swansont said:

If your equation for period doesn’t have the form T^2/R^3 = constant, you’ve done it wrong.

Rooted gives
k=T/(r^{^1.5}).

T=(2*Pi/((G*M)^(1/2)))*(r^(3/2)).

 Wt=F_g
 m*g=G*M*m/(R^{^2}), /m
 g=G*M/(R^{^2}).

NO big deal. Right?

On 12/8/2023 at 8:56 PM, swansont said:

Indeed they do. 

If you are standing on earth you are not in orbit. You are not in freefall.

Naturally, NOT falling.
Standing, was (ONLY) an analogy
 (to an orbit equivalent).
Take it or leave it.
I find it useful.

On 12/8/2023 at 8:56 PM, swansont said:

An orbit has more conditions than a circular motion. 

Yes. Ellipse, etc.
That'( i)s why I kept it simple KIS
 to start with ONLY circular_motion.
Once the basics have been established correctly,
 the (complexity) details may be (included later &) improved upon.

On 12/8/2023 at 8:56 PM, swansont said:

Any object moving in circular motion has an acceleration toward the center of the circle

That'( i)s an interesting point.

On 12/8/2023 at 8:56 PM, swansont said:

The vertical fall is observable. But there is an equal amount of sideways motion as well,

I DON'T deny it.
(I mean I will have to consider & deal with it, later, of course.)

On 12/8/2023 at 8:56 PM, swansont said:

which is why the path is circular, and why vertical (y) and horizontal (x) aren't the most useful descriptions.

There you go preferring
 a (particular, alternative) coordinate system;
 when I simply convert
 (if needed)
 as option.

On 12/8/2023 at 8:56 PM, swansont said:

In a circular coordinate system you use radial and tangential.

Why is that (suppose to be) better?
I assume you mean simpler.?

On 12/8/2023 at 8:56 PM, swansont said:

An object in orbit is not free;

Thus it can NOT fall.?

On 12/8/2023 at 8:56 PM, swansont said:

it must be in a bound state.

What does that mean (more exactly, please)?

On 12/8/2023 at 8:56 PM, swansont said:

You have to add energy to get it to an arbitrary distance

Yes. Faster (orbit_speed) is a larger radius.
So if things like a sattelite
 go faster,
 then they automatically (ascend)
 & go up
 to a higher radius.

By the same token
 if they slow (down)
 their orbit_speed vc=cir/T
 then they will (automatically)
 decrease their (orbit_)radius.

On 12/8/2023 at 8:56 PM, swansont said:

Freefall just means you are acceleration

accelerating(?)(y/n?)

On 12/8/2023 at 8:56 PM, swansont said:

at the local gravitational acceleration.

?
I assume you mean to say:

Freefall just means
you are accelerating
at the local gravitational acceleration.

But I am still NOT clear
 on what you mean by local
 & how you measure
 that acceleration,
 e.g. what reference
 do "you" use
 (to measure, with).
How is such a measurement done=performed?

On 12/8/2023 at 8:56 PM, swansont said:

The acceleration is centripetal (center-seeking)

Well done!

On 12/8/2023 at 8:56 PM, swansont said:

It's not expressed as an area per time squared.

But its units are so.
Energy too.
(Kilogram) Meters_squared per second_squared.
Should that mean otherwise.
Orthogonality is NOT stated mathematically for a rectangle.

On 12/8/2023 at 8:56 PM, swansont said:

Weightless (especially in this context) means no weight.

Yes, that is (also) my intended meaning for this thread.
(But I also tolerate a NON_zero decrease as well,
 NOT that its usage is wrong,
 because it is NOT,
 but because I am (lazy &) accustomed
 to using inappropriate usage.
E.g. I have bad habits.

On 12/8/2023 at 8:56 PM, swansont said:

Again, we should

Please fill me in
 as to the necessity!,
 if any?

On 12/8/2023 at 8:56 PM, swansont said:

speak of radial, and there is a radial acceleration.

