Does gravity become infinity where the center of gravity of a mass lies?This equation doesn't work to calculate acceleration due to gravity, because you're using it in a case where nothing is being accelerated. There is no 'm' in your calculations, therefore there is no acceleration. You can see this because:
Gravity pulls towards the center of mass. Simple Example;
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
Does gravity become infinity where the center of gravity of a mass lies?No.
Gravity pulls towards the center of mass. Simple Example;
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
How are two objects attracted to their center of mass with F=GMm/d^^{2} which is less than infinity?
Does gravity become infinity where the center of gravity of a mass lies?This equation doesn't work to calculate acceleration due to gravity, because you're using it in a case where nothing is being accelerated. There is no 'm' in your calculations, therefore there is no acceleration. You can see this because:
Gravity pulls towards the center of mass. Simple Example;
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
g = GM/R^^{2}=nothing
There is no R, since there's no 'distance between' anything, or 'radius between' anything if you like, since there's only one object. Also, even if R=0, anything divided by 0 is nothing, not infinity.
So no, gravity is not infinite at the centre of mass, nor is gravity necessarily infinite at the centre of black holes. The Earth won't get sucked into itself, that doesn't make sense. For the Earth's mass to become a black hole, the Earth would have to be crushed to the size of a peanut. Then the density is so great it collapses in on itself to become a black hole.
tl;dr - your equation is wrong
In the centre of a black hole is a gravitational singularity, a one-dimensional point which contains a huge mass in an infinitely small space, where density and gravity become infinite and space-time curves infinitely, and where the laws of physics as we know them cease to operate.This is a possibility, but not necessarily true; it raises a number of problems such as all our known laws of physics being wrong, since they cease to operate at this point.
In the centre of a black hole is a gravitational singularity, a one-dimensional point which contains a huge mass in an infinitely small space, where density and gravity become infinite and space-time curves infinitely, and where the laws of physics as we know them cease to operate.This is a possibility, but not necessarily true; it raises a number of problems such as all our known laws of physics being wrong, since they cease to operate at this point.
Also not seeing how this ties in at all with the OP.
Does gravity become infinity where the center of gravity of a mass lies?This equation doesn't work to calculate acceleration due to gravity, because you're using it in a case where nothing is being accelerated. There is no 'm' in your calculations, therefore there is no acceleration. You can see this because:
Gravity pulls towards the center of mass. Simple Example;
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
g = GM/R^^{2}=nothing
There is no R, since there's no 'distance between' anything, or 'radius between' anything if you like, since there's only one object. Also, even if R=0, anything divided by 0 is nothing, not infinity.
So no, gravity is not infinite at the centre of mass, nor is gravity necessarily infinite at the centre of black holes. The Earth won't get sucked into itself, that doesn't make sense. For the Earth's mass to become a black hole, the Earth would have to be crushed to the size of a peanut. Then the density is so great it collapses in on itself to become a black hole.
tl;dr - your equation is wrong
In the centre of a black hole is a gravitational singularity, a one-dimensional point which contains a huge mass in an infinitely small space, where density and gravity become infinite and space-time curves infinitely, and where the laws of physics as we know them cease to operate....’a bit like the inside of shifters head, but I thought you hated gravity!
Does gravity become infinity where the center of gravity of a mass lies?This equation doesn't work to calculate acceleration due to gravity, because you're using it in a case where nothing is being accelerated. There is no 'm' in your calculations, therefore there is no acceleration. You can see this because:
Gravity pulls towards the center of mass. Simple Example;
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
g = GM/R^^{2}=nothing
There is no R, since there's no 'distance between' anything, or 'radius between' anything if you like, since there's only one object. Also, even if R=0, anything divided by 0 is nothing, not infinity.
So no, gravity is not infinite at the centre of mass, nor is gravity necessarily infinite at the centre of black holes. The Earth won't get sucked into itself, that doesn't make sense. For the Earth's mass to become a black hole, the Earth would have to be crushed to the size of a peanut. Then the density is so great it collapses in on itself to become a black hole.
tl;dr - your equation is wrong
In the centre of a black hole is a gravitational singularity, a one-dimensional point which contains a huge mass in an infinitely small space, where density and gravity become infinite and space-time curves infinitely, and where the laws of physics as we know them cease to operate....’a bit like the inside of shifters head, but I thought you hated gravity!
Just because your primitive knowledge doesn't understand what happens doesn't mean it is unknowable. Terms like 'laws of physics break down' or your 'cease to operate' is just short hand for 'I have no fucking idea what I'm talking about' and 'Our understanding of physics is incomplete and wrong'
You said it yourself...This is a possibility, but not necessarily true; it raises a number of problems such as all our known laws of physics being wrong, since they cease to operate at this point.
Citation please....just borrowing one of your lines’
In the centre of a black hole [...] the laws of physics as we know them cease to operate.
The "g" of an earth decreases above the surface of the earth while increases below its surface till it reach infinity. Try at a point which is 0.000000000000000000000000000000000000000000001 mm just above the center of the earth. Plug it in the formula. This is just an example.It has been demonstrated to you precisely how and why the formula is wrong. That isn't a scientific formula, it's a poor corruption of one.
You said it yourself...This is a possibility, but not necessarily true; it raises a number of problems such as all our known laws of physics being wrong, since they cease to operate at this point.
Citation please....just borrowing one of your lines’In the centre of a black hole [...] the laws of physics as we know them cease to operate.
Well, I mean you didn't say it, but the website you ripped it from did.
(https://preview.ibb.co/jLgNrS/lone_try_to_shill_harder_please.png) (https://ibb.co/ewcvBS)
If you want to disprove physics, you really need to do more research than the first result on a Google search...
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}Actually, the gravity at the center of the earth is zero because all of the mass around it pulls equally in all directions and cancels out.
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
Why?Why is it problematic to only use the first Google search result in an argument about an incredibly complex scientific topic? Gee, I'm not sure.
Does gravity become infinity where the center of gravity of a mass lies?No, it is not. The gravitation field at a point inside a sphere, at radius r from the centre, is due only to the mass inside radius r.
Gravity pulls towards the center of mass. Simple Example;
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
How are two objects attracted to their center of mass with F=GMm/d^^{2} which is less than infinity?
Does gravity become infinity where the center of gravity of a mass lies?This equation doesn't work to calculate acceleration due to gravity, because you're using it in a case where nothing is being accelerated. There is no 'm' in your calculations, therefore there is no acceleration. You can see this because:
Gravity pulls towards the center of mass. Simple Example;
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
g = GM/R^^{2}=nothing
There is no R, since there's no 'distance between' anything, or 'radius between' anything if you like, since there's only one object. Also, even if R=0, anything divided by 0 is nothing, not infinity.
