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**Flat Earth General / Re: Is "g = 9.8 m/s/s = GM/d^2 bona fide on the surface of earth?**

« **on:**August 17, 2021, 03:22:14 AM »

Quote from: E E K 58

I suspect the reason you failed to answer my question is because of the answer : No, you cannot provide an intelligible explanation in English physics as to how multiplication is one of the natural fundamental issues that the gravity model has.Quote from: Amoranemix 57I can't find an intelligible explanation of that. Can you provide that explanation in English physics ?I would suggest using Google translate.

I am not seeing any service on Google translate to explain your claim intelligibly in English physics. Why don't you let Google do that and then paste the explanation here ?

Quote from: E E K

You forgot to answer my questions.Quote from: Amoranemix 571. Why would the person fall down ?[no response]

2. I don't understand the setup.

3. Why would the table fall ?

Quote from: E E K

I don't have a lower grade student available.Quote from: Amoranemix 57Consult a local student of lower graders/3rd grader is fine!!!!Quote from: E E K 46Counting like 10 p * 3 o/p = 30 o is wrong - sorry about misrepresentaion of orange.personSo you claim, but can you prove that ?

Ask if items of different categories are multiplied? For example pencils, trees, stars, aliens etc.

Suppose such student would tell me what you want them to, which I doubt. Then what ? Are you suggesting that the authority of lower grade students in matters of physics is so great that would should accept their claims as proof ?

Quote from: E E K

[9] Communication relies on clear, unambiguous conventions. One meaning being conveyed in more ways makes study of the conventions more laborious and the choice more error prone. For example, someone following the prevalent convention might falsely think that with s/m you mean seconds per meter.Quote from: Amoranemix 57Why would it be confusing? Its just you are not used to it.[9]Quote from: E E K 53And why do the scientists stick to the definition of velocity as a meter per second when it can be written as a second/meter “time /distance”. And acceleration as changes in time /meter/meter.[7][7] Writing a velocity, which has the dimension of distance / time, as time / distance would be confusing.

So what would be the new form of F=GMm/d^2 where g =GM/d^2?[8]

[8] The form would be unchanged. Formulas are independent of how the units of the variables and constants are written.

Using the second concept, the form of F=GMm/d^2 is invalidated due to the unit of g=GM/d^2 (s/m^2).[10] For, G is asserted just to get the unit of g in the first concept, not natural.

The unit of g in the first concept is m/s/s or m/s^2. Inverse is s^2/m

The unit of g in the second concept is s/m/m or s/m^2[11]

So s^2/m is not equivalent to s/m^2[12]

[10] What is that second concept you are referring to ?

In the convention used by almost all scientists the unit of g is written as m/s^2, not s^2/m.

[11] I am confused. Does the problem lie with the second concept or with the convention you use ?

[12] You are the only one who seems to have suggested otherwise.

Quote from: E E K 58

[a] Those beings must have phenomenal measurement equipement in order to 'see' that.Quote from: Amoranemix 57[1] You are claiming that a falling feather would be accellerated differently than a falling hammer. In what way ? What evidence can you present that the earth's gravitational accelleration depends on the mass that is being accellerated ?No offense but below is only for your clarification.

[2] Greater masses mean greater accelleration, which reduces the distance faster, which again increases accelleration. However, when talking about g (of earth), the mass of the falling object (A) is assumed tiny compared to earth's. Hence, the earth's accelleration is tiny.

[3] In the case of an object falling on earth one microsecond later the accellation would be exactly the same. I am sure Newtin knew that.

There are three accelerations due to gravities involved. The gm of moon, the gh of hammer, and the gf of feather.

The gm >>>>>>>>>> gh >>>>gf

A tiny being on the surface of hammer sees moon falling on hammer due to gh

A tiny being on the surface of feather sees moon falling on feather due to gf [a]

A normal being on the surface of moon sees falling both hammer and feather on moon due to gm

Note down the striking time of moon and hammer if hammer is dropped from height h. let it be t1

Note down the striking time of moon and feather if feather is dropped from height h. let it be t2 [13]

You know t2 would be >>>> t1 – Reasons; The gm >>>>>>>>>> gh >>>>gf [14]

Fall both the hammer and feather simultaneously from h

All strike one another at t1. The falling acceleration of the feather is disturbed by the falling acceleration of moon due to gh. Moon falls on feather at gh not gf.[15]

Here striking is ignorable but it cannot be ignored if the sizes of the hammer increase to a notable size.