I'm still trying to picture that.
E.g. From what perspective.
You talk about the (circle's) center.(?)

But that tends to erase (some) things
 (perhaps definitions?)
 in my head.

On 12/8/2023 at 8:56 PM, swansont said:

I refer you again to Newton's first law. If there was no

(center seeking)

On 12/8/2023 at 8:56 PM, swansont said:

acceleration, the object would travel in a straight line.

Yes, most likely (I think, perhaps?).

On 12/8/2023 at 8:56 PM, swansont said:

Do you deny the correctness of Newton's laws?

NO. I DON'T think so.
I suspect they are accurate (observations=laws).

(Sidetrack:
 My only beef (=complaint),
 is he (=Newton) did NOT always use them (3 laws),
 although they were the best ((universal) observations).
E.g. for the (~12 h) tides.)

Newton's 1st law
 of inertia
 is, objects
 in motion
 (tend to) stay in motion
 (thus (they) stay moving in a straight line);
 & objects at rest
 (tend to) stay at rest

 (in other words
 they do NOT change
 (what they were doing
 or NOT doing);

 til (Newton's 2nd law)
 they are accelerated
 or decelerated;
 by a (repelling) collision recoil-Newton's 3rd law.

There is a particular detail
 concerning
 the in_line (NO_angle) affect
 which I have forgotten.

Newton's 1st law says (indirectly)
 that there must be a reason
 for any (speed_)change
 that happens;
 a speed_change does NOT happen automatically
 on its own
 for NO reason.
The 1st law is basically
 conservation
 of (average_)momentum.
Or the law of constant(=consistent)_speed.

The 2nd law
 is the law of acceleration.
Particularly linear_acceleration.
I interpret it (=The 2nd law)
 as the average_momentum squared.

On 12/8/2023 at 8:56 PM, swansont said:

No, I'm using physics definitions. If you're going to use physics terminology, you have to use the same definitions.

If you make up your own definitions you can't communicate ideas.

I CAN'T communicate to Physicists
 because they will be the last to understand, otherwise.
(They seem to me,
 to be on their own (isolated) island,
 NOT always good.)

A normal (NON_physicist) person usually gets my drift (faster, easier).

On 12/8/2023 at 8:56 PM, swansont said:

At the location under discussion

Yes, but a ("local") location
 is usually typically
 wrt to some reference
 that is meaning
 2 (different) points,
 NOT just 1;
 otherwise why a distinction.

On 12/8/2023 at 8:56 PM, swansont said:

Why NOT?
Orbit is ZERO vertical_acceleration (observed).

On 12/8/2023 at 8:56 PM, swansont said:

Patently untrue

Then I must conclude (=interpret, centripetal_acceleration)
 zero_seeking
 is (instead) deceleration.

Disclaimer:

I can NOT explain it (=that (radial), vertical orbit position, constant height)
 any other way.

How otherwise
 can you get more
 from less?

How (else)
 can you get a (linear, vertical) "acceleration"
 from a constant (circular_)"speed" v_c=cir/T, cir=2*Pi*r.

Normally (=Typically)
 it is the other way around.
An acceleration will produce a speed
 (on a resting object).
Newton's 2nd law.

On 12/8/2023 at 8:56 PM, swansont said:

Acceleration is a change in velocity.

Yes.

On 12/8/2023 at 8:56 PM, swansont said:

It works in more than one dimension.

Yes, but NOT always.
Sometimes it works in ONLY 1 direction.
But yes 3D automatically includes all 3 dimensions
 (for any "thing",
 in the universe).

On 12/8/2023 at 8:56 PM, swansont said:

Velocity is a vector; it has a magnitude (the speed) and a direction

=angle.

Let us say
 wrt the x_axis,
 e.g. (x,y,z)=(1,0,0).
But that still (also) needs y, too.
?

On 12/8/2023 at 8:56 PM, swansont said:

. If you change direction, there is an acceleration, even if speed is constant.

Yes. (Good example (last phrase).)

On 12/8/2023 at 8:56 PM, swansont said:

If the discussion is about circular orbits, you can't be looking at this in one dimension.