So no, gravity is not infinite at the centre of mass, nor is gravity necessarily infinite at the centre of black holes. The Earth won't get sucked into itself, that doesn't make sense. For the Earth's mass to become a black hole, the Earth would have to be crushed to the size of a peanut. Then the density is so great it collapses in on itself to become a black hole.
tl;dr - your equation is wrong
In the centre of a black hole is a gravitational singularity, a one-dimensional point which contains a huge mass in an infinitely small space, where density and gravity become infinite and space-time curves infinitely, and where the laws of physics as we know them cease to operate....’a bit like the inside of shifters head, but I thought you hated gravity!
Just because your primitive knowledge doesn't understand what happens doesn't mean it is unknowable. Terms like 'laws of physics break down' or your 'cease to operate' is just short hand for 'I have no fucking idea what I'm talking about' and 'Our understanding of physics is incomplete and wrong'
Ok smart ass spill your beans....
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}Actually, the gravity at the center of the earth is zero because all of the mass around it pulls equally in all directions and cancels out.
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
https://www.pbslearningmedia.org/resource/oer08.sci.phys.maf.gravitynsn/gravity-at-earths-center/
Does gravity become infinity where the center of gravity of a mass lies?This equation doesn't work to calculate acceleration due to gravity, because you're using it in a case where nothing is being accelerated. There is no 'm' in your calculations, therefore there is no acceleration. You can see this because:
Gravity pulls towards the center of mass. Simple Example;
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
g = GM/R^^{2}=nothing
There is no R, since there's no 'distance between' anything, or 'radius between' anything if you like, since there's only one object. Also, even if R=0, anything divided by 0 is nothing, not infinity.
So no, gravity is not infinite at the centre of mass, nor is gravity necessarily infinite at the centre of black holes. The Earth won't get sucked into itself, that doesn't make sense. For the Earth's mass to become a black hole, the Earth would have to be crushed to the size of a peanut. Then the density is so great it collapses in on itself to become a black hole.
tl;dr - your equation is wrong
In the centre of a black hole is a gravitational singularity, a one-dimensional point which contains a huge mass in an infinitely small space, where density and gravity become infinite and space-time curves infinitely, and where the laws of physics as we know them cease to operate....’a bit like the inside of shifters head, but I thought you hated gravity!
Just because your primitive knowledge doesn't understand what happens doesn't mean it is unknowable. Terms like 'laws of physics break down' or your 'cease to operate' is just short hand for 'I have no fucking idea what I'm talking about' and 'Our understanding of physics is incomplete and wrong'
Ok smart ass spill your beans....
Firstly, I have tried to articulate the workings of the universe here prior, but either no one listens, cares or just parrots shit they think is correct and will take no other thinking seriously. Even if they are wrong and what they deny is the truth. Secondly, is it even wise to impart this knowledge to people that are far too young to deal with it? All I can do is guide you on a path. You must walk it.
What I will reveal is you need to think of black holes as not operating on a linear time dimension. If you wish to learn more about the nature of the universe you need to shake off the notion that time runs like a line, that has a beginning, middle and end. Time is a point (no not a spatial dimension style). Everything is, was and will be. The universe is both a planck hot singularity and an infinitely vast void of nothingness what you describe as its heat death as well as everything and every 'moment' in between all together. Your life experience and consciousness may seem linear, but for the universe it's not. What does this have to do with mass and black holes you ask? Well, what you perceive as the universe, IS the inside of a blackhole, or more apt, a singularity
You think you have free will, but do you? Everything post you make, has already been written ;)
Does gravity become infinity where the center of gravity of a mass lies?This equation doesn't work to calculate acceleration due to gravity, because you're using it in a case where nothing is being accelerated. There is no 'm' in your calculations, therefore there is no acceleration. You can see this because:
Gravity pulls towards the center of mass. Simple Example;
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
g = GM/R^^{2}=nothing
There is no R, since there's no 'distance between' anything, or 'radius between' anything if you like, since there's only one object. Also, even if R=0, anything divided by 0 is nothing, not infinity.
So no, gravity is not infinite at the centre of mass, nor is gravity necessarily infinite at the centre of black holes. The Earth won't get sucked into itself, that doesn't make sense. For the Earth's mass to become a black hole, the Earth would have to be crushed to the size of a peanut. Then the density is so great it collapses in on itself to become a black hole.
tl;dr - your equation is wrong
In the centre of a black hole is a gravitational singularity, a one-dimensional point which contains a huge mass in an infinitely small space, where density and gravity become infinite and space-time curves infinitely, and where the laws of physics as we know them cease to operate....’a bit like the inside of shifters head, but I thought you hated gravity!
Just because your primitive knowledge doesn't understand what happens doesn't mean it is unknowable. Terms like 'laws of physics break down' or your 'cease to operate' is just short hand for 'I have no fucking idea what I'm talking about' and 'Our understanding of physics is incomplete and wrong'
Ok smart ass spill your beans....
Firstly, I have tried to articulate the workings of the universe here prior, but either no one listens, cares or just parrots shit they think is correct and will take no other thinking seriously. Even if they are wrong and what they deny is the truth. Secondly, is it even wise to impart this knowledge to people that are far too young to deal with it? All I can do is guide you on a path. You must walk it.
What I will reveal is you need to think of black holes as not operating on a linear time dimension. If you wish to learn more about the nature of the universe you need to shake off the notion that time runs like a line, that has a beginning, middle and end. Time is a point (no not a spatial dimension style). Everything is, was and will be. The universe is both a planck hot singularity and an infinitely vast void of nothingness what you describe as its heat death as well as everything and every 'moment' in between all together. Your life experience and consciousness may seem linear, but for the universe it's not. What does this have to do with mass and black holes you ask? Well, what you perceive as the universe, IS the inside of a blackhole, or more apt, a singularity
You think you have free will, but do you? Everything post you make, has already been written ;)
So if that’s the case you will have read this before...
Your totally full of shit, but slightly funny with it.
Actually, the gravity at the center of the earth is zero because all of the mass around it pulls equally in all directions and cancels out.
Only as far as objects external to the earth are concerned. Inside the earth is a different story.QuoteActually, the gravity at the center of the earth is zero because all of the mass around it pulls equally in all directions and cancels out.
All the masses of the earth don't act separately when attracting things but act jointly at the center of the earth where the center of gravity of spherical earth is.
Isaac Newton proved the shell theorem[1] and stated that:
A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.
If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.
A corollary is that inside a solid sphere of constant density, the gravitational force varies linearly with distance from the centre, becoming zero by symmetry at the centre of mass.
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}Actually, the gravity at the center of the earth is zero because all of the mass around it pulls equally in all directions and cancels out.
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
https://www.pbslearningmedia.org/resource/oer08.sci.phys.maf.gravitynsn/gravity-at-earths-center/
At this point, outside forces come into play, like the sun, or the jiggling jugs of an alien babe on the far end of the Triangulum galaxy. There is no point in the universe where this 'gravity' is non existent. Everything that has mass is attracted to everything in the universe. It is measurable. No exceptions
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}Actually, the gravity at the center of the earth is zero because all of the mass around it pulls equally in all directions and cancels out.
