Do M1 and M2 really need gravitational force F=GM1M2/d^2 in b/t them if both M1 and M2 can fall towards each other w/o any force?
Gravitating mass doesn't depend upon the mass of the falling object.
Yes, it’s Newton who came up with idea first after the statement of Galileo that all objects fall at the rate.Do M1 and M2 really need gravitational force F=GM1M2/d^2 in b/t them if both M1 and M2 can fall towards each other w/o any force?
Gravitating mass doesn't depend upon the mass of the falling object.
Who says objects are pulled without any force?
There is a force, and we measure that force in Newtons.
In Relativity mass bends space-time and causes objects to follow this new bent path.
In both theories, the end result is a force pulls mass together.
In Relativity mass bends space-timeEinstein comes after Newton and Galileo. Einstein used the same gravitational constant “G” which was derived from the concept of newton’s law of gravitation. So his theories (though wrong) are also in question too.
It doesn't.
Dr. Erik Verlinde (2019):
General Relativity remains just a description of the force we call gravity. It leaves unanswered the key question of exactly how matter affects space and time.
The magnitude of gravitational force, which exists in between falling mass and gravitating mass
It cannot be attractive.
Please explain how a water molecule from lake Ontario which emits gravitons is attracted by the iron/nickel core which releases gravitons.
The magnitude of the acceleration at which the falling mass is accelerated can be found if we know the mass and radius of the gravitating mass in equation of g=GM/R^2 formulated by Newton. It doesn’t depend upon
In Relativity mass bends space-time
It doesn't.
Dr. Erik Verlinde (2019):
General Relativity remains just a description of the force we call gravity. It leaves unanswered the key question of exactly how matter affects space and time.
The magnitude of gravitational force, which exists in between falling mass and gravitating mass
It cannot be attractive.
Please explain how a water molecule from lake Ontario which emits gravitons is attracted by the iron/nickel core which releases gravitons.
In Relativity mass bends space-timeNo graviton.
...
Please explain how a water molecule from lake Ontario which emits gravitons is attracted by the iron/nickel core which releases gravitons.
The general formula of acceleration due to gravity “g” of any object of mass M and radius R is = g = GM/R^2, where R is just a distance from the center of gravity of the object to its outer surface if the object is a sphere.The magnitude of the acceleration at which the falling mass is accelerated can be found if we know the mass and radius of the gravitating mass in equation of g=GM/R^2 formulated by Newton. It doesn’t depend upon
Why are you talking about the radius? That's not part of the equation and we don't need to know it to calculate the attraction between two masses using that formula.
G = Gravitational constant.
M = The two masses, usually represented as M1 and M2.
R = Distance between the two masses.
There is nothing about radius in there.
Do you not think it's more likely you are misunderstanding these equations, rather than hundreds of years worth of scientists, engineers and theorists have been wrong, and every bridge, skyscraper and technology that relies on them just somehow worked even though everyone is wrong?
Do you think you discovered a flaw that thousands of scientists and mathematicians have missed?
Yes, it’s Newton who came up with idea first after the statement of Galileo that all objects fall at the rate.
The magnitude of the acceleration at which the falling mass is accelerated can be found if we know the mass and radius of the gravitating mass in equation of g=GM/R^2 formulated by Newton. It doesn’t depend upon
1- The magnitude of gravitational force, which exists in between falling mass and gravitating mass
2- The mass of the object, which is accelerated by the gravitating mass towards its center.
Did you see any role of the gravitational force in finding the magnitude of acceleration? Even newton didn’t know about it. Isn’t it weird? So if
M2 falls on M1 @ rate of g1=GM1(R1+H1)^2
M1 falls on M2 @ rate of g2=GM2(R2+H2)^2
Where is role gravitational force b/w M1 and M2 when both can fall w/o force and its magnitude?
Removing is called something ELIMINATION in math – F and m are eliminated so that we can solve for more simplified form. No idea what your discipline is but ELIMINATION is taught at school level.Yes, it’s Newton who came up with idea first after the statement of Galileo that all objects fall at the rate.
The magnitude of the acceleration at which the falling mass is accelerated can be found if we know the mass and radius of the gravitating mass in equation of g=GM/R^2 formulated by Newton. It doesn’t depend upon
1- The magnitude of gravitational force, which exists in between falling mass and gravitating mass
2- The mass of the object, which is accelerated by the gravitating mass towards its center.
Did you see any role of the gravitational force in finding the magnitude of acceleration? Even newton didn’t know about it. Isn’t it weird? So if
M2 falls on M1 @ rate of g1=GM1(R1+H1)^2
M1 falls on M2 @ rate of g2=GM2(R2+H2)^2
Where is role gravitational force b/w M1 and M2 when both can fall w/o force and its magnitude?
I tried to explain this in your other thread.
The equation you quoted is derived from F=ma and F=GMm/r2. In doing so the terms F and m are removed from the resulting equation.
All that means is that whatever value for m (and therefore F) we input into the equation, the result is the same.
It doesn’t mean that there is no force F or mass m in reality, and each equation must still work separately.
Removing is called something ELIMINATION in math – F and m are eliminated so that we can solve for more simplified form. No idea what your discipline is but ELIMINATION is taught at school level.
So once you eliminate F and m then it means then the resulting equation of g =GM/R^2 doesn't depend upon F and m. This equation of Newton which is independent of falling mass "m" satisfies Gallelio's statement of falling of all objects at the same rate which people still believe (including you) for many 100 years.
Removing is called something ELIMINATION in math – F and m are eliminated so that we can solve for more simplified form. No idea what your discipline is but ELIMINATION is taught at school level.
So once you eliminate F and m then it means then the resulting equation of g =GM/R^2 doesn't depend upon F and m. This equation of Newton which is independent of falling mass "m" satisfies Gallelio's statement of falling of all objects at the same rate which people still believe (including you) for many 100 years.
Knowing the name of something doesn't mean you're using it right.
If you really think that tens of thousands of scientists for hundreds of years are wrong about objects falling at the same rate, prove it. If you think you know better than millions of engineers who use those equations to design, build and fly things all around the world then lets see your experiments.