If the theory is not satisfactory for small masses then how it can be true for bigger masses/celestial–solar systems.[16] Sorry it might hurt you but its true.

[13] You failed to mention the time at which the hammer and feather were released. If they were both released at the same time t0, then t1 = t2.

[14] You are mistaken, for I don't know that. gh and gf are irrelevant due to gh, gf <<<<<<<<<< gm.

Jackblack did the math in post 59. He calculated t as a function of x and h. Translated to your example, t stands for arrival times t1 and t2, x stands for h (the height) and a stands for accelleration gm.

So, we have

t1 = sqrt(2*h/gm) = t2

[15] That is correct and the impact on t1 and t2 is the same and negligible.

[16] For one thing, truth and satisfaction may not be related. Relevance ? I haven't seen anyone claiming the theory is not satisfactory for small masses. I certainly haven't.

Quote from: E E K 58

[17] I take it that what you mean is that G contributes to g and that since G has a unit of time, so does g. OK.Quote from: Amoranemix 57- g got its unit due to the presence of G in g=GM/d^2. There is no actual variable of time or unit of time in equation of F=GMm/d^2 or g=GM/d^2. G is used just to get the unit of g.[17]Quote from: E E K 36You know all the variables of M, m G, g, F and d but still can’t figure out nascent velocities, d, F and g. This means either the theory is incomplete or there is something wrong with it.

The unit of time used in the unit of g is borrowed from G but we need the actual variable of time for calculation of distance covered which this theory lacks.

How is the unit of time of g borrowed from G ?

What do you mean with actual variable of time ? There is only one variable of time in Newton's theory of gravity and only one is needed.

This formula doesn’t work in the second concept as G is not natural but fitted for the unit of g.[18]

Read the following; there is no consistency in the unit of mass. How do we know that masses of atomic particles are correct?

#13 - Is the definition of UNIT [gram, kilogram] of MASS absolute or iffy?

https://en.wikipedia.org/wiki/Talk:Gram#Is_definition_of_UNIT_[gram,_kilogram]_of_MASS_absolute_or_iffy?[19]

[18] So what ? No one claims that formula works in the second concept.

[19] What are you talking about ? Are you annoyed by the fact that not everyone agrees on what a gram is ? What does that have to do with the theory of gravity or with the shape of the earth ?

You claimed that the theory of gravity is either incomplete or that their is something wrong with it. You are right. Even the general theory of relativity is not a complete theory. As for a calculating distances covered, whether the ways to do that are considered to be part of the theory of gravity is nit-picking. It is inappropriate to blame a theory for not doing what is not supposed to do. If you dislike that omission, then just include the methods to do that in the theory.

Quote from: E E K 58

[20] So you claim, but can you prove that ?Quote from: Amoranemix 57[4] The existence of gravitational potential wells is independent of their worthwhileness.No second mass, no gravitational wells for a lone object in the universe.[20] Again d is o/c distance b/t two masses in the equation of g=GM/d^2[21], g=0 when d=0.[22] Here d is not the radius of the earth. There is potential for g to exit upon the presence of second no matter what its mass is – This should be the case. I just favor you

[21] There is only one mass in g = GM/d/d. That is not a coincidence.

[22] You are mistaken, g would be infinite if d = 0.

Quote from: E E K 58

You forgot to answer my question.Quote from: Amoranemix 57What does any of that have to do with the earth allegedly being flat ?[no response]

Quote from: E E K 60

Indeed. Contrary to what you claimed, a single mass generates a gravitational potential well, but in order for that well to affect a second mass, a second mass must exist.Quote from: JackBlack 59Repeatedly asserting the same nonsense won't magically make it correct.I made it clear earlier that g = GM/d^2 is independent of falling therefore one can calculate g at any height w/o the presence of a second mass. It doesn’t mean that lone mass has all those values in reality. The potential word is used for these values of g. The activation of these requires the presence of a second mass.