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
https://www.pbslearningmedia.org/resource/oer08.sci.phys.maf.gravitynsn/gravity-at-earths-center/
At this point, outside forces come into play, like the sun, or the jiggling jugs of an alien babe on the far end of the Triangulum galaxy. There is no point in the universe where this 'gravity' is non existent. Everything that has mass is attracted to everything in the universe. It is measurable. No exceptions
Yeah, he should have said "the gravity at the center of the earth due to the mass of the earth is zero because all of the mass around it pulls equally in all directions and cancels out."
Better?
BTW, gravity from masses in the Triangulum galaxy (whatever they are, titillating as your suggestion about what is important might be) is insignificant, and what is there is mostly balanced by gravity from similar (or otherwise) masses in galaxies in the opposite and all other directions, anyway, so it can be considered to be zero for most computations.
Only as far as objects external to the earth are concerned. Inside the earth is a different story.
And I provided a reference showing why that doesn't apply inside a solid sphere. Here it is again. This time please make an effort to read it.QuoteOnly as far as objects external to the earth are concerned. Inside the earth is a different story.
The weight of a mass of 1 kg at the base of theoretical hole whose base is just 1m above the center of earth would be = w = mg = 1 (GM/(d)^^{2 }= 3.8x10^1^{4} N = I am just using mathematical equation w = mg
And I provided a reference showing why that doesn't apply inside a solid sphere....There is a difference between force on the particle and force between the particles. I mentioned this in one of my previous posts.
If the gravity of earth is zero at its center then this means the value of gravitational constant "G" is zero at the center of gravity of all masses.That might be true if the formula you are using is scientific and accurate. But as has been demonstrated, it is not.
The "g" of an earth decreases above the surface of the earth while increases below its surface till it reach infinity. Try at a point which is 0.000000000000000000000000000000000000000000001 mm just above the center of the earth. Plug it in the formula. This is just an example.Your example is many trillions times smaller than the Planck length.
Your example is many trillions times smaller than the Planck length.- i said that was just for your example. you can try this
QuoteYour example is many trillions times smaller than the Planck length.- i said that was just for your example. you can try this
The weight of a mass of 1 kg at the base of theoretical hole whose base is just 1m above the center of earth would be = w = mg = 1 (GM/(d)^2 = 3.8x10^14 N = I am just using mathematical equation w = mg
Details matter.QuoteYour example is many trillions times smaller than the Planck length.- i said that was just for your example. you can try this
The weight of a mass of 1 kg at the base of theoretical hole whose base is just 1m above the center of earth would be = w = mg = 1 (GM/(d)^2 = 3.8x10^14 N = I am just using mathematical equation w = mg
Yes, but the force of gravitation is a vector and so the gravitational forces of all points outside the radius of your "test particle" cancel.QuoteAnd I provided a reference showing why that doesn't apply inside a solid sphere....There is a difference between force on the particle and force between the particles. I mentioned this in one of my previous posts.
Gravity is always attractive, never cancel. As gravity pulls towards the center of mass, therefore, the power of gravity of all masses of earth concentrated at its center. If the gravity of earth is zero at its center then this means the value of gravitational constant "G" is zero at the center of gravity of all masses.
Actually, the gravity at the center of the earth is zero because all of the mass around it pulls equally in all directions and cancels out.Or read Wikipedia, Shell theorem (https://en.m.wikipedia.org/wiki/Shell_theorem)
https://www.pbslearningmedia.org/resource/oer08.sci.phys.maf.gravitynsn/gravity-at-earths-center/
At the center of the Earth g is zero -FINE i have no problem with that but it will make gravitational constant "G" zero at the center of earth as said before.
Your formula assumes that the entire mass of the earth is 1 meter away from your 1 kg mass.First its not my formula. second the whole mass of earth concentrated at its center. Try at that place, plumb bob suspended from the top surface of the earth in a thought experiment.
regarding shell theorem -i went through it many times but just keep in mind force of gravitatioon is BETWEEN two mass not ON masses - Newton says
So you believe Newton's general formula for gravity, but you don't believe the shell theorem that he proved to be correct for spherical objects? Sorry, I can't help.QuoteAnd I provided a reference showing why that doesn't apply inside a solid sphere....There is a difference between force on the particle and force between the particles. I mentioned this in one of my previous posts.
Gravity is always attractive, never cancel. As gravity pulls towards the center of mass, therefore, the power of gravity of all masses of earth concentrated at its center. If the gravity of earth is zero at its center then this means the value of gravitational constant "G" is zero at the center of gravity of all masses.
So you believe Newton's general formula for gravity, but you don't believe the shell theorem that he proved to be correct for spherical objects? Sorry, I can't help.
The whole mass of the Earth is not concentrated at its center in the formula you used. (Your formula. Knock off the pedantry.) Your thought experiment is flawed.QuoteYour formula assumes that the entire mass of the earth is 1 meter away from your 1 kg mass.First its not my formula. second the whole mass of earth concentrated at its center. Try at that place, plumb bob suspended from the top surface of the earth in a thought experiment.
Acceleration due to the gravity of earth on its surface; g = GM/R^^{2}Actually, the gravity at the center of the earth is zero because all of the mass around it pulls equally in all directions and cancels out.
Acceleration due to gravity of earth at its center; g = GM/R^^{2} = INFINITY, where R=0
https://www.pbslearningmedia.org/resource/oer08.sci.phys.maf.gravitynsn/gravity-at-earths-center/
At this point, outside forces come into play, like the sun, or the jiggling jugs of an alien babe on the far end of the Triangulum galaxy. There is no point in the universe where this 'gravity' is non existent. Everything that has mass is attracted to everything in the universe. It is measurable. No exceptions
Yeah, he should have said "the gravity at the center of the earth due to the mass of the earth is zero because all of the mass around it pulls equally in all directions and cancels out."
Better?
BTW, gravity from masses in the Triangulum galaxy (whatever they are, titillating as your suggestion about what is important might be) is insignificant, and what is there is mostly balanced by gravity from similar (or otherwise) masses in galaxies in the opposite and all other directions, anyway, so it can be considered to be zero for most computations.
I don't care if there is a Googolplex of zeros after a decimal point before a simple '1'. It is measurable at some point. Sure, our clumsy senses might not know the difference or feel ourselves being pushed and pulled from all directions but at some point, you can measure a number for anything that has mass in the universe.
The whole mass of the Earth is not concentrated at its center in the formula you used. (Your formula. Knock off the pedantry.) Your thought experiment is flawed.