Build or buy a vacuum chamber, film some objects dropping and prove that heavier ones always fall faster.
W=mg, which is a flat earth equation
In Relativity mass bends space-time
It doesn't.
Dr. Erik Verlinde (2019):
General Relativity remains just a description of the force we call gravity. It leaves unanswered the key question of exactly how matter affects space and time.
The magnitude of gravitational force, which exists in between falling mass and gravitating mass
It cannot be attractive.
Please explain how a water molecule from lake Ontario which emits gravitons is attracted by the iron/nickel core which releases gravitons.
Sandokhan, do you know what a Taylor expansion is?Removing is called something ELIMINATION in math – F and m are eliminated so that we can solve for more simplified form. No idea what your discipline is but ELIMINATION is taught at school level.
So once you eliminate F and m then it means then the resulting equation of g =GM/R^2 doesn't depend upon F and m. This equation of Newton which is independent of falling mass "m" satisfies Gallelio's statement of falling of all objects at the same rate which people still believe (including you) for many 100 years.
Knowing the name of something doesn't mean you're using it right.
If you really think that tens of thousands of scientists for hundreds of years are wrong about objects falling at the same rate, prove it. If you think you know better than millions of engineers who use those equations to design, build and fly things all around the world then lets see your experiments.
Build or buy a vacuum chamber, film some objects dropping and prove that heavier ones always fall faster.
Cut the crap.
Engineering calculations for arches:
(https://concretedomestructures.weebly.com/uploads/2/4/0/9/24096046/1981918.jpg?704)
W=mg, which is a flat earth equation
This thread makes very clear the following facts: F=GMm/r^2 is an artificial equation which cannot be used until and unless one proves first that the shape of the Earth is spherical and that gravity is attractive; g=Gm/r^2 is the correct equation, which can only take place on a flat surface of the Earth.
Removing is called something ELIMINATION in math – F and m are eliminated so that we can solve for more simplified form. No idea what your discipline is but ELIMINATION is taught at school level.Yes, it’s Newton who came up with idea first after the statement of Galileo that all objects fall at the rate.
The magnitude of the acceleration at which the falling mass is accelerated can be found if we know the mass and radius of the gravitating mass in equation of g=GM/R^2 formulated by Newton. It doesn’t depend upon
1- The magnitude of gravitational force, which exists in between falling mass and gravitating mass
2- The mass of the object, which is accelerated by the gravitating mass towards its center.
Did you see any role of the gravitational force in finding the magnitude of acceleration? Even newton didn’t know about it. Isn’t it weird? So if
M2 falls on M1 @ rate of g1=GM1(R1+H1)^2
M1 falls on M2 @ rate of g2=GM2(R2+H2)^2
Where is role gravitational force b/w M1 and M2 when both can fall w/o force and its magnitude?
I tried to explain this in your other thread.
The equation you quoted is derived from F=ma and F=GMm/r2. In doing so the terms F and m are removed from the resulting equation.
All that means is that whatever value for m (and therefore F) we input into the equation, the result is the same.
It doesn’t mean that there is no force F or mass m in reality, and each equation must still work separately.
So once you eliminate F and m then it means then the resulting equation of g =GM/R^2 doesn't depend upon F and m. This equation of Newton which is independent of falling mass "m" satisfies Gallelio's statement of falling of all objects at the same rate which people still believe (including you) for many 100 years.
So now electrons exist?May be or may be not because I don't understand why was the mass of every electron created exactly the same when there were chances of more than one type by the natural selection in random during the early stages of the universe in chaos/ big bang?
So now electrons exist?May be or may be not because I don't understand why was the mass of every electron created exactly the same when there were chances of more than one type by the natural selection in random during the early stages of the universe in chaos/ big bang?
Mass of one electron = 9.10938356 × 10-31 kg
Other type could have = 8.10938356 × 10-30 kg – this is just an example
it could have zillion types of masses (Ranges in Kg)
The same is applied to all atomic particles, not just electron.
Sandokhan doesn’t believe in electrons yet he used them as evidence.So now electrons exist?May be or may be not because I don't understand why was the mass of every electron created exactly the same when there were chances of more than one type by the natural selection in random during the early stages of the universe in chaos/ big bang?
Mass of one electron = 9.10938356 × 10-31 kg
Other type could have = 8.10938356 × 10-30 kg – this is just an example
it could have zillion types of masses (Ranges in Kg)
The same is applied to all atomic particles, not just electron.
An electron is an elementary particle. It came about due to physics, which would be due to the properties of energy. You are making it out to be more complex than it is.So now electrons exist?May be or may be not because I don't understand why was the mass of every electron created exactly the same when there were chances of more than one type by the natural selection in random during the early stages of the universe in chaos/ big bang?
Mass of one electron = 9.10938356 × 10-31 kg
Other type could have = 8.10938356 × 10-30 kg – this is just an example
it could have zillion types of masses (Ranges in Kg)
The same is applied to all atomic particles, not just electron.
No; in relativity there is no 'force' 'pulling' anything.Do M1 and M2 really need gravitational force F=GM1M2/d^2 in b/t them if both M1 and M2 can fall towards each other w/o any force?
Gravitating mass doesn't depend upon the mass of the falling object.
Who says objects are pulled without any force?
There is a force, and we measure that force in Newtons.
In Relativity mass bends space-time and causes objects to follow this new bent path.
In both theories, the end result is a force pulls mass together.
No; in relativity there is no 'force' 'pulling' anything.Do M1 and M2 really need gravitational force F=GM1M2/d^2 in b/t them if both M1 and M2 can fall towards each other w/o any force?
Gravitating mass doesn't depend upon the mass of the falling object.
Who says objects are pulled without any force?
There is a force, and we measure that force in Newtons.
In Relativity mass bends space-time and causes objects to follow this new bent path.
In both theories, the end result is a force pulls mass together.
