Super Polymath Posted April 29, 2017 Report Share Posted April 29, 2017 The article does not support your claim. Why do you suppose that after having been propelled into the void Is this "void" effected by gravity?? Quote Link to comment Share on other sites More sharing options...
JMJones0424 Posted April 29, 2017 Report Share Posted April 29, 2017 There is no area of the universe that is not effected by gravity Quote Link to comment Share on other sites More sharing options...
JMJones0424 Posted April 29, 2017 Report Share Posted April 29, 2017 Any point of liberation. An accelerated craft can be decelerated by gravitational attraction by, say, the mass of a large asteroid, can it not? Yes, and it must be. But, the point of escape velocity, is that an object that has achieved escape velocity has achieved the velocity necessary to ensure that gravity is insufficient to bring the object back. I don't understand why we are arguing over this. If you would expound upon your previous claim, that "I'm pretty sure spatial distortions have a gravitational range", then I might have a better understanding of what you are claiming. exchemist 1 Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted April 29, 2017 Report Share Posted April 29, 2017 (edited) Yes, and it must be. But, the point of escape velocity, is that an object that has achieved escape velocity has achieved the velocity necessary to ensure that gravity is insufficient to bring the object back. I don't understand why we are arguing over this. If you would expound upon your previous claim, that "I'm pretty sure spatial distortions have a gravitational range", then I might have a better understanding of what you are claiming.How is escape velocity possible without making gravitational range a variable? For instance, a larger mass will have a greater gravitational influence. Equalizing gravity equalizes gravitational range (given the laws of physics as defined by this cosmological constant) meaning that distance will forbid two atoms from gravitationally attracting. Why doesn't the moon fall into the earth? Distance, space. Which atom or orbits the other? The smaller one. It's a simply conceptualized idea that would state that the earth orbits the sun, not the other way around. Edited April 29, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted April 29, 2017 Report Share Posted April 29, 2017 (edited) Basically this thread asks if space experiences gravity. My answer is, not without mass. And you're saying that's wrong. Which goes against everything. Edited April 29, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
JMJones0424 Posted April 29, 2017 Report Share Posted April 29, 2017 For instance, a larger mass will have a greater gravitational influence. Equalizing gravity equalizes gravitational range (given the laws of physics as defined by this cosmological constant) meaning that distance will forbid two atoms from gravitationally attracting. Why doesn't the moon fall into the earth? Distance, space. No. The moon doesn't fall into the Earth because it is traveling too fast to fall into the Earth. What equation for gravity are you using that requires that at a specific distance, "that distance will forbid two atoms from gravitationally attracting?" The moon doesn't fall into the Earth because it is traveling on average faster relative to the Earth than it is falling towards the Earth. At times during its orbit, the moon is falling towards the Earth. At other times, it is falling away from the Earth. I am firmly convinced that you don't understand how orbits work and how gravity works. Please, you've maintained many times that there exists some distance that negates gravitational attraction between two bodies. This is obviously false. What I want to know is why you believe this to be the case. How can I show you that this assertion is false? exchemist 1 Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted April 29, 2017 Report Share Posted April 29, 2017 (edited) I mean, ****, range is just amount of curvature, you're saying that when curvature occurs at all it's area must be infinite. Idk if that's true, you certainly haven't given any citation. The further away you get from a mass the less curvature you must experience, right? So even if curvature is nonterminal, greater curvature (given equal area) must mean greater spatial imprint. Meaning that, given enough space, gravity eventually gets neutralized by other masses. Edited April 29, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
JMJones0424 Posted April 29, 2017 Report Share Posted April 29, 2017 Basically this thread asks if space experiences gravity. My answer is, not without mass. And you're saying that's wrong. Which goes against everything.Well, yes, if I had claimed such a ludicrous thing, it would go against everything. Quote Link to comment Share on other sites More sharing options...