QuoteThe whole mass of the Earth is not concentrated at its center in the formula you used. (Your formula. Knock off the pedantry.) Your thought experiment is flawed.
come to the question - what is the gravity of earth at its own center - zero or infinity
QuoteThe whole mass of the Earth is not concentrated at its center in the formula you used. (Your formula. Knock off the pedantry.) Your thought experiment is flawed.
come to the question - what is the gravity of earth at its own center - zero or infinity
the velocity is at a maximum and the acceleration is zero- no "g" means no velocity as acceleration is the change in speed
Zero. If you stood there, the mass to your left would cancel the mass to your right. And so on.
Hypothetically if we could drill a hole through the Earth, traversing the centre, and dropped an object into the hole, then we would expect to see simple harmonic motion. The acceleration is at a maximum at the surface, and the velocity is equal to zero. At the centre, the velocity is at a maximum and the acceleration is zero.
Quotethe velocity is at a maximum and the acceleration is zero- no "g" means no velocity as acceleration is the change in speed
QuoteZero. If you stood there, the mass to your left would cancel the mass to your right. And so on.
Hypothetically if we could drill a hole through the Earth, traversing the centre, and dropped an object into the hole, then we would expect to see simple harmonic motion. The acceleration is at a maximum at the surface, and the velocity is equal to zero. At the centre, the velocity is at a maximum and the acceleration is zero.
The mass of earth above the said object pulls it upward while the mass of the earth below the object pulls it downward at any point during its fall via the said hole. An object starts losing its acceleration "g" slowly till becomes zero at to center due bto equal pull in all direction on it.
QuoteAt the center of the Earth g is zero -FINE i have no problem with that but it will make gravitational constant "G" zero at the center of earth as said before.
Nope.- Great
It would make internal part of the mass zero.
Every force between two bodies is actally resutant of all forces between atoms of the two objects.
QuoteEvery force between two bodies is actally resutant of all forces between atoms of the two objects.
I think Newton didn't discuss his law of gravity at atomic level however if you want then you can calculate the gravitational acceleration (g = GM/R^^{2}) of proton or neutron at their centers; where M is the mass of proton or neutron and R is the radius of proton or neutron. At their centers, R=0 but not M.
QuoteEvery force between two bodies is actally resutant of all forces between atoms of the two objects.
I think Newton didn't discuss his law of gravity at atomic level however if you want then you can calculate the gravitational acceleration (g = GM/R^^{2}) of proton or neutron at their centers; where M is the mass of proton or neutron and R is the radius of proton or neutron. At their centers, R=0 but not M.
I know what would you like things to be, but I can't help you with that.
Newton discussed his laws for gravitational fields outside of objects.
Newton discussed his laws for gravitational fields outside of objects.- Any source
Newton and all the objects he was using in his experiments were outside the earth.QuoteNewton discussed his laws for gravitational fields outside of objects.- Any source
But Newton's universal law of gravitation states that any two masses attract each other with a force equal to a constant (constant of gravitation) multiplied by the product of the two masses and divided by the square of the distance between them - I don't see anything regarding gravitational fields outsides objects.
Why can't you believe what you are told by people who obviously know far more than you do? Read this again:Actually, the gravity at the center of the earth is zero because all of the mass around it pulls equally in all directions and cancels out.Or read Wikipedia, Shell theorem (https://en.m.wikipedia.org/wiki/Shell_theorem)
https://www.pbslearningmedia.org/resource/oer08.sci.phys.maf.gravitynsn/gravity-at-earths-center/
Or Physics Forums, Gravity inside a solid sphere. (https://www.physicsforums.com/threads/gravity-inside-a-solid-sphere.148579/)
how would you handle an object inside a half-spherical shell
how would you handle an object inside a half-spherical shellWith great difficulty.
It would be different. Why do you ask?
In the general case you need to sum all the individual forces between each small piece of one object with each small piece of the other object. The problem becomes a 3-D vector integration and a "bit above my pay-grade", which doesn't say much as my pay-grade is zero.QuoteIt would be different. Why do you ask?
Please explain how would you apply universal law gravitation in an aforesaid situation (an object inside a half-spherical shell)? - I wanna know
Edit: Would the praxis of shell theorem be amenable to hemispherical shell or quasi-cross-section, disorderly holes in the shell and asymmetrical shell?
In the general case you need to sum all the individual forces between each small piece of one object with each small piece of the other object. The problem becomes a 3-D vector integration and a "bit above my pay-grade", which doesn't say much as my pay-grade is zero.QuoteIt would be different. Why do you ask?
Please explain how would you apply universal law gravitation in an aforesaid situation (an object inside a half-spherical shell)? - I wanna know
It is a little simpler if you just want the gravitational field at one point near a fairly regular object.
Often that object can be broken up into simpler parts, like rings etc, and these summed.
In general the only practical solution is to sum all individual contributions numerically, see:
Physics Stack Exchange, Gravitational Field from Irregular Object. (https://physics.stackexchange.com/questions/69514/gravitational-field-from-irregular-object)
And to show the complexity of the problem, there is: Physics Forums, Gravitational field strength for irregular object. (https://www.physicsforums.com/threads/gravitational-field-strength-for-irregular-object.629942)Quote from: E E KEdit: Would the praxis of shell theorem be amenable to hemispherical shell or quasi-cross-section, disorderly holes in the shell and asymmetrical shell?
I don't think so, but maybe someone more qualified than I can tell you more.
In the general case you need to sum all the individual forces between each small piece of one object with each small piece of the other object. The problem becomes a 3-D vector integration and a "bit above my pay-grade", which doesn't say much as my pay-grade is zero.QuoteIt would be different. Why do you ask?
Please explain how would you apply universal law gravitation in an aforesaid situation (an object inside a half-spherical shell)? - I wanna know
It is a little simpler if you just want the gravitational field at one point near a fairly regular object.
Often that object can be broken up into simpler parts, like rings etc, and these summed.
In general the only practical solution is to sum all individual contributions numerically, see:
Physics Stack Exchange, Gravitational Field from Irregular Object. (https://physics.stackexchange.com/questions/69514/gravitational-field-from-irregular-object)
And to show the complexity of the problem, there is: Physics Forums, Gravitational field strength for irregular object. (https://www.physicsforums.com/threads/gravitational-field-strength-for-irregular-object.629942)Quote from: E E KEdit: Would the praxis of shell theorem be amenable to hemispherical shell or quasi-cross-section, disorderly holes in the shell and asymmetrical shell?
I don't think so, but maybe someone more qualified than I can tell you more.
Someone who doesn't think satellites sit on tables maybe?
https://www.theflatearthsociety.org/forum/index.php?topic=73925.msg2024607#msg2024607
QuoteNewton discussed his laws for gravitational fields outside of objects.- Any source
But Newton's universal law of gravitation states that any two masses attract each other with a force equal to a constant (constant of gravitation) multiplied by the product of the two masses and divided by the square of the distance between them - I don't see anything regarding gravitational fields outsides objects.