I’m not making it more complex but I'm talking about the availability of chances in early stages of the big bang if natural selection had availed in selecting of different masses of each type of atomic particle.An electron is an elementary particle. It came about due to physics, which would be due to the properties of energy. You are making it out to be more complex than it is.So now electrons exist?May be or may be not because I don't understand why was the mass of every electron created exactly the same when there were chances of more than one type by the natural selection in random during the early stages of the universe in chaos/ big bang?
Mass of one electron = 9.10938356 × 10-31 kg
Other type could have = 8.10938356 × 10-30 kg – this is just an example
it could have zillion types of masses (Ranges in Kg)
The same is applied to all atomic particles, not just electron.
The properties of energy will answer your problems.
There is evidence.(evidence is the correct word)There is no such instruments which can either see or measure space-time. So what do you mean by evidence?
So now electrons exist?May be or may be not because I don't understand why was the mass of every electron created exactly the same when there were chances of more than one type by the natural selection in random during the early stages of the universe in chaos/ big bang?
Mass of one electron = 9.10938356 × 10-31 kg
Other type could have = 8.10938356 × 10-30 kg – this is just an example
it could have zillion types of masses (Ranges in Kg)
The same is applied to all atomic particles, not just electron.
This one is easy. The standard model doesn't predict electrons with differing masses.
But more importantly, in all the experiments and atom smashing and evidence and data that has been collected... we never found an electron with a different mass.
We do not know WHY that is so, but there is no evidence or data to suggest otherwise.
If the LHC starts spitting out electrons with twice their weight, well that will be exiting and interesting and then you can be right. But until the, there is no proof, nor any hints that they exist.
There is evidence.(evidence is the correct word)There is no such instruments which can either see or measure space-time. So what do you mean by evidence?
Experiments have been performed.There is evidence.(evidence is the correct word)There is no such instruments which can either see or measure space-time. So what do you mean by evidence?
We have seen that the Sun bends light that passes very close to it.All matter is composed of atoms. Nuclei are about 100,000 times smaller than the atoms they’re housed in. Since Atoms are the building blocks of matter therefore it is said the human body is almost empty spaces (99.99999%). If this is true then aren’t all these celestial bodies are also almost empty spaces as well if yes then how do they bend space-time at such a large scale as quoted when atomic particles don’t have the ability of bend space-time unless in a collection?
QuoteWe have seen that the Sun bends light that passes very close to it.All matter is composed of atoms. Nuclei are about 100,000 times smaller than the atoms they’re housed in. Since Atoms are the building blocks of matter therefore it is said the human body is almost empty spaces (99.99999%). If this is true then aren’t all these celestial bodies are also almost empty spaces as well if yes then how do they bend space-time at such a large scale as quoted when atomic particles don’t have the ability of bend space-time unless in a collection?
You are also wrong that atomic particles can't bend space-time. They can, they just bend it VERY VERY VERY slightlyThen why don t you consider these slight distortion of space-time individually instead of taking a celestial body as a whole. it will limit the bending of space time to the actual size of celestial body.
QuoteYou are also wrong that atomic particles can't bend space-time. They can, they just bend it VERY VERY VERY slightlyThen why don t you consider these slight distortion of space-time individually instead of taking a celestial body as a whole. it will limit the bending of space time to the actual size of celestial body.
Here is the correct stress-energy tensor equation:
https://pdfs.semanticscholar.org/d92d/7f8b7771e0e3c4df0a25b712d7de2274ed59.pdf
You get the same answer either way.The bending of the space-time is outward if the whole mass of earth is considered.
If he has two phds why did he simplify E=mc2?K.E = 0.5 mv^2
QuoteYou get the same answer either way.The bending of the space-time is outward if the whole mass of earth is considered.
The bending of the space-time is inside the whole mass of earth (99% empty as explained) if the mass of the each individual atomic particle is considered.QuoteIf he has two phds why did he simplify E=mc2?K.E = 0.5 mv^2
P.E= mgh
E =mc^2
Now
E = mgh=mc^2
"m" cancel out;
We get gh=c^2 ----- Eq 1
Similarly, 0.5mv^2 = mc^2
"m" cancel out;
we get 0.5v^2=c^2 ….. Eq 2
What is the meaning of Eq 1 and eq 2
Do you know what energy is?QuoteYou get the same answer either way.The bending of the space-time is outward if the whole mass of earth is considered.
The bending of the space-time is inside the whole mass of earth (99% empty as explained) if the mass of the each individual atomic particle is considered.QuoteIf he has two phds why did he simplify E=mc2?K.E = 0.5 mv^2
P.E= mgh
E =mc^2
Now
E = mgh=mc^2
"m" cancel out;
We get gh=c^2 ----- Eq 1
Similarly, 0.5mv^2 = mc^2
"m" cancel out;
we get 0.5v^2=c^2 ….. Eq 2
What is the meaning of Eq 1 and eq 2
No meaning at all.
The E in E=mc^2 is the energy that would be released if you completely annihilate a mass.
It has nothing to do with gravitational potential energy or kinetic energy for a given situation.
You’ve just slapped some equations together randomly and got total nonsense.
QuoteYou get the same answer either way.The bending of the space-time is outward if the whole mass of earth is considered.
The bending of the space-time is inside the whole mass of earth (99% empty as explained) if the mass of the each individual atomic particle is considered.
If you consider the mass as a whole then the bending of the space time happens outside the whole mass.QuoteYou get the same answer either way.The bending of the space-time is outward if the whole mass of earth is considered.
The bending of the space-time is inside the whole mass of earth (99% empty as explained) if the mass of the each individual atomic particle is considered.
What do you mean space-time is bending outward?
Again, consider a one pound piece of lead and a one pound piece of foam. Which is heavier? One is mostly empty space, the other is very dense.
If you consider the mass as a whole then the bending of the space time happens outside the whole mass.QuoteYou get the same answer either way.The bending of the space-time is outward if the whole mass of earth is considered.
The bending of the space-time is inside the whole mass of earth (99% empty as explained) if the mass of the each individual atomic particle is considered.
What do you mean space-time is bending outward?