Yes.

On 12/8/2023 at 8:56 PM, swansont said:

A circle has two dimensions

.
Yes.

Edited December 9, 2023 by Capiert

[swansont]

2 hours ago, Capiert said:

That'( i)s new to me.
What then is your definition
 of motion?
Mine is, a change of position
 with time.

A change in position with respect to time is velocity.

 

2 hours ago, Capiert said:

Please explain.

I can NOT imagine acceleration
 without a change
 of position
 (wrt time).

Please fill me in, there.

Toss something up in the air. It's moving up. Later, it moves down. There will be a point in time where it is at the top of its arc, and is motionless. v=0

 

2 hours ago, Capiert said:

(=That (statement) does NOT make sense
 to me.
E.g. I can NOT have a chicken
 without the (prerequisite=)egg, 1st.
I can NOT build(=continue)
 upon NO foundation.)

You can tell if you are accelerating. Even blindfolded. You don't need a point of reference.

That's not true of velocity, which is relative to some frame of reference.

 

2 hours ago, Capiert said:

As I said (=implied),
 I need help there.

So it (=g) is wrt radius r. (y/n?)

g is only an acceleration
(it's called free_fall acceleration),
& it is vertical.

It's vertical in a cartesian coordinate system, where you have a y axis and an x axis.

 

2 hours ago, Capiert said:

(That means:)
It has magnitude, & direction.
(So it must (also) be a vector, too. ?)

It depends. g can be a scalar, 9.8 m/s^2, or you can use it as a vector if you acknowledge the direction, toward the center of the earth

2 hours ago, Capiert said:

That sentence is a head_twister
 for me.
Would you mind explaining
 a bit better.
I assume your 1st "g" is a symbol,
 but your 2nd "g" makes NO sense
 to me.

g is 9.8 m/s^2

If you are not at the surface of the earth, your gravitational acceleration will be different than this value.

It's sloppy to use g as a generic acceleration

2 hours ago, Capiert said:

My examples stated how (my) g varied,
 e.g. wrt height,
 because they were wrt to a reference.
At least I knew (exactly) what & where
 I was dealing with.
But with your definitions
 I go off
 into nirvana
 (because they seem not_founded
 e.g. NOT specified enough,
 or arbitrarily ambiguous).

That's why one should use GM/r^2

2 hours ago, Capiert said:

I DON'T mind
 an extra complication
 if it helps me understand clearly,
 instead of get lost.

The problem being that you are posting this for others to read, and it just begs the question of why you would do calculations one way when you can do it in far fewer steps. It's confusing, and points to a lack of understanding of the concepts.

2 hours ago, Capiert said:

Unnecessary? complicated? maybe for you;
 but NOT for me.]

Seeing as how often you get a wrong answer, that's not readily apparent

2 hours ago, Capiert said:

Considering it (=my syntax)
 is only algebra
 I should NOT (even) need
 to state "average_" every time;
 NOR delta, etc.
 for simple differences.
But I do (have to)
 (just to make the (algebraic) distinction.)
It'( i)s that simple!
 but has become
 that complicated.

Most of the time you use average it's because you've decided to make things more complicated by insisting on using an average value. I might have something to do with your disdain of calculus.

2 hours ago, Capiert said:

 

Naturally, NOT falling.
Standing, was (ONLY) an analogy
 (to an orbit equivalent).
Take it or leave it.
I find it useful.

Orbit defines a specific set of conditions. Standing on the earth is not an orbit.

If you make up your own definitions of terminology you can't communicate with others

2 hours ago, Capiert said:

Yes. Ellipse, etc.

You miss the point. A circular orbit has more conditions than moving in a circle.

 

2 hours ago, Capiert said:

That'( i)s why I kept it simple KIS
 to start with ONLY circular_motion.
Once the basics have been established correctly,
 the (complexity) details may be (included later &) improved upon.

Then get the basics right.

An orbit means the gravitational force is what is keeping it moving in a circle. Not standing on the ground.