I don't think so, but maybe someone more qualified than I can tell you more.- You seem to be an honest guy
Ok, apply your own claim on explanation WHY homogenous infinite plane doesn't change attraction on object with object's distance from the plane.- I don’t have any personal claim. I just follow the laws of physics. Can you rephrase your question I didn't understand
Also, WHY is gravity force inside shell zero?- I never said gravity force inside shell is zero – Again G = 0, when gravity force or “g” = 0
Drive an equation for the acceleration due to gravity at any depth within the earth’s surfaceIs it derived by Newton?
Also, how one can determine the acceleration due to gravity “g =GM/R^2” of a spherical shell,The spherical shell is easy. Inside g = 0 and outside g =GM/R^2.
hemispherical shell, disorderly holes in the shell and asymmetrical shell etc - Let the mass of each = 1 kg. and Radius = 1 meter - Just wonderingNot at all easy, each has to be down by a complicated integration process. Often this is done numerically.
QuoteOk, apply your own claim on explanation WHY homogenous infinite plane doesn't change attraction on object with object's distance from the plane.- I don’t have any personal claim. I just follow the laws of physics. Can you rephrase your question I didn't understand
QuoteAlso, WHY is gravity force inside shell zero?- I never said gravity force inside shell is zero – Again G = 0, when gravity force or “g” = 0
Let an object located inside the spherical shell. The acceleration due to gravity “g” of an object is much greater (say 1000 times) than the “g” of a spherical shell. The radius of the object is much smaller than the radius of spherical shell. An object may be solid sphere, semi-solid spherical shell etc - What would you think about the aforesaid?
Also, how one can determine the acceleration due to gravity “g =GM/R^2” of a spherical shell, hemispherical shell, disorderly holes in the shell and asymmetrical shell etc - Let the mass of each = 1 kg. and Radius = 1 meter - Just wondering
Yes.QuoteDrive an equation for the acceleration due to gravity at any depth within the earth’s surfaceIs it derived by Newton?
You didn't say it, but physics did, and you say you follow laws of physics.Let's try again
Inside homogenous hollow sphere acceleration g and gravitational force F is zero.
I think that a lot of your confusion is trying to assign g's to every object. In my opinion the idea of assigning a gravitational acceleration to an object is useful only for massive objects like the earth or the moon.QuoteYou didn't say it, but physics did, and you say you follow laws of physics.Let's try again
Inside homogenous hollow sphere acceleration g and gravitational force F is zero.
Imagine a small solid sphere “A” located inside a very large hollow sphere “B” but their centers don’t coincide. The gravitational acceleration “g” of “A” is much greater than “g” of “B”.
There are two possibilitiesThere is no gravitational field inside the "very large hollow sphere B" due to B. So there is no force on A or B.
• Gravitational Force between two masses: Center of both “A” and “B” coincide due to universal law of gravitation
• Gravitational Force on mass: Part of “B” located near “A” fall on the solid sphere due to less distance and greater accelerations due to gravities. The greater is the distance them the smaller is the “g”
The force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth but what makes the difference are accelerations due to gravitiesNo, the accelerations are due to the forces. One way to calculate "g" is to calculate the gravitational force on a test mass of 1 Kg.
You seem to sum all the individual forces and make the resultant zero but ignore their accelerations due to gravities.There are no separate "accelerations due to gravities". The acceleration of each object is simply dye to the one force acting on each object.
I asked a question about gravity at the center of earth but you said: “it would make internal part of the mass zero” by limiting the value of R to zero in g=GM/R^^{2}.As I and others have said, there is no gravitational field inside a spherical shell, due to that shell. This is true for any thickness shell.
Gravity at the center of earth with a radius of R is not equal to zero. This is the center of gravity of earth or this is the center where gravity of whole mass of earth appears to act.No, I'm afraid that "Gravity at the center of earth with a radius of R is" equal to zero.
QuoteYou didn't say it, but physics did, and you say you follow laws of physics.Let's try again
Inside homogenous hollow sphere acceleration g and gravitational force F is zero.
Imagine a small solid sphere “A” located inside a very large hollow sphere “B” but their centers don’t coincide. The gravitational acceleration “g” of “A” is much greater than “g” of “B”.
There are two possibilities
• Gravitational Force between two masses: Center of both “A” and “B” coincide due to universal law of gravitation
• Gravitational Force on mass: Part of “B” located near “A” fall on the solid sphere due to less distance and greater accelerations due to gravities. The greater is the distance them the smaller is the “g”
The force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth but what makes the difference are accelerations due to gravities
You seem to sum all the individual forces and make the resultant zero but ignore their accelerations due to gravities.
I asked a question about gravity at the center of earth but you said: “it would make internal part of the mass zero” by limiting the value of R to zero in g=GM/R^^{2}.
Gravity at the center of earth with a radius of R is not equal to zero. This is the center of gravity of earth or this is the center where gravity of whole mass of earth appears to act.
The force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth but what makes the difference are accelerations due to gravities
Inside solid hollow homogenous sphere resultant g is zero, not because G is zero.
Total pull is zero.Most physicists agreed that gravities do not cancel each other even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero.
No, as has been presented numerous times and with proof.QuoteInside solid hollow homogenous sphere resultant g is zero, not because G is zero.
Case #1: Resultant "g" is zero at the center of solid hollow homogenous sphere (HS) when there is no mass M at all inside HS
Case #2: Presence of M inside HS, however, M can be more in numbers too
There are two gravitational fields inside HS
1- Due to HS
2- Due to M
Now there are three conditionsSince the “g” of HS inside the HS is everywhere zero, obviously the only case is “g” of HS < M.
The “g” of HS < M, The “g” of HS = M, The “g” of HS > M
Also, the size, mass, and shape of M can be varied and its posture & location inside HS as wellThere is no question at all the total pull is zero.
All above is self-explanatory however I can explain in there is any question
Where did you drag this up from, "Most physicists agreed that gravities do not cancel each other even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero."QuoteTotal pull is zero.Most physicists agreed that gravities do not cancel each other even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero.
Mass closer to one side of the wall of HS: it depends upon the “g” of M and "g" of the mass of HS closer to the M. We know “g” increases with a decrease in distance.
QuoteInside solid hollow homogenous sphere resultant g is zero, not because G is zero.
Case #1: Resultant "g" is zero at the center of solid hollow homogenous sphere (HS) when there is no mass M at all inside HS
Case #2: Presence of M inside HS, however, M can be more in numbers too
There are two gravitational fields inside HS
1- Due to HS
2- Due to M
Now there are three conditions
The “g” of HS < M, The “g” of HS = M, The “g” of HS > M
Also, the size, mass, and shape of M can be varied and its posture & location inside HS as well
All above is self-explanatory however I can explain in there is any questionQuoteTotal pull is zero.Most physicists agreed that gravities do not cancel each other even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero.
Mass closer to one side of the wall of HS: it depends upon the “g” of M and "g" of the mass of HS closer to the M. We know “g” increases with a decrease in distance.