Again, consider a one pound piece of lead and a one pound piece of foam. Which is heavier? One is mostly empty space, the other is very dense.
if you consider the masses of all atomic particles individually of the whole mass as then the bending of the space time happens within the wholw mass not outside the whole mass
Do you know what energy is?
All (above) three are ENERGY equations.
As pointed out, you are wrong. E=mc2 Is rest mass energy. It’s not the full equation.Do you know what energy is?QuoteYou get the same answer either way.The bending of the space-time is outward if the whole mass of earth is considered.
The bending of the space-time is inside the whole mass of earth (99% empty as explained) if the mass of the each individual atomic particle is considered.QuoteIf he has two phds why did he simplify E=mc2?K.E = 0.5 mv^2
P.E= mgh
E =mc^2
Now
E = mgh=mc^2
"m" cancel out;
We get gh=c^2 ----- Eq 1
Similarly, 0.5mv^2 = mc^2
"m" cancel out;
we get 0.5v^2=c^2 ….. Eq 2
What is the meaning of Eq 1 and eq 2
No meaning at all.
The E in E=mc^2 is the energy that would be released if you completely annihilate a mass.
It has nothing to do with gravitational potential energy or kinetic energy for a given situation.
You’ve just slapped some equations together randomly and got total nonsense.
All (above) three are ENERGY equations.
B3Do you know what energy is?
All (above) three are ENERGY equations.
No shit, Sherlock.
That doesn’t change the fact that you are using them wrong.
Your equations would only be valid, if you were converting one type of energy into another. You can do this with kinetic energy and potential energy for example when you drop something. You can work out the potential energy difference between two heights and that gives you the kinetic energy gained over that distance. Although if you drop through a medium like air, there will also be some energy loss from drag.
But when you drop something, you don’t convert any of the mass into energy, do you? So why do you think Einstein’s equation is relevant?
That’s twice in this thread where you’ve lashed out at my attempts to explain by patronising me. Well, allow me to respond in kind.
BAD NEWS, SUNSHINE. The ability to look up equations on Google does not equal knowing how to apply them in any meaningful way. You apparently think your TINY and completely INADEQUATE level of knowledge on the subject is superior to all the physicists and engineers who bothered to actually study this, because of course you do.
I don’t suppose it’s ever occurred to you that you could have spent the time it takes writing your LAUGHABLE NONSENSE doing some basic mechanics tutorials, and you’d probably understand where you’re going wrong by now?
But no, I expect you’ll carry on declaring you’ve found a fatal flaw in ABSOLUTELY FUNDAMENTAL mechanics that has somehow been missed by everyone who has used these equations to build all the technology you take for granted over the last few hundred years.
LOL
B3
ABSOLUTELY FUNDAMENTAL mechanics that has somehow been missed by everyone who has used these equations to build all the technology you take for granted over the last few hundred years.
Cut the crap.
The jet engine was invented by Viktor Schauberger, using ether theory.
The B-2 bomber flies using the Biefeld-Brown effect.
ABSOLUTELY FUNDAMENTAL mechanics that has somehow been missed by everyone who has used these equations to build all the technology you take for granted over the last few hundred years.
Cut the crap.
The jet engine was invented by Viktor Schauberger, using ether theory.
The B-2 bomber flies using the Biefeld-Brown effect.
I must have missed those modules, along with every other mechanical engineering student at every university.
These effects appear in no standard texts that engineers learn. Newton’s laws of motion and universal gravitation on the other hand are in the first few pages of all the relevant books (a great many of them).
I wonder how engineers working at the likes of Boeing, Airbus, GE, Rolls-Royce, etc. manage to design functional aircraft and engines if they’re using all the wrong physics as you claim?
ABSOLUTELY FUNDAMENTAL mechanics that has somehow been missed by everyone who has used these equations to build all the technology you take for granted over the last few hundred years.
Cut the crap.
The jet engine was invented by Viktor Schauberger, using ether theory.
The B-2 bomber flies using the Biefeld-Brown effect.
I must have missed those modules, along with every other mechanical engineering student at every university.
These effects appear in no standard texts that engineers learn. Newton’s laws of motion and universal gravitation on the other hand are in the first few pages of all the relevant books (a great many of them).
I wonder how engineers working at the likes of Boeing, Airbus, GE, Rolls-Royce, etc. manage to design functional aircraft and engines if they’re using all the wrong physics as you claim?
His airplanes claims are mind boggling.
Are you going to call Hermann Weyl insane?
Why do airplanes need to reach takeoff speed? Why is takeoff speed not zero?
Why do airplanes need to reach takeoff speed? Why is takeoff speed not zero?Why would airplanes need a take-off angle if the earth is round? Do they adjust for curvature when flying parallel to the ground?
Why do airplanes need to reach takeoff speed? Why is takeoff speed not zero?Why would airplanes need a take-off angle if the earth is round? Do they adjust for curvature when flying parallel to the ground?
This stuff is taught in high school.They teach good stuff but not all are correct. I asked "Do pilots adjust for curvature when they fly parallel to the ground"? just like moving jet on the runway
I didn’t mentions angle. I mentioned takeoff speed.Why do airplanes need to reach takeoff speed? Why is takeoff speed not zero?Why would airplanes need a take-off angle if the earth is round? Do they adjust for curvature when flying parallel to the ground?
QuoteThis stuff is taught in high school.They teach good stuff but not all are correct. I asked "Do pilots adjust for curvature when they fly parallel to the ground"? just like moving jet on the runway
you asked him a question and i asked you the question related to your question.I didn’t mentions angle. I mentioned takeoff speed.Why do airplanes need to reach takeoff speed? Why is takeoff speed not zero?Why would airplanes need a take-off angle if the earth is round? Do they adjust for curvature when flying parallel to the ground?
How often in ref to distance?QuoteThis stuff is taught in high school.They teach good stuff but not all are correct. I asked "Do pilots adjust for curvature when they fly parallel to the ground"? just like moving jet on the runway
You asked two questions, I answered one, as the other has been answered many times. I'll answer it again.
Yes, planes tilt as they follow the curve. A plane that has gone halfway around the world will have tilted downward 180 degrees. If it makes it back to where it started it will have tilted a full 360.