The gravitational well exists regardless of if another object exists to be acted upon by it.

d is the distance to the object.

This applies for any point, regardless of if an object is there or not.

Repeatedly ignoring that fact will not change it.

Quote from: E E K 60

JackBlack has already calculated it for you :Quote from: JackBlack 59No, it is simply a different unit.So kids are being misled.

There are 10 students in a class. Each student has 3 oranges. I can say 1 student per 3 oranges instead of saying 3 oranges per student - 1 student for 3 oranges/ 1 student to 3 oranges/ 1 student for every 3 oranges.

How many oranges are there in total? - Please.

10 students times 3 oranges per student = 30 oranges.

In this case 3 oranges per student must converted to mathematics as 3 oranges divided by one student.

Quote from: E E K 60 to JackBlack

Come to the real picture on which universal law is based for two objects A and B of discernable sizes in order to address the issue of misunderstanding.[23] Higher types of motion ? What are those ?

As soon as both objects A and B start moving towards each other simultaneously, their inceptive accelerations “gs=gees” start facing challenges due to latent change in o/c distance “d” (equal or unequal on either side) b/w them. Both A and B have to face this change in d every moment till they strike each other. Both A and B adopt higher types of motion right at their starting points due to such changes in d.[23] Of course, g is there in gravitational jerk,/jolt, gravitational snap/jounce, etc. So g in the equation of F = GMm/d^2 where g = GM/d^2 and = Gm/d^2 are not best fit due to the involvement of higher types of motion.[24]

What evidence can you present that A and B adopt those right at their starting points ?

[24] So what ? No one claimed g are the best fit.

Quote from: E E K 65

The injustice of gravity exposed.Quote from: JackBlack 64A better way to determine mass is inertia.How?

This inertia property of the falling mass is also ignored in gravity.

A falling body has never got a chance to use its inertia against g.

Quote from: E E K 87 to JackBlack

All above shows center to center distance in F=GMm/d^2 doesn’t work – Right[25][25] How so ?

he concept of the gravitational acceleration of the earth came in by the idea of rolling a ball on an inclined plane. Newton tried to grasp it and finally formulated it. Einstein didn’t introduce new things but shapes the same idea of acceleration. He used the same G and other basics fundamentals not different. So if newton is wrong then Einstein as well.[26]

[26] How is that supposed to follow ?

Quote from: E E K 98

Furthermore, it is said that the gravitational field inside shell is zero. This means G is zero inside the shell.[27] If G is zero then how come a mass of shell is concentrated at its center and attracts things towards its center.[27] The gravitational field inside a hollow sphere is 0. That implies g, not G, is 0 inside the sphere.

Doesn't an object inside the shell fall on the gravitating mass located outside the shell?[28]

Gravitational law fails when an object achieves escape velocity.[29]

[28] Indeed, it doesn't.

[29] What evidence can you present to support that claim ?

Quote from: E E K 106

I assumed the escape velocity was constant forever but anyway, how about if we let the escape velocity/speed (11,186 m/s. or greater) constant with hypothetical auto adjustment of thrust but not forever as a point will come in the future where it requires almost or congruent to zero thrusts.[30] At great enough distance the pull of earth's gravity (and thus the impact of G) becomes negligible. At infinite distance the pull is 0.

Does G still affect the above escaping object if yes then the object is not escaping?[30]

In a nutshell, G is zero for an object in escaping mode. It may be for some time or forever.[31]

[31] Below the escape velocity, eventually gravity (and thus G) wins. Above escape velocity, the fleeing object wins. Earth is still pulling, but not hard enough.

Quote from: E E K 109

[32] Apparently most scientists have a different view than yours on what constitutes escaping. They would consider an inmate to have escaped when he left the prison compound and authorities are unaware of his whereabouts. Even though it may bother the escapee that the authorities are still looking for him and even though he may have keep a low profile to keep his freedom, he still has escaped.Quote from: JackBlack 108No, there is a single escape velocity.in your case all escaping velocities even if greater than the min are also influenced by gravity.

This velocity is the minimum velocity required. Any velocity above that velocity will allow the object to continue moving away.

I understand what you say but no influence of gravity means, no influence of G on escaping objects. The escaping object should be free from gravitational worries otherwise it is not escaping.