Mass on one side is closer, but there is more mass on another side.There are three masses and three centers of gravities “cg”
Case #1: Resultant "g" is zero at the center of solid hollow homogenous sphere (HS) in the absence of mass both inside and outsideIt's no point you saying the same ting over and over again. Whether you accept it or not,
Gravity Inside a Spherical ShellYou could also look at: Wikipedia, Shell theorem (https://en.m.wikipedia.org/wiki/Shell_theorem) or Quora.com Why is gravity inside a spherical shell considered to be zero? (https://www.quora.com/Why-is-gravity-inside-a-spherical-shell-considered-to-be-zero)(http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/imgmech/sphshellin.gif)For application of the law of gravity inside a uniform spherical shell of mass M, a point is chosen on the axis of a circular strip of mass. The problem is envisioned as dividing an infinitesemally thin spherical shell of density σ per unit area into circular strips of infinitesemal width. The choice of such a point involves no loss of generality because for any point inside the shell, the mass elements could be chosen so that the point is on their symmetry axis.<< I'll let you look up the calculations if you feel qualified. >>From: Gravity Force Inside a Spherical Shell (http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html)
Most physicists agreed that gravities do not cancel each otherNo! That is completely untrue. Gravitation has direction and so the gravitation from two object's can cancel.
even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero.The bit you are ignoring is that the closer part is smaller in area than the farther part by exactly the correct ratio for their gravitational forces to be equal and opposite.
Mass closer to one side of the wall of HS: it depends upon the “g” of M and "g" of the mass of HS closer to the M. We know “g” increases with a decrease in distance.
Your link to Quora points to pretty clear image of distribution of forces:- This is one way where is particle pull?
QuoteYour link to Quora points to pretty clear image of distribution of forces:- This is one way where is particle pull?
Following is the edit that I had done in the previous post. here it is if missed
Particle mass is Gravitating mass
Closer mass of HS is Falling mass
More mass of HS is Falling mass
But the value of “g” of particle at height h or at the ”cg” of closer mass is greater than the value of “g” of particle at height h1 or at the “cg” of more mass
Here is another attempt at explaining that there is no "gravity" inside a spherical shel, due to thst shell:QuoteYour link to Quora points to pretty clear image of distribution of forces:- This is one way where is particle pull?
Following is the edit that I had done in the previous post. here it is if missed
Particle mass is Gravitating mass
Closer mass of HS is Falling mass
More mass of HS is Falling mass
But the value of “g” of particle at height h or at the ”cg” of closer mass is greater than the value of “g” of particle at height h1 or at the “cg” of more mass
Gravitational field intensity inside a hollow sphere
It is quite easy to derive the gravitational field intensity at a point within a hollow sphere. However, the result is quite surprising. The field intensity at any point within a hollow sphere is zero.
Answer from Hypnosifl
One intuitive way I've seen to think about the math is that if you are at any position inside the hollow spherical shell, you can imagine two cones whose tips are at your position, and which both lie along the same axis, widening in opposite direction. Imagine, too, that they both subtend the same solid angle, but the solid angle is chosen to be infinitesimal. Then you can consider the little chunks of matter where each cone intersects the shell, as in the diagram on this page:
(https://i.stack.imgur.com/CHcrm.gif)
One intuitive way I've seen to think about the math is that if you are at any position inside the hollow spherical shell, you can imagine two cones whose tips are at your position, and which both lie along the same axis, widening in opposite direction. Imagine, too, that they both subtend the same solid angle, but the solid angle is chosen to be infinitesimal. Then you can consider the little chunks of matter where each cone intersects the shell, as in the diagram on this page:
You still need to do a bit of geometric math, but you can show that the area of each red bit is proportional to the square of the distance from you (the blue point) to it--and hence the mass of each bit is also proportional to the square of the distance, since we assume the shell has uniform density. But gravity obeys an inverse-square law, so each of those two bits should exert the same gravitational pull on you, but in opposite directions, meaning the two bits exert zero net force on you. And you can vary the axis along which the two cones are drawn so that every point on the surface of the shell ends up being part of a pair like this, which leads to the conclusion that the entire spherical shell exerts zero net force on you.
Physics Stack Exchange, Gravitational field intensity inside a hollow sphere (https://physics.stackexchange.com/questions/150238/gravitational-field-intensity-inside-a-hollow-sphere)
If feather and hammer fall together, they hit Earth together, because they have combined pull and Earth intercepts them together.
If they fall in succession, one and then another, hammer would in deed fall a tiny bit faster, but that tiny bit is many times tinier than our measuring abilities.
Mathematically it can be shown, but there is no way (not even slightest chance) to measure such small difference in reality.
If feather and hammer fall together, they hit Earth together, because they have combined pull and Earth intercepts them together.- i think you missed one of my previous posts regarding hammer and feather
If they fall in succession, one and then another, hammer would in deed fall a tiny bit faster, but that tiny bit is many times tinier than our measuring abilities.
Mathematically it can be shown, but there is no way (not even slightest chance) to measure such small difference in reality.
Since g_{e } > g_{h} > g_{f} therefore earth falls on hammer at greater accelaration than falls on feather - quite simpleWould you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?
Earth moves towards hammer if both hammer and feather drop at the same time from the same altitude on antipodes
Since g_{e } > g_{h} > g_{f} therefore earth falls on hammer at greater accelaration than falls on feather - quite simpleWould you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?
Earth moves towards hammer if both hammer and feather drop at the same time from the same altitude on antipodes
All you talk of separate g_{e }, g_{h} and g_{f} is a totally ridiculous waste of time because the typical masses would be:If you have any concept of relative values you might get the message.
- Earth: 5.972 × 10^{24} kg,
- Hammer: 2 kg and
- Pigeon Tail Feather: 50 mg.
PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.
Since g_{e } > g_{h} > g_{f} therefore earth falls on hammer at greater accelaration than falls on feather - quite simpleWould you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?
Earth moves towards hammer if both hammer and feather drop at the same time from the same altitude on antipodes
All you talk of separate g_{e }, g_{h} and g_{f} is a totally ridiculous waste of time because the typical masses would be:If you have any concept of relative values you might get the message.
- Earth: 5.972 × 10^{24} kg,
- Hammer: 2 kg and
- Pigeon Tail Feather: 50 mg.
PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.
Whatever you claim Newton's Shell Theorem is correct
EEK thinks he has discovered something- Neither claim nor discovery but its just a truth
PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.What do you think if the "g" of M >>>>>>> the "g" of HS? For Example, Earth (ideally sphere) inside a homogeneous HS. The "g" of earth >>>>>>>> the "g" HS
I did ask E E K, "Would you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?"Since g_{e } > g_{h} > g_{f} therefore earth falls on hammer at greater accelaration than falls on feather - quite simpleWould you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?
Earth moves towards hammer if both hammer and feather drop at the same time from the same altitude on antipodes
All you talk of separate g_{e }, g_{h} and g_{f} is a totally ridiculous waste of time because the typical masses would be:If you have any concept of relative values you might get the message.