How often in ref to distance?QuoteThis stuff is taught in high school.They teach good stuff but not all are correct. I asked "Do pilots adjust for curvature when they fly parallel to the ground"? just like moving jet on the runway
You asked two questions, I answered one, as the other has been answered many times. I'll answer it again.
Yes, planes tilt as they follow the curve. A plane that has gone halfway around the world will have tilted downward 180 degrees. If it makes it back to where it started it will have tilted a full 360.
Ref pls.How often in ref to distance?QuoteThis stuff is taught in high school.They teach good stuff but not all are correct. I asked "Do pilots adjust for curvature when they fly parallel to the ground"? just like moving jet on the runway
You asked two questions, I answered one, as the other has been answered many times. I'll answer it again.
Yes, planes tilt as they follow the curve. A plane that has gone halfway around the world will have tilted downward 180 degrees. If it makes it back to where it started it will have tilted a full 360.
1 degree every 70 miles.
you asked him a question and i asked you the question related to your question.I didn’t mentions angle. I mentioned takeoff speed.Why do airplanes need to reach takeoff speed? Why is takeoff speed not zero?Why would airplanes need a take-off angle if the earth is round? Do they adjust for curvature when flying parallel to the ground?
Ref pls.How often in ref to distance?QuoteThis stuff is taught in high school.They teach good stuff but not all are correct. I asked "Do pilots adjust for curvature when they fly parallel to the ground"? just like moving jet on the runway
You asked two questions, I answered one, as the other has been answered many times. I'll answer it again.
Yes, planes tilt as they follow the curve. A plane that has gone halfway around the world will have tilted downward 180 degrees. If it makes it back to where it started it will have tilted a full 360.
1 degree every 70 miles.
how do they bend space-time at such a large scale as quoted when atomic particles don’t have the ability of bend space-time unless in a collection?They can bend space time. Everything can.
Then why don t you consider these slight distortion of space-time individually instead of taking a celestial body as a whole. it will limit the bending of space time to the actual size of celestial body.Because doing it all at once is a much simpler calculation.
I asked "Do pilots adjust for curvature when they fly parallel to the ground"? just like moving jet on the runwayThey do so automatically by maintaining altitude.
W=mg, which is a flat earth equationNo, it is a simplification, based upon the assumption that the value of g varies insignificantly for the object.
Mass is what bends space-timeYes, it doesn't describe how matter bends spacetime, just that it does, and the results of it doing so.
Cut the crap you troll.
General Relativity remains just a description of the force we call gravity. It leaves unanswered the key question of exactly how matter affects space and time.
The B-2 bomber flies using the Biefeld-Brown effect.Are you sure it isn't using the wings for aerodynamic lift, along with engines to provide thrust?
In many ways, this is quite analogous to electrostatic interactions.One word – not impressive! Although its very clear but I’m writing again. I would suggest reading the posts carefully again and again not just you – no rude at all as it is not about the win or loss.
G is the proportionality constant. It depends upon the force b/w 2 masses and the square of o/c distance b/t them as F =Gm1m2/d^2 =mg where g = Gm1/d^2It doesn't depend on the force, it determines it.
If you say, “there is clearly a force” then there is clearly a second mass (falling one) as well. Then all objects do not fall at the same rate.No. There being a force doesn't mean that all objects don't fall at the same rate.
So how does force F is determine from G=gd^2/M?G is the proportionality constant. It depends upon the force b/w 2 masses and the square of o/c distance b/t them as F =Gm1m2/d^2 =mg where g = Gm1/d^2It doesn't depend on the force, it determines it.
So how does force F is determine from G=gd^2/M?F=GMm/d^2.
So you unable to determine F from G=gd^2/M - Right?So how does force F is determine from G=gd^2/M?F=GMm/d^2.
i.e. the force is dependent upon G.
No, I provided how.So you unable to determine F from G=gd^2/M - Right?So how does force F is determine from G=gd^2/M?F=GMm/d^2.
i.e. the force is dependent upon G.
I’m still wanting to make a real post but until then, is it just me or is it supposed to be r2 not d2?It can be either, as long as you know what it stands for, and I have seen it presented in many ways.
QuoteIn many ways, this is quite analogous to electrostatic interactions.One word – not impressive! Although its very clear but I’m writing again. I would suggest reading the posts carefully again and again not just you – no rude at all as it is not about the win or loss.
G is the proportionality constant. It depends upon the force b/w 2 masses and the square of o/c distance b/t them as F =Gm1m2/d^2 =mg where g = Gm1/d^2,
Considering “g” separately from force F as g = Gm1/d^2, means G is considered separately from force F which is impossible. Therefore I would say “G” is wrongly placed in g=GM/d^2.
Two objects are must for the presence of gravitational force “F” [F=GMm/d^2, where g=GM/d^2] and gravitational constant “G”. Gravitational force “F“ should generate gravitational acceleration “g” but “g” which is equal to GM/d^2 independent of second mass (falling mass) and gravitational force “F”.
As said, gravitational constant “G” is wrongly placed in g=GM/d^2 as explained above.
If you say, “there is clearly a force” then there is clearly a second mass (falling one) as well. Then all objects do not fall at the same rate.
QuoteIn many ways, this is quite analogous to electrostatic interactions.One word – not impressive! Although its very clear but I’m writing again. I would suggest reading the posts carefully again and again not just you – no rude at all as it is not about the win or loss.
G is the proportionality constant. It depends upon the force b/w 2 masses and the square of o/c distance b/t them as F =Gm1m2/d^2 =mg where g = Gm1/d^2,
Considering “g” separately from force F as g = Gm1/d^2, means G is considered separately from force F which is impossible. Therefore I would say “G” is wrongly placed in g=GM/d^2.
Two objects are must for the presence of gravitational force “F” [F=GMm/d^2, where g=GM/d^2] and gravitational constant “G”. Gravitational force “F“ should generate gravitational acceleration “g” but “g” which is equal to GM/d^2 independent of second mass (falling mass) and gravitational force “F”.
