# Gravity g's, missing term(?)

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STOP!
Under Construction

g=2*h/(t^2)-2*vi/t.

It looks (to me)
like that term -2*vi/t

has been missing
for a long time.

Surely
know it;
but do you use it
?

For

(fallen) height h
time(_difference) t
initial_speed vi.

The free_fall acceleration
g~-(Pi^2) [m/(s^2)]+ac
g=~-9.8 [m/(s^2)]
is (instantaneous,) linear, & negative;

(the bothersome details:)

(=but it) (is) slightly) reduced
by the (Earth's daily_rotational)
centrifugal_acceleration
ac=(vc^2)/r

having the (squared)
Earth's (surface) circumferential_speed
vc=cir/T

with (the Earth's) circumference
cir=2*Pi*r

using ((your) Earth's) per radius r (position)

& (sidereal_)day (time_)period T~23 56 [min] 4 [sec]
in seconds;

---

Disclaimer:

there is to it (=g, overview),
when (also) observing
& position.

(Air friction
would have to be considered;
but I'm NOT going to bother
for in(side) a vacuum.)

E.g. Naturally
(all) those (extra) details
(for the centrifugal_acceleration ac, etc.)
can be determined.

However,
I'( a)m concerned (here)
with (mostly

only) the initial_speed's vi term.

(E.g. A term that would (also) occur (for you)
in the ((total) work_)energy
WE=F*d
if you ((were to) also) included the (integration) limits
from zero to the initial_speed;
as a different reference (frame, choice)).

(Which can be determined, graphically
with a plot:
speed versus time.

The (plot's) area is the distance.)

It's easier (for me)
to use positive

---

Freefall:

If I let a pebble (stone or ball) fall
(in a vacuum)
(from rest, (with) initial_speed vi=0 [m/s], at time t0=0)

---

CAUTION:
From here on
this website's software
has WRONGLY Strikedthrough
the rest of my text, automatically.

until someone
helps me undo the Strikedthough.

Please would someone help me fix this?

Marking & reclicking Strikethrugh
does NOT undo Strikethrough
like it would for Bold, italics or underline.

What is wrong with your Software?

---

then in (time(_difference) t1=)1 second
it will (have) fall(en) the height(_difference) h1=-4.9 [m]
& have the final_speed vf1=-9.8 [m/s].

(I could also reverse that
& say:
If I let a stone (pebble) fall
(from a height h1=)4.9 [m] (wrt the ground),
-4.9 [m]
(to the ground),
then it will hit the ground
in (time t1=)1 second
& impact (ruffly) at a (final_) speed vf=-9.8 [m/s].

We know freefall
is "linear" acceleration,
so in time(_difference) t2=2
fallen_height h2=-9.8 [m]
& final_speed vf2=-19.6 [m/s].

etc.

E.g.

we could have used
the 1st final_speed (vf1)
(from the 1st (time_(difference)) second (t1))
as the 2nd initial_speed vi2=-9.8 [m/s]
(e.g. as if thrown)
& allow the pebble to restart (accelerating)
as if from zero in initial_speed vi1=0 [m/s],
but for: (only) the 2nd (time_(difference) t(2-1)=1) second;
its (=the pebble's) additional fallen_height h(2-1)=-4.9 [m];

So, the (total) fallen_height h2=h1+h(2-1)=-4.9 [m]+(-4.9 [m])=-9.8 [m];

& the (total) final_speed vf2=vf1+vf(2-1)=-9.8 [m/s]+(-9.8 [m/s])=-19.6 [m/s].

The fallen height (taken from a graph),
h=vi*t+g*(t^2)/2
is only 2 terms:
the constant (initial_)speed term vi*t;
& the accelerating term g*t*t/2.

The speed(_difference)
v=g*t
is the (freefall) acceleration g
multiplied by time t.

But
the accelerating term
g*t*t/2=(v/2)*t
is "half"
of that speed(_difference) v(=vf-vi)
multiplied by time t

h=vi*t+(v/2)*t
h=(vi+v/2)*t, & va=(vi+v/2) is the average_speed
h=va*t.