Again, any velocity if reduced or turn into a declaration should not be considered as escaping velocity from the grip of gravity.[32]

Escape velocity is more authentic if constant forever.[33]

The escape velocity is independent of the escaping mass. It doesn't work vice versa.

Smaller masses may escape from bigger but bigger ones can’t escape from smaller masses. For example, the earth would take the smaller mass along with it if ever tries to escape from a smaller mass.[34]

[33] Scientists prefer to use useful defenitions. Definitions about things that (can) exist tend to be more useful than those about things that can't. There is no velocity with the property of staying constant forever.

[34] That idea seems to lack potential applications, but the earth's escape volocity, assuming the small object is not fixed, would also be 11,186 m/s.

Quote from: E E K 112

Similarly, if Newtonian gravity were really true then the phenomenon of temperature, which is related to the average kinetic energy of atoms and molecules in a system, would not happen. And the same is applied to pressure.[35][35]What evidence can you present to support those claims ?

How do these atoms and molecules of a system gain motion (K.E) when each individual particle of the system is accelerated @ GM/d^2 towards the center of the earth?[36]

[36] You tell us. No one has claimed these atoms and molecules of a system gain motion.

Atoms can bounce of each other, like a ball doesn't keep accellerating until it reaches the earth's center, because usually it bounces of the surface first.

Quote from: E E K 116

Similarly, elasticity is to bounce therefore why would they be bouncing in the first place (not the real reason) in a random direction?[37] Any object if thrown upward in any direction comes back to the surface of the earth and is finally settled.[38] Therefore all those atoms and molecules should settle now even if they were bouncing. Enough spaces among similar charge particles are included in the gravitational settlement. Keep in mind all masses fall at the rate. Regarding pressure, doesn’t liquid transmit pressure equally in all directions?[39] The g of individual particles nullifies pressure.[40][37] That is due to elektron clouds repelling each other due to the elektromagnetic force.

[38] What evidence can you present to support that claim ?

[39] Indeed it does and so does a gas. However, the pressure at the top of a body is lower than at the bottom.

[40] It is true that in an atmosphere gravity and atmospheric pressure compensate each other for the athmospheric molecules.

Quote from: E E K 116

When A collides with B. A loss some of its momentum to B due to the rate of change of momentum, however, it is said the total momentum of the system remains constant – Both declaration of A and acceleration of B are ignored in the equation of m1v1 = m2v2 but anyway,[41] I have not understood your explanation, but you claimed that momentum is not conserved. Can you demonstrate that ? Do the math. Calculate the momentum change.

Rate of change of momentum = force applied OR F = mv/t

This is not true when the difference in masses is mammoth. For example if m1 <<<<<<<<<< m2 like a dust particle and moon. B doesn’t move at all due to its greater inertia when A collides with B at greater m1v1 where v1 is very high velocity. Object A rests on B due to the greater size, inertia, and gravitational strength of B. There is a continuous force of A on B (may be less, =, and or greater than gravitational force b/t A and B) but no change of momentum as t is yet to be known for the change in momentum and this unknown t leads to force ultimately tends to zero (mv/t) and hence momentum of the system isn’t conserved.[41] We can say B absorbs the shock of A. Gravitational force may exits b/t A and B but this is irrelevant to the momentum mentioned previously.

What does that have to do with gravity or the earth's shape ?

Quote from: E E K 116

Bigger masses are harder to move as well as stops as compared to lighter ones and you know that.That phenomenon is known as gravity. Learn about it here : www.britannica.com/science/gravity-physics

So if A can’t move B due to its greater inertia then how come an apple accelerates earth with such a small force.

Quote from: E E K 122

Since A rests on B in either case the momentum therefore in the foregoing case is disappeared though there are gravitational forces but they have nothing to do with m1v1.What does that mean, “the momentum is disappeared” ? Momentum does not disappear. It is conserved. It may be transferred.

Quote from: E E K 125 to JackBlack

Make it simpler. Here on earth, anything if throws upward it comes downward. The momentum of that thing is not conserved.Indeed. Conservation of momentum applies to systems that do not exchange momentum with the rest of the world.

What does that have to do with the shape of the earth or the alleged problems with the theory of gravity ?