As said, gravitational constant “G” is wrongly placed in g=GM/d^2 as explained above.
If you say, “there is clearly a force” then there is clearly a second mass (falling one) as well. Then all objects do not fall at the same rate.
Still stuck on this I see.
Seems the only hope is for you to start plugging some numbers into the equations and see what happens.
I suggest this.
1. Open up a spreadsheet as it’s the easiest way to see a bunch of different values side by side. (Download Open Office if you don’t have one).
2. First Row: Mass m. Write a bunch of different masses for the falling object. Perhaps 1kg, 10kg, 100kg, etc but it doesn’t matter.
3. Second row: Force F. Use F=GMm/d^2 with M as mass of Earth, d as radius of Earth (for gravity at sea level) and m taken from above row.
4. Third row: Acceleration a. Use a = F/m using the numbers from rows 1 and 2.
5. Report back. What do you see?
As F=GMm/d^2 .......Eq #1, And g=GM/d^2....Eq #2No, the 2 equations are equivalent.
From Eq #1: G=Fd^2/Mm ...Eq #3 And From Eq #2: G=gd^2/M .....Eq #4
Gravitational constant "G" depends upon M and m in Eq #3 while the same "G" depends only on M in Eq#4 so isn't "G" wrongly placed in Eq #2 and #4? You know "G" requires two masses M and m and F as well.
A single mass (if present alone in the universe) used in Eq #2 and #4 can't generate gravitational force "F" and determine gravitational "G". Both "G" and "F" require two masses M and m.The other way to think about it is g is the gravitational field.
No, the 2 equations are equivalent.I understand what you say about F=ma. Here “F” and “a” depend upon “m” on which the force is applied but “g” in the Eq of “g=GM/d^2” doesn’t depend on the mass (second or falling mass) on which the gravitational force is applied.
g is an acceleration.
Remember that F=ma.
With g, this is F=mg, which can also be written as g=F/m
So the F in eq #3 is actually mg.
Or conversely, in eq#4, g is actually F/m.
The 2 equations are equivalent, but written with different units.
The other way to think about it is g is the gravitational field.Similarly "G" in the Eq of g=GM/d^2 requires the presence of a second mass "m" for the determination of its constant value. No second mass "m" means no "G". No "G" means no g=GM/^2 of a single mass existed if lonely present in the universe.
A mass in that field will experience a force which is the product of the field and its mass (the mass of the object).
So that single object in the universe will generate a gravitational field
but not a force as it requires another object to have that force.But Newton determines it to be “g=GM/d^2” for any object of mass "M" when he says “g” is independent of second or falling mass “m” after formulating the universal law of gravitation of F=GMm/d^2. Here independent of second mass means "in the absence of a second mass".
With a single object alone in the universe, there would be no way to determine g.
I understand what you say about F=ma. Here “F” and “a” depend upon “m” on which the force is applied but “g” in the Eq of “g=GM/d^2” doesn’t depend on the mass (second or falling mass) on which the gravitational force is applied.The point is that the "a" is g. i.e. in the equation F=GMm/d^2, that IS F=m*g
Similarly "G" in the Eq of g=GM/d^2 requires the presence of a second mass "m" for the determination of its constant value. No second mass "m" means no "G". No "G" means no g=GM/^2 of a single mass existed if lonely present in the universe.Only for the determinisation of the value, not the values existence.
As "G" requires gravitational force "F" and "F" requires the presence of two masses (M and m) therefore the presence of "G" in g=GM/^2 means that "g" in the foregoing Eq depend upon second mass "m" too. Thus "g" of any gravitating mass depends upon the mass of the falling object.
But Newton determines it to be “g=GM/d^2” for any object of mass "M" when he says “g” is independent of second or falling mass “m” after formulating the universal law of gravitation of F=GMm/d^2. Here independent of second mass means "in the absence of a second mass".Here independent of the second mass means it does not depend on the second mass.
It means gravitational field g=GM/d^2 is the innate property of any mass "M".
What is the role of gravitational force F=GMm/d^2 between objects A and B whenFirst a note on when m is 0, if rest mass is 0, and you have a massless object then Newton's Laws break down. But note that even "massless" particles like light will be accelerated.
Object A falls on B @ rate of gb=GMb/d^2.
Here gb is innate to Mb and doesn’t depend on the magnitude of Ma even if it is zero and hence F as well.
Object B falls on A @ rate of ga=GMa/d^2.
Here ga is innate to Ma and doesn’t depend on the magnitude of Mb even if it is zero and hence F as well.
Both accelerations (ga and gb) produced don’t depend upon the F between them.
So we have the equal and opposite force.Acceleration needs force for its generation - Right
The other way aroundThis is where you start going into philosophy.QuoteSo we have the equal and opposite force.Acceleration needs force for its generation - Right
How do accelerations produce when both equal and opposite forces of gravitating and falling cancel each other. There is no net force at all, which can accelerate either of the objects due to their cancellation. This net force is always zero no matter what the o/c distance is in between A and B -RightThe net force on the AB system is 0. That means the centre of mass of the AB system will not move.
The net force on A is the same as the net force on B which is GMaMb/d^2.But in opposite direction - Right. It is said
Lone point mass can’t have its gravitational field = g = GM/d^2 because “d” in the preceding Eq is not just a length but it is the o/c center distance between A and B.It is the distance from the object.
This means that there is always exists ONE gravitational force F=GMm/d^2 between A and B which has to accelerate both A and B towards one another simultaneously.No, it means there are 2 equal and opposite forces, just like you always find forces paired up.
[/b]I don’t understand why are the equations of F=ma and F=GMm/d^2 are compared with each other.Because the force is the same.
It is the distance from the object.Distance from the object means when no second mass, d=0, and hence g = GM/d^2 = 0. For a person like you, it's not that difficult to work out.
No, it means there are 2 equal and opposite forces, just like you always find forces paired up.This means gravitational doesn’t exist anymore or always zero between the objects when these two forces cancel each other.
There is the force A exerts on B which draws B towards A and there is the force B exerts on A which draws A towards B.