The average_speed
va=h/t
is the fallen height h(=d distance)
per time(_difference) t.

(The final_speed
vf=vi+v
vf=((vi^2)+g*h*2)^0.5
)

The (linear) average_acceleration
ga=h/(t^2). <----That'( i)s Important! So simple, NO other terms!
ga=va/t
ga=

The freefall gravitational "average_(linear)_acceleration", (&) is
ga=vi/t+g/2, swap sides
vi/t+g/2=ga, -vi/t
g/2=ga-vi/t, *2

then the (instantaneous) linear freefall gravitational acceleration, is
g=2*(ga-vi/t).

Note: Syntax
(my) ga=g_a (yours).

Those are the conversions

to & from linear
instantaneous versus average:

speeds; & accelerations.

Again:

The average_speed
va=h/t
va=vi+(v/2).

The (linear) average_acceleration
ga=h/(t*t)=va/t
ga=vi/t+g/2.

The (average_)time
t=h/va
is the height h
per average_speed va; or
t=(h/ga)^0.5
the rooted:
height h;
per (linear) average_acceleration ga.

It'( i)s that simple.

(Algebra.)

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55 minutes ago, Capiert said:

g=2*h/(t^2)-2*vi/t.

It looks (to me)
like that term -2*vi/t

has been missing
for a long time.

Missing from what? This not an equation for g. It comes from a different equation, rearranged, which you have not shared.(i.e. g is typically not an unknown in kinematics problems)

IOW, you are skipping some necessary steps in the discussion.

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16 minutes ago, swansont said:

Missing from what?

From g.
You obviously disagree,
otherwise you would NOT have asked.

Quote

This not an equation for g.

Why NOT?
It works.

Quote

It comes from a different equation, rearranged, which you have not shared.

(As far as I know) I have shared the equation:
the fallen height (taken from a graph),
h=vi*t+g*(t^2)/2.

Quote

(i.e. g is typically not an unknown in kinematics problems)

Typical is NOT all cases,
but instead most (e.g. a majority).

For me it was an unknown
that I wanted to find,
e.g. experimentally.

with unstrikethrough my text (above)?

Edited by Capiert
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23 minutes ago, Capiert said:

From g.
You obviously disagree,
otherwise you would NOT have asked.

Why NOT?
It works.

(As far as I know) I have shared the equation:
the fallen height (taken from a graph),
h=vi*t+g*(t^2)/2.

Typical is NOT all cases,
but instead most (e.g. a majority).

For me it was an unknown
that I wanted to find,
e.g. experimentally.

with unstrikethrough my text (above)?

I'm not quite sure what this is about but regarding the effect on g of the Earth's rotation, this once came up elsewhere and I worked out that even at the equator, the reduction in apparent g due to the Earth's rotation was only about 0.3%. It went as follows:

- The radius of the Earth is ~4000miles, so its circumference, say at the equator is 2 x π x 4000 which comes to ~25,000miles.

- The tangential speed at the equator is therefore 25000miles/day, which is close to 1000mph.

- The centripetal acceleration needed to keep an object moving with tangential velocity v, in a circle of radius r, is v²/r. So, for an object at the equator, the apparent centrifugal acceleration it experiences, counter to the acceleration of gravity, will be 1,000,000/4000 =  250 miles/hr².

What we need to know is how this acceleration compares with g, the acceleration due to gravity, which is about 10m/sec². To do that, we need to get this result into the same units as g is quoted in:-

1 mile is ~1600m. And 1hr is 3600 seconds. So 250 miles/hr² becomes 250 x 1600/(3600)² = 25 x 16/(360 x 36) = 25 x 4/(360 x 9) = 100/3240 = ~ 0.03 m/sec².

So the centrifugal force at the equator, as a proportion of the force of gravity, is of the order of 0.03/10 =0.003, or 0.3%.

At higher latitudes it will be less, presumably by a factor of the cosine of the latitude, falling to zero at the poles.