- Earth: 5.972 × 10^{24} kg,
- Hammer: 2 kg and
- Pigeon Tail Feather: 50 mg.
PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.
Where is it falling? Obviously must be gravitationally affected by something which in turn means the object with heavier mass is pulling on that object more than the object with a lighter mass. Also the object with a lighter mass will be gravitationally affected by the falling object that is heavier. An infinitesimal amount sure but enough that God himself would consider it a variable in the experiment. The idea of experiments is to eliminate the variables
If the earth, of mass say M_{e}, were inside and concentric with a huge HS of any mass, say m_{HS}, then:QuoteWhatever you claim Newton's Shell Theorem is correctQuoteEEK thinks he has discovered something- Neither claim nor discovery but its just a truthQuotePS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.What do you think if the "g" of M >>>>>>> the "g" of HS? For Example, Earth (ideally sphere) inside a homogeneous HS. The "g" of earth >>>>>>>> the "g" HS
I did ask E E K, "Would you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?"
But the difference is so small that there not the slightest chance of evenQuoteI did ask E E K, "Would you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?"
Earth and hammer will strike first if aforementioned both hammer and feather drop at the same time from the same altitude but on antipodes - It's all about the concept, not calculation
As said in above Edit:There is no "natural tendency"
All objects may free fall at the same rate regardless of their mass if it is the natural tendency of smaller objects to drive towards a massive object in the region.
Barycenter
Rather than appearing to orbit a common center of mass with the smaller body, the larger will simply be seen to wobble slightly. This is the case for the Earth–Moon system, where the barycenter is located on average 4,671 km (2,902 mi) from the Earth's center, well within the planet's radius of 6,378 km (3,963 mi).
(https://upload.wikimedia.org/wikipedia/commons/5/59/Orbit3.gif)
Two bodies with a major difference
in mass orbiting a common
barycenter internal to one body
(similar to the Earth–Moon system)
distances not to scale
The moon is big enough to have a significant and the earth-moon system orbits a common centre, the barycentre,Freedom of choice - Choose what is right for You
But that doesn't make your choice correct, though in this Trump era, with it's fake-fax and alternate-fax, who knows?QuoteThe moon is big enough to have a significant and the earth-moon system orbits a common centre, the barycentre,Freedom of choice - Choose what is right for You
If feather and hammer fall together, they hit Earth together, because they have combined pull and Earth intercepts them together.
If they fall in succession, one and then another, hammer would in deed fall a tiny bit faster, but that tiny bit is many times tinier than our measuring abilities.
Mathematically it can be shown, but there is no way (not even slightest chance) to measure such small difference in reality.
Wrong
That's like saying someone with an open parachute and someone without a parachute will land at the same time.
Aerodynamics and the much lighter weight of the feather could result in the feather being picked up by the wind and not landing somewhere else a long time after.
The moon is big enough to have a significant and the earth-moon system orbits a common centre, the barycentre,Gravitational force between sun and moon; F_{1} = GMm/d^^{2}; where M = mass of sun R = radius of sun and m= mass of moon
That was what Apollo astronaut did on the Moon.It might be true if they really went to the moon but the mass or the gravity of the moon is still unknown according to the mainstream science. Had you missed one of my posts? Here
QuoteThe moon is big enough to have a significant and the earth-moon system orbits a common centre, the barycentre,Gravitational force between sun and moon; F_{1} = GMm/d^^{2}; where M = mass of sun R = radius of sun and m= mass of moon
Gravitational force between earth and moon; F_{2} = GMm/d^^{2}; where M = mass of earth R = radius of earth, and m = mass of moon
Since F_{1} > F_{2} after calculation therefore shouldn’t moon revolve the sun in its separate orbit?
Gravitational force between earth and sun; F = GMm/d^^{2}; where M = mass of sun, R = radius of sun and m= mass of earth
Gravitational acceleration of sun; g_{s} = GM/R^^{2} ;
Earth revolves in its orbit due to g_{s} = GM/R^^{2},
Doesn’t earth change its acceleration (in reference to the g_{s}=GM/d^^{2} of sun) due to the barycenter of earth and moon in its orbit around the sun?
Incorrect. The mass and gravity of the moon were calculated approximately long ago from the tidal effect on earth and from the position of the earth-moon barycentre.QuoteThat was what Apollo astronaut did on the Moon.It might be true if they really went to the moon but the mass or the gravity of the moon is still unknown according to the mainstream science. Had you missed one of my posts? Here
https://www.theflatearthsociety.org/forum/index.php?topic=74238.0 (https://www.theflatearthsociety.org/forum/index.php?topic=74238.0)
The Soviet Union sent the first spacecraft to the vicinity of the Moon, the robotic vehicle Luna 1, on January 4, 1959. It passed within 6,000 kilometres (3,200 nmi; 3,700 mi) of the Moon's surface, but did not achieve lunar orbit. Luna 3, launched on October 4, 1959, was the first robotic spacecraft to complete a circumlunar free return trajectory, still not a lunar orbit, but a figure-8 trajectory which swung around the far side of the Moon and returned to the Earth. This craft provided the first pictures of the far side of the Lunar surface.
The Soviet Luna 10 became the first spacecraft to actually orbit the Moon in April 1966. It studied micrometeoroid flux, and lunar environment until May 30, 1966.
The first United States spacecraft to orbit the Moon was Lunar Orbiter 1 on August 14, 1966. The first orbit was an elliptical orbit, with an apolune of 1,008 nautical miles (1,867 km; 1,160 mi) and a perilune of 102.1 nautical miles (189.1 km; 117.5 mi).[5] Then the orbit was circularized at around 170 nautical miles (310 km; 200 mi) to obtain suitable imagery. Five such spacecraft were launched over a period of thirteen months, all of which successfully mapped the Moon, primarily for the purpose of finding suitable Apollo program landing sites.
You are using the same letter "d" for distance in all formulas.- I presumed you all know that at this stage.
incorrect. The mass and gravity of the moon were calculated approximately long ago from the tidal effect on earth and from the position of the earth-moon barycentre.
According to what link, this one, Does "g =GM/d^2" best in situ in mathematical equation of "F = GMm/d^2 = mg"? (https://www.theflatearthsociety.org/forum/index.php?topic=74238.0)?Quoteincorrect. The mass and gravity of the moon were calculated approximately long ago from the tidal effect on earth and from the position of the earth-moon barycentre.
Your claim is unbeknownst to me but none of the celestial bodies mass or gravity is known as per aforesaid link.
It is comparitively easy to calculate the mass of any body with one or more orbiting satellitesHow or Any reference
If the size of the body is known, its mass can also by calculated if an object can be observed falling into it.
Both of these methods were used to determine the mass of the moon even before any unmanned craft soft landed on the moon..