These forces are equal in magnitude and opposite in direction.
Because the force is the same.Nope they are two different forces
You could just write it like this:
F=ma=GMm/d^2=mg
They are 2 ways to write out the same force.
No, the field is defined everywhere.QuoteIt is the distance from the object.Distance from the object means when no second mass, d=0, and hence g = GM/d^2 = 0. For a person like you, it's not that difficult to work out.
No, it doesn't.QuoteNo, it means there are 2 equal and opposite forces, just like you always find forces paired up.This means gravitational doesn’t exist anymore or always zero between the objects when these two forces cancel each other.
There is the force A exerts on B which draws B towards A and there is the force B exerts on A which draws A towards B.
These forces are equal in magnitude and opposite in direction.
As asked, what is the role of gravitational force then b/t A and B when Fa and Fb cancel out?
It is the same force, no matter how much you don't want it to be.QuoteBecause the force is the same.Nope they are two different forces
You could just write it like this:
F=ma=GMm/d^2=mg
They are 2 ways to write out the same force.
means there is a force on m but m doesn’t exert force on anything.This is physically impossible. If there is a force on m, m MUST exert a force equal in magnitude and opposite in direction on something else.
Similarly, if g = GM/d^2 doesn’t depend on falling mass then why does a greater mass settle more than light mass on the soft surface of the ground?Because g is the acceleration, not the force.
Both lighter and greater masses have the same g = GM/d^2
QuoteIn many ways, this is quite analogous to electrostatic interactions.One word – not impressive! Although its very clear but I’m writing again. I would suggest reading the posts carefully again and again not just you – no rude at all as it is not about the win or loss.
G is the proportionality constant. It depends upon the force b/w 2 masses and the square of o/c distance b/t them as F =Gm1m2/d^2 =mg where g = Gm1/d^2,
Considering “g” separately from force F as g = Gm1/d^2, means G is considered separately from force F which is impossible. Therefore I would say “G” is wrongly placed in g=GM/d^2.
Two objects are must for the presence of gravitational force “F” [F=GMm/d^2, where g=GM/d^2] and gravitational constant “G”. Gravitational force “F“ should generate gravitational acceleration “g” but “g” which is equal to GM/d^2 independent of second mass (falling mass) and gravitational force “F”.
As said, gravitational constant “G” is wrongly placed in g=GM/d^2 as explained above.
If you say, “there is clearly a force” then there is clearly a second mass (falling one) as well. Then all objects do not fall at the same rate.
Still stuck on this I see.
Seems the only hope is for you to start plugging some numbers into the equations and see what happens.
I suggest this.
1. Open up a spreadsheet as it’s the easiest way to see a bunch of different values side by side. (Download Open Office if you don’t have one).
2. First Row: Mass m. Write a bunch of different masses for the falling object. Perhaps 1kg, 10kg, 100kg, etc but it doesn’t matter.
3. Second row: Force F. Use F=GMm/d^2 with M as mass of Earth, d as radius of Earth (for gravity at sea level) and m taken from above row.
4. Third row: Acceleration a. Use a = F/m using the numbers from rows 1 and 2.
5. Report back. What do you see?
As F=GMm/d^2 .......Eq #1, And g=GM/d^2....Eq #2
From Eq #1: G=Fd^2/Mm ...Eq #3 And From Eq #2: G=gd^2/M .....Eq #4
Gravitational constant "G" depends upon M and m in Eq #3 while the same "G" depends only on M in Eq#4 so isn't "G" wrongly placed in Eq #2 and #4? You know "G" requires two masses M and m and F as well.
A single mass (if present alone in the universe) used in Eq #2 and #4 can't generate gravitational force "F" and determine gravitational "G". Both "G" and "F" require two masses M and m.
G is a universal constant, it doesn’t depend on anything. But you can calculate G if you have all the terms for either of the two equations. There is no problem here.
Did you try my suggestion of actually putting some numbers into the equations to see what happens with different values? It should make everything much clearer.
QuoteG is a universal constant, it doesn’t depend on anything. But you can calculate G if you have all the terms for either of the two equations. There is no problem here.
Did you try my suggestion of actually putting some numbers into the equations to see what happens with different values? It should make everything much clearer.
Have you gone through my last reply where I showed that second mass m matter in the equation of acceleration due to gravity g=GM/d^2? This will answer your all questions.
The position of both Ma and Mc are located on the opposite sides of the globe of Mb as shown. The point e and f are the antipodal points.No, they don't.
Galileo and Newton say that both Ma and Mc fall at the same rate
So the falling mass matter to g=GM/d^2 - RightNo, because what you actually have now when you consider object A falling is:
Have you gone through my last reply where I showed that second mass m matter in the equation of acceleration due to gravity g=GM/d^2? This will answer your all questions.You didn't.
They discuss 2 objects dropped from the same location, not antipodal points.Reference, please!!!!!
One example would be this:QuoteThey discuss 2 objects dropped from the same location, not antipodal points.Reference, please!!!!!
Small objects do fall at the same rate even if dropped from the antipodal points simultaneously because the acceleration of the earth is ignored.Under the assumption that the only significant gravitational field is that from Earth. Then all other masses are ignored and the motion of Earth is ignored.
This changes its falling rate theoretically.No, it doesn't. It means Earth falls towards it.
Question: Why do cooler /denser molecules of air take the position of warmer molecules on or above the surface of the earth when “g” is independent of temperature and pressure?Because pressure is dependent upon density.
Because pressure is dependent upon density.Great!! Gravity depends upon the density (gamma = mass per unit volume) of the object. The mass of the denser molecules has greater acceleration due to gravity while the mass of the warmer molecules has lesser acceleration due to gravity. So it means denser objects fall faster than lighter objects. You yourself choose period.
No, it doesn't. It means Earth falls towards it.Reduction in the distance means changes in the rate of falling - it starts falling with higher types of motion
That reduces the distance. It doesn't make the object fall faster.
If it can be used in g=GM/d^2 then why do we ignore the acceleration of earth if we increase the size of one of the two falling masses to a greater size or close to the earth?We don't.