 

2 hours ago, Capiert said:

 

There you go preferring
 a (particular, alternative) coordinate system;
 when I simply convert
 (if needed)
 as option.

The choice of coordinate system seems logical given the problem.

You can use x and y if you want, but then there is motion in both the x and y directions, which varies over time. The speed remains the same; v^2 =v_x^2 + v_y^2

(gosh, that's the equation of a circle!)

 

2 hours ago, Capiert said:

Why is that (suppose to be) better?
I assume you mean simpler.?

Simpler and more descriptive. If you are trying t describe a satellite's orbit, vertical only works when it's directly overhead. At any other time "vertical" doesn't work.

 

2 hours ago, Capiert said:

Thus it can NOT fall.?

What does that mean (more exactly, please)?

"free" and "bound" have definitions in physics. You would do well to learn such terminology.

A free particle is one that could get infinitely far away without being subject to new force. A bound one cannot.

In terms of energy, KE + PE < 0 for a bound particle

 

2 hours ago, Capiert said:

Yes. Faster (orbit_speed) is a larger radius.

No, it's not. For a circular orbit, v = sqrt(GM/r)

As r gets larger, v decreases

 

2 hours ago, Capiert said:

So if things like a sattelite
 go faster,
 then they automatically (ascend)
 & go up
 to a higher radius.

The dynamics of it is more complicated than that, actually.

You add energy, but this goes into increasing your potential energy (it's negative, but gets smaller in magnitude) and your speed decreases. KE goes down, but PE increases twice as fast.

2 hours ago, Capiert said:

 

accelerating(?)(y/n?)

?
I assume you mean to say:

Freefall just means
you are accelerating
at the local gravitational acceleration.

Yes

2 hours ago, Capiert said:

But I am still NOT clear
 on what you mean by local
 & how you measure
 that acceleration,
 e.g. what reference
 do "you" use
 (to measure, with).
How is such a measurement done=performed?

Local means where you are.

You can measure it with a spring scale. If you know your mass, the spring scale tells you the apparent weight, and from that you can get the acceleration.

If you want to know the rotation (which is another form of acceleration), you can use a Foucalt pendulum.

2 hours ago, Capiert said:

 

But its units are so.

The SI units of acceleration are meters/ second^2

There's no area

2 hours ago, Capiert said:

Energy too.
(Kilogram) Meters_squared per second_squared.

The m^2 in Joules does not represent an area

 

2 hours ago, Capiert said:

 

I'm still trying to picture that.
E.g. From what perspective.
You talk about the (circle's) center.(?)

A satellite is oin circular motion. The center of that circle is the center of the object it's orbiting.

 

2 hours ago, Capiert said:

The 2nd law
 is the law of acceleration.
Particularly linear_acceleration.
I interpret it (=The 2nd law)
 as the average_momentum squared.

Force is tied in with momentum, not momentum squared.

 

2 hours ago, Capiert said:

 

Then I must conclude (=interpret, centripetal_acceleration)
 zero_seeking
 is (instead) deceleration.

The acceleration in circular motion does not change the speed, it only changes the direction (velocity is a vector. Changing the direction of motion requires an acceleration)

 

2 hours ago, Capiert said:

Disclaimer:

I can NOT explain it (=that (radial), vertical orbit position, constant height)
 any other way.

How otherwise
 can you get more
 from less?

There is no more, no less. The speed is constant.

 

2 hours ago, Capiert said:

How (else)
 can you get a (linear, vertical) "acceleration"
 from a constant (circular_)"speed" v_c=cir/T, cir=2*Pi*r.

You don't. It's not a linear (i.e. one-dimensional) system, and the direction is not vertical. The direction is "toward the center of the circle" (i.e., it's radial) 

 

2 hours ago, Capiert said:

Normally (=Typically)
 it is the other way around.
An acceleration will produce a speed
 (on a resting object).
Newton's 2nd law.

If the force is perpendicular to the velocity, it will only change the direction. No work is done, so there is no change in KE