Easy, peasy: How do scientists measure or calculate the weight of a planet? (https://www.scientificamerican.com/article/how-do-scientists-measure/)QuoteIt is comparitively easy to calculate the mass of any body with one or more orbiting satellitesHow or Any reference
If the size of the body is known, its mass can also by calculated if an object can be observed falling into it.
Both of these methods were used to determine the mass of the moon even before any unmanned craft soft landed on the moon..
I'll edit your equations to make it easier for me to follow. I hope it keeps your intent:QuoteThe moon is big enough to have a significant and the earth-moon system orbits a common centre, the barycentre,Gravitational force between sun and moon; F_{1} = GMm/d^^{2}; where M = mass of sun R = radius of sun and m= mass of moon
Gravitational force between earth and moon; F_{2} = GMm/d^^{2}; where M = mass of earth R = radius of earth, and m = mass of moon
Since F_{1} > F_{2} after calculation therefore shouldn’t moon revolve the sun in its separate orbit?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(https://www.wired.com/images_blogs/wiredscience/2012/12/earthmoonpath.jpeg) This just shows half a month. If I wanted to show a longer time period, the motion of the Earth and moon around the Sun would make it super-difficult to see the motion of the moon relative to the Earth. |
easy peasyThe only way we can measure a planet's mass is through its gravity as per your link - This needs a gravity model free of mathematical errors.
Since Fsm > Fem after calculation therefore shouldn’t moon revolve the sun in its separate orbit?
Quoteeasy peasyThe only way we can measure a planet's mass is through its gravity as per your link - This needs a gravity model free of mathematical errors.
QuoteSince Fsm > Fem after calculation therefore shouldn’t moon revolve the sun in its separate orbit?
There are 4 possible case according to current model - impov
1- When the earth is in between the sun and moon: Shouldn't the combined effect of earth and sun alter the orbit of the moon and its speed as well
2- when the moon is in between earth and sun: Either Moon should fall to the sun instead of earth or at least there should be a reduction in its speed and the change in its orbit around the earth
3- Moon should revolve the sun in its separate orbit instead of orbiting the earth
4- The current orbit of the earth around the sun doesn't fit for both earth and moon if there is a combined effect of binary plates (earth and moon) on sun
Errors limit the precision we're capable of obtaining, but if the errors (or unknowns) are reasonably small, the precision can be reasonably high.
There are no known mathematical errors, apart from using Newtonian Gravitation in lieu of General Relativity. Apart from a minute discrepancy in the precession of the perihelion of the planet Mercury GR is not necessary.Quoteeasy peasyThe only way we can measure a planet's mass is through its gravity as per your link - This needs a gravity model free of mathematical errors.
This was my answer to that question from the previous post:QuoteSince Fsm > Fem after calculation therefore shouldn’t moon revolve the sun in its separate orbit?
This bit is a very good question and is more than enough for one answer.
Part of the answer is that, as you say, the sun's gravitational field (gravity) near earth is greater than the gravitation field of the earth at the moon.
The sun's gravitational field (gravity) near the earth, however, changes only slightly over the radius of the moon's orbit around the earth.
In other words, the whole earth-moon system is attracted by the sun by almost the same amount, so the moon orbits the earth.
But, the earth-moon system travels around the sun at a much higher tangential velocity than the moon's velocity around the earth.
So the motion is an orbit around the sun at the same radius as the earth with just a little wriggle, with a period on one month, to make it orbit the earth.
This diagram, to scale, shows just half a month.Read in detail in: WIRED, Does the Moon Orbit the Sun or the Earth? (https://www.wired.com/2012/12/does-the-moon-orbit-the-sun-or-the-earth/)
(https://www.wired.com/images_blogs/wiredscience/2012/12/earthmoonpath.jpeg)
This just shows half a month. If I wanted to show a longer
time period, the motion of the Earth and moon around the
Sun would make it super-difficult to see the motion of the
moon relative to the Earth.
There are many more references on this, as at first it seems a little puzzling.
Physics Stack Exchange, Why does the moon not revolve around the sun directly? (https://physics.stackexchange.com/questions/92465/why-does-the-moon-not-revolve-around-the-sun-directly)
Then one that brings in Gravity Wells, EXPLAIN xkcd, Gravity Wells (https://www.explainxkcd.com/wiki/index.php/681:_Gravity_Wells).
And there are plenty more. Just search for "Why does moon orbit the earth and not the sun?".[/b]
There will be some effect, but what matters is not the absolute value os the sun's gravitational field near the earth, but on how much it changes during the moon's orbit.
There are 4 possibilities in the current model - impov
1- When the earth is in between the sun and moon: Shouldn't the combined effect of earth and sun alter the orbit of the moon and its speed as well
2- when the moon is in between earth and sun: Either Moon should fall to the sun instead of earth or at least there should be a reduction in its speed and the change in its orbit around the earth
3- Moon should revolve the sun in its separate orbit instead of orbiting the earthNo it shouldn't! That has been shown in all the above references.
Out here, at the distance we orbit the sun, the gravitational pull of the sun is only 0.0006 of the strength of the earth's gravity on the surface of the earth.or about 0.006 m/s^{2}.
4- The current orbit of the earth around the sun doesn't fit for both earth and moon if there is a combined effect of binary plates (earth and moon) on sunIt does fit perfectly well, though I can't follow your "combined effect of binary plates (earth and moon) on sun" is meant to mean.
It does fit perfectly well, though I can't follow your "combined effect of binary plates (earth and moon) on sun" is meant to mean.
What "combined effect"? Between the earth and moon, there is one gravitational force, F_{em} = G x M_{e} x m[/i]_{m}/d_{em}^{2}, where d_{em} is the distance between the "centres of gravity" (simply the centres if they are spherically symetrical).QuoteIt does fit perfectly well, though I can't follow your "combined effect of binary plates (earth and moon) on sun" is meant to mean.
We are going off the topic but shouldn't the radius of the orbit of binary planets (earth and moon) be greater than the greater current orbit of earth due to the combined effect.
Although binary planets (earth and moon) orbit around the sun but half of the moon cycle is in opposite direction while half of its cycle is along the direction of motion of barycentre in its orbit around the sun. So my question is why the orbital motion of the moon around earth doesn’t affect the speed as well as the position of barycentre around the sun in its orbit if the moon is influenced by both the “g” of sun and earth?For the earth-moon system what matters is not the absolute value of the sun's gravity near earth, but only the changes in that field around the moon's orbit.
moon-sun (km) Grav_{m-s} Change in Grav_{m-s} Grav_{m-e} min 149.3 x 10^{6} 0.00595 m/s^{2} 0.000031 m/s^{2} 0.0027 m/s^{2} avg 149.7 x 10^{6} 0.00592 m/s^{2} 0.000000 m/s^{2} 0.0027 m/s^{2} max 150.1 x 10^{6} 0.00589 m/s^{2} -0.000030 m/s^{2} 0.0027 m/s^{2} |