The mass of the denser molecules has greater acceleration due to gravity while the mass of the warmer molecules has lesser acceleration due to gravity. So it means denser objects fall faster than lighter objects.No, it doesn't.
Technically pressure (gamma x height =F/A) is not even a force.And I never said it was.
No, it doesn't. It means Earth is moving as well.QuoteNo, it doesn't. It means Earth falls towards it.Reduction in the distance means changes in the rate of falling - it starts falling with higher types of motion
That reduces the distance. It doesn't make the object fall faster.
No, it doesn't. It means Earth is moving as well.In other words to the offensive/nonsense: Light mass enters quickly than it is supposed to be into the gravitational field of the earth. The quick increase in its velocity means there is a quicker change in its velocity as well than usual.
You continually asserting the same nonsense will not make it true.
Air molecules push outwards to the sides equally, but in my understanding, their weights due to gravity indeed push down harder than they push upwards.For a gas, there is a pressure gradient.
Hot air can have a lower density but the same pressure as cold air. Therefore cold air sinks due to gravity, not because of pressure.It is not simply the pressure, but the pressure gradient.
In other words to the offensive/nonsense: Light mass enters quickly than it is supposed to be into the gravitational field of the earth.No, that is the nonsense.
No, that is the nonsense.Similarly,
Earth moving towards the object doesn't mean the object is accelerating faster.
Lets take the example of footing settlement.And the actual load.
Contact pressure: Response of soil against applied pressure of footing. It depends upon
1- Stiffness of footing (rigid or flexible)
2- Stiffness of soil (loose or hard)
3- Type of load (point or uniform)
Let g1=9.8 m/s/s,g2=9.7m/s/s..................g5=9.4m/s/s and g6=9.3 m/s/s are zones of gravitational field above earth.That is the point though, it is incorrect.
Case #1: Erath moves upward while mass m is stationary: Mass m is at rest at a height where the value of g of earth is g6. The mass “m” seems to enter into g5 zone of the earth slowly as soon as the earth starts accelerating towards “m” @ the rate of “g” of “m”. After some time it crosses the zone of g5 and then g4,g3,g2, and finally, the earth hits the mass “m”.
Case #2; Erath is stationary while mass m is moving: Mass m is at rest at a height where the value of g of earth is g6. The mass “m” starts entering into g5 zone of earth slowly as soon as it starts accelerating towards “earth” @ the rate of “g6” of “earth” initially. After some time it crosses the zone of g5 and then g4,g3,g2 and finally hit the earth @ the rate of g1.
Case #3: Now imagine both start accelerating towards each other simultaneously. They experience the same gravitational force but the acceleration of m towards earth is greater than the acceleration of earth towards “m”. Since mass “m” enters the gravitational field of earth faster than before therefore there will be an increase in the velocity of “m” and acceleration as well.
Conclusion: Although it's incorrect but for simplicity we can say case #3 = case #1 + case #2. We can add the velocities involved in case #1 and case #2.
That is the point though, it is incorrect.The motion is negligible for small masses but it can't be ignored when the size of the falling mass is increased. Also, we are discussing theory.
Earth accelerating towards the object changes the distance and thus changes the rate of acceleration.
It is not the object itself accelerating towards Earth faster, it is Earth moving towards the object.
And for most things, that motion is negligible.
And the actual load.This is what I’m trying to explain that greater the density of mass is, the greater will be the settlement in ground but you say it doesn’t support my claim of “ g depends on the falling mass” or “all objects don’t fall at the same rate”
This actual load depends on the mass of the object.
It isn't helping you support your prior claims.
The motion is negligible for small masses but it can't be ignored when the size of the falling mass is increased. Also, we are discussing theory.But that doesn't mean that the mass is making it accelerate faster.
Case #3 = Case #1 +case#2
We know g = GM/d^2 is the gravitational strength of any point mass. As soon as the on-center distance “d” b/t the two objects starts reducing from both ends simultaneously, the “g “ of earth and the “g” of mass start increasing at the same times than it were considered to be like in case #1 or case #2. Both masses are gravitating and falling at the same time therefore they enter into the gravitational field of one another very quickly not like in case#1 or Case#2.
As “g” neither depends on falling mass nor gravitational force therefore if all objects would fall at the same rate then those objects should have shown at least equivalency/ similarity in settlements. For example:No, it shouldn't.
But that doesn't mean that the mass is making it accelerate faster. The change in distance to the other object, by the other object moving is what does. It doesn't show any fault with the theoryYes, it makes it faster; don’t forget it is a gravitating mass as well, which accelerates the falling masses. Reduction in the “d” w/o increased speed in less time is impossible.
All objects fall at the same speed. This means an object’s weight will not change its falling speed. Both identical spheres with different masses (A and B) act as two point loads. The rule is the rule.The rule is g=GM/d^2.
Why would A feel weightless in a state of free fall?For the simple reason that gravity acts on all parts of the object. But people don't feel a force applied to them, nor velocity, nor acceleration. Instead they feel their body transmitting a force. And the same effectively applies to all objects.
The force of gravity doesn’t exit alone. There is always an equal opposing force of gravity to it.Acting on the other object.
Call it a non-contact in air or space is also wrongA contact force is not saying the 2 objects are connected by a force.
The force of gravity pulls the body of A towards its center if stands on the ground. There is always an equal but opposite force called the normal force “N” that pushes A in a direction opposite to the force of the earth. It is this force due to which A perceives the force of gravity as weight.I would not say always. That is the case when A is standing stationary on the ground.
Both normal forces of equal magnitude but opposite in direction cancel other. So neither A nor earth feels force of gravity of one another.They feel the normal force, not gravity.
As explained above, there is no force that exists in b/t the A or B (identical spheres of different masses) and the earth then how come A and B settled in ground.That was not explained above.
how come one settled greater than the other when both A and B have the same speed?For a similar reason that throwing a lightweight ball at a wall will have it bounce off, while a much heavier ball smashes the wall.
So does "g=9,8 m/s/s=GM/d^2 valid on the surface of earth?Yes, F=mg.