Edited by exchemist
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1 hour ago, Capiert said:

(As far as I know) I have shared the equation:
the fallen height (taken from a graph),
h=vi*t+g*(t^2)/2.

1. No, you did not post this equation

2. vi*t is clearly present in it.

So what is missing?

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1 hour ago, swansont said:

1. No, you did not post this equation

That surprises me
because
I can read the following (from above)

The fallen height (taken from a graph),
h=vi*t+g*(t^2)/2
is only 2 terms:
the constant (initial_)speed term vi*t;
& the accelerating term g*t*t/2.

& also in my posted (above, 2nd of 2) .pdf's (above).

Surely you will be able
to determine
when that (equation) was posted.

1 hour ago, swansont said:

2. vi*t is clearly present in it.

Good! (for that product).

But now for the quotient -2*vi/t
in  g (alone).

1 hour ago, swansont said:

So what is missing?

Someone to assist me
in getting
the WRONGLY strikedthrough text (above)
to be NOT strikedthrough, anymore.

Would you help?

Again, clearly -2*vi/t was NOT present in g (before my threads).
I have NOT seen that term in books;
but algebraically
it is correct
(as to how I gave it to you).

Science should confirm itself,
but it seems
(to me)
you (might) probably prefer NOT
that the g equation be intact,
when rearranged
(e.g. if you leave out that term -2*vi/t)?

(Is that possible?)

I have only moved the acceleration g
to the left side,
& all other terms to the right side.

That should give
the correct g (equation).

(Otherwise, the (general) equality
is either destroyed,
or distorted.)

It'(=What I have done i)s NOT complicated.

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28 minutes ago, Capiert said:

That surprises me
because
I can read the following (from above)

The fallen height (taken from a graph),
h=vi*t+g*(t^2)/2
is only 2 terms:
the constant (initial_)speed term vi*t;
& the accelerating term g*t*t/2.

Sorry, you didn’t post it before the other equation. I only read the first section, because of the nonsense value. Plus I would’ve skipped the strikethrough text anyway.

28 minutes ago, Capiert said:

Again, clearly -2*vi/t was NOT present in g (before my threads).
I have NOT seen that term in books;
but algebraically
it is correct
(as to how I gave it to you).

Clearly this is bollocks.

h=vi*t+g*(t^2)/2 (or an equivalent equation) is standard fare in any physics textbook that presents kinematics equations.

top left of the “big 4” https://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations

and so on...

(you can also notice that nobody is presenting this as an equation for acceleration)

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11 hours ago, swansont said:

Swansont:
Sorry, you didn’t post it before the other equation.

Sorry, that'( i)s NOT true!
It's in other threads before this 1
Is half a year too early for you?

Swansont: (Kinetic_Energy is already relative, 2020 11 23 16:26)

What valid physics principle is this based on?

Capiert:  (Kinetic_Energy is already relative, 2020 11 23 16:28)

Mechanics:
Gravity's (free_fall, linear_acceleration a=g)
fallen_height
h=hf-hi=vi*t+g*t*t/2
for time t
& initial_speed vi
gives
final_speed
vf=((vi^2)+g*h*2)^0.5

11 hours ago, swansont said:

I only read the first section, because of the nonsense value.

What value
do you mean;
& why?

Maybe you missed something (important)?

If you mean
g=2*h/(t^2)-2*vi/t

You do NOT sound
completely connected.

11 hours ago, swansont said:

Plus I would’ve skipped the strikethrough text anyway.

That is quite a natural reaction.

But you have NOT made a suggestion
how I can fix
No response to my question (request)
as neither: yes; NOR NO. =NO!
Makes it (=my request) sound (=seem) rhetoric.

11 hours ago, swansont said:
11 hours ago, Capiert said:

Again, clearly -2*vi/t was NOT present in g (before my threads).
I have NOT seen that term in books;
but algebraically
it is correct
(as to how I gave it to you).

Swansont:

Clearly this is bollocks.

h=vi*t+g*(t^2)/2 (or an equivalent equation) is standard fare in any physics textbook that presents kinematics equations.

Yes, we (all) know that.

11 hours ago, swansont said:

top left of the “big 4” https://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations

and so on...

(you can also notice that nobody is presenting this as an equation for acceleration)

So (I guess) I am (1 of) the 1st!
to present
the gravitational_acceleration (formula)
g=2*h/(t^2)-2*vi/t
as exactly so
(=in its corrected form).
P.S. (But) I can NOT believe
I am the ONLY 1
who has ever done that,
or tried.

You (only) confirm
that I can NOT find (exactly) this ("g=2*h/(t^2)-2*vi/t") in books.

I have NOT seen that term
(-2*vi/t, exactly in that position)
in books.

thus I want to present (either) something new,
or an improvement against a flaw
or weakness.)

My complaint is (from),
Capiert: (Kinetic_Energy is already relative, 2020 11 23 16:28)
How do we know that acceleration a=x/(t^2)?

Better said: If I let (distance) x=h (height), instead;
then:
How do we know that acceleration a is (suppose to be) h/(t^2)?

We do NOT know that (at all)!
I do NOT get that formula, that you claim. (It is NOT possible (for me).!)
Especially when the height h=d distance NEEDS a factor "2".

g=2*h/(t^2)-2*vi/t".

will get
g#h/(t^2)
h#g*(t^2).

But

a~2*x/(t^2) is the minimum requirement. E.g.
x~a*(t^2)/2.

& where is the vi*t term
in that?
NOWHERE!
So (thus) it is NOT the (EXACT) general (equality) formula

of "standard fare in any physics textbook.."
NOR "an equivalent equation", thereof;
BUT instead a distortable (mere) approximation.

(I can NOT rely on it for (my) other derivations.)

You may be happy with such carelessness;
but I can NOT be.

It (=That formula: whether e.g. a#x/(t^2), or e.g. x~a*(t^2)/2)
is either an equation (=equality);
or (else) it is NOT.

& there it is definitely NOT (an equation).

11 hours ago, swansont said:

Swansont: (Kinetic_Energy is already relative, 2020 11 23 16:26)
If I drop a mass from some height, how fast will it be moving when it hits the floor?

Capiert: (Kinetic_Energy is already relative, 2020 11 23 16:28)
the final_speed is
vf=((vi^2)+g*h*2)^0.5 .

But you did NOT like that (answer),
perhaps because
its result
is independent
of mass m?

(It (=That answer) does NOT NEED mass.)

(But it is also new to me
that algebra
is NOT suppose to be math,
if it is?

Perhaps you wanted some number( value)s
of something
you can easily do yourself;
but doubted my ability. ?
E.g. Mistakes happen. Typos.)

I was only trying
, (preparing)
while you closed.

That (=your impatience) does NOT seem fair.
(Especially when you said there is NO time_limit.)
(Mordred also stated to (someone) NOT to answer
if you do NOT know.
I also wanted to check some details
Thus my hesitation=delay.
There are still (some) things I need to clear.
However:)

For that equation
vf=((vi^2)+g*h*2)^0.5
we typically set the initial_speed vi=0 [m/s]
if (=when) we only drop,
g=9.8 [m/(s^2)],
& e.g. let the height h=1 [m]

vf=((0^2)+9.8 [m/(s^2)]*1 [m]*2)^0.5
vf=4.4 [m/s].

E.g. let the height h=2 [m]

vf=((vi^2)+g*h*2)^0.5
vf=((0^2)+9.8 [m/(s^2)]*2 [m]*2)^0.5
vf=6.3 [m/s].

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6 minutes ago, Capiert said:

So (I guess) I am (1 of) the 1st!
to present
the gravitational_acceleration (formula)
g=2*h/(t^2)-2*vi/t
as exactly so
(=in its corrected form).
P.S. (But) I can NOT believe
I am the ONLY 1
who has ever done that,
or tried.

Don’t injure yourself patting yourself on the back. Rearranging the equation is algebra, so any high-schooler should be able to manage that.

I am focused solely on your claim that vi is missing from the equation, when that’s clearly not true.

11 minutes ago, Capiert said:

Capiert: (Kinetic_Energy is already relative, 2020 11 23 16:28)

If you’re going to reference an earlier thread, you should post an actual link, rather than give a time/date.

17 minutes ago, Capiert said:

How do we know that acceleration a is (suppose to be) h/(t^2)?

We do NOT know that (at all)!
I do NOT get that formula, that you claim. (It is NOT possible (for me).!)

You should learn calculus. a = dv/dt

Integrate twice, applying boundary conditions, and you get that result (as would many thousands of people who can do the math)

That you can’t do it in no way implies that the equation is wrong.

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3 minutes ago, swansont said:

Don’t injure yourself patting yourself on the back.

I'll do my best!  Thanks.
(I got a good chuckle (out of that).)

3 minutes ago, swansont said:

Rearranging the equation is algebra, so any high-schooler should be able to manage that.

Yes! That is my intention.
It's NOT complicated!
It's easy.

3 minutes ago, swansont said:

I am focused solely on your claim that vi is missing from the equation, when that’s clearly not true.

vi included: is the general equation.
But without vi: is a limited (specific example) formula,
that can NOT work for all cases.

3 minutes ago, swansont said:

If you’re going to reference an earlier thread, you should post an actual link, rather than give a time/date.

Maybe (other_options) sharing can help linking, a bit.?

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4 hours ago, Capiert said:

You should know that requiring this is not in compliance with the rules.

4 hours ago, Capiert said:

vi included: is the general equation.
But without vi: is a limited (specific example) formula,
that can NOT work for all cases.

Yes. That’s what you do - write down the general equation and eliminate the terms that are zero in a particular problem. In the other thread I expressly said dropping from rest, making vi = 0. There’s no reason to continue to include it for that problem. There’s no implication that it applies to all cases once you have imposed such restrictions

4 hours ago, Capiert said:

Locked threads also aren’t supposed to be resurrected

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• 1 month later...

On 5/25/2021 at 6:42 PM, Capiert said:
Quote

I am focused solely on your claim that vi is missing from the equation, when that’s clearly not true.

vi included: is the general equation.
But without vi: is a limited (specific example) formula,
that can NOT work for all cases.

On 5/25/2021 at 7:55 PM, swansont said:

Yes. That’s what you do - write down the general equation and eliminate the terms that are zero in a particular problem. In the other thread I expressly said dropping from rest, making vi = 0. There’s no reason to continue to include it for that problem. There’s no implication that it applies to all cases once you have imposed such restrictions

What I mean is:
(vi was missing from the "general" equation,
(NOT (just a) formula.)

How do I know
when I have
the general equation
,
(if or when I start
with only (limited (specific example) formula) fragments)?

I'm searching for the general formula.

I (attempt(ed) to) maintain vi (initial_speed)
to try
to NOT get lost
((&) for other things (=projects, concepts, extrapolations)).

(You (may) think)
you do NOT need it (=vi), (?) fine(!);
but I do.

E.g.
vi remains
an invisible (hidden (excluded)) term
for you(r)
speed(_difference)
v=vf-(vi).

It's always there.

Edited by Capiert
An example.
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2 hours ago, Capiert said:

What I mean is:
(vi was missing from the "general" equation,
(NOT (just a) formula.)

I provided three links that shows this to be in error. Have you provided any examples of it being "hidden"?

Quote

How do I know
when I have
the general equation
,
(if or when I start
with only (limited (specific example) formula) fragments)?

I'm searching for the general formula.

Usually it's derived with few constraints to it, and no initial conditions.

e.g. start with a = dv/dt, which is the definition of acceleration (always true)

Add in the condition that acceleration is constant, and integrate the equation and you get v = v0 +at

Apply the definition of velocity, v = dx/dt

Integrate again and you get s = v0t + 1/2 at2

The initial velocity is there (as it is in so many textbooks, if you'd only bother to look), and the only restriction is that you must have a constant acceleration

Quote

vi remains
an invisible (hidden (excluded)) term

Only if you forget to include it, but that's on you

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