JMJones0424 Posted April 29, 2017 Report Share Posted April 29, 2017 I mean, ****, range is just amount of curvature, you're saying that when curvature occurs at all it's area must be infinite. Idk if that's true, you certainly haven't given any citation.https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation#Modern_form I understand that GR refines this approximation, but I don't understand that as you claim there exists a distance at which two bodies can be separated in order to negate gravitational attraction. In spite of numerous requests, you haven't provided support for your claim. You haven't even bothered to explain it other than asserting that it must be obvious, while I understand that it must be obviously false. Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted April 29, 2017 Report Share Posted April 29, 2017 (edited) https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation#Modern_form I understand that GR refines this approximation, but I don't understand that as you claim there exists a distance at which two bodies can be separated in order to negate gravitational attraction. In spite of numerous requests, you haven't provided support for your claim. You haven't even bothered to explain it other than asserting that it must be obvious, while I understand that it must be obviously false. Noone says that in space there is no gravity. The force of gravity is F=G*m1*m2/r^2 therefore gravitational forces just decreases reaching zero at infinity--> hence never.Not so, inflation creates horizons that are intrinsically isolated, even from one another's gravitational fields, outside of each other's influence. The spatial imprint of masses collapses at the particle horizons because gravitational waves lack the velocity to pass the particle horizon. Additionally, there are parts of space, many gravitational fields can interrupt one another and create langrange points where, if a mass was sitting there, it wouldn't experience even infinitesimal amounts of gravitational tug. I tried to explain that earlier when I said "gravitational neutralization". Notice how all langrange points are sort of a zero g space - welcome to the inspiration for anti-gravity propulsion - the gravitational fields of the sun, moon, and earth do not effect these langrange points but seem to distort around them, as the waves interfere with one another. Add these facts together, and space has zero g between two cosmological horizons, as inflation is expanding that gap pushing both sides of the horizon apart FTL. Gravitational waves have been outraced by spatial inflation, thus they will never again catch up to inflationary velocities. So all gravitational fields do have a finite range. And langrange points are gravationally neutralized zero g space. Edited April 30, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
billvon Posted May 1, 2017 Report Share Posted May 1, 2017 Check most sources/books and they prefer to gravity in space as Zero-G (which is clearly wrong).Lately books have been using the more accurate term "microgravity." Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted May 1, 2017 Report Share Posted May 1, 2017 (edited) Lately books have been using the more accurate term "microgravity."Not even that. Space by itself has no mass, and thus, no gravity. Edited May 1, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
OceanBreeze Posted May 1, 2017 Report Share Posted May 1, 2017 Not even that. Space by itself has no mass, and thus, no gravity. Sure, but space does not exist “by itself”. Quote Link to comment Share on other sites More sharing options...
Darky Posted May 1, 2017 Author Report Share Posted May 1, 2017 Hi, Let me clear some things out. ☆ Super Polymath You're very wrong. There's nothing known as empty space since the happening of the Big Bang. However, let's imagine there is a space craft going through infinite empty space. In that case the space craft will have its own gravitational field. Thus, still giving off gravity. You've been proven wrong.Also, the escape velocity is entirely different from gravitation change. Gravity never ****ing changes. It's a proven fact (unless mass changes). When you reach an escape velocity, you're slowly drifting out of the specific range of the planet's gravitational pull. As the range decreases, the actual pull of gravity decreases, however, gravity doesn't. Please correct your statements. ☆ Billvon Hmm.. let me see. Do you have any books suggested to support yourself? ☆ JMJones The reason for the moon not following into earth is because it's stuck in the centrofield orbital space. Where at an extent, the gravity of the Sun cancels the gravity of earth. To be more precise, the gravity of earth candles the gravity of the Sun to an extent as at that range, the gravity of sun is larger pulling the moon outwards. As it rotates around, it reaches a point where the sun's gravity is in the same direction as the earth's. This is called the oval-orbital space. It's a visual deficiency that in books and stuff, the orbit of the moon is shown round. Nontheless, there you go. sanctus 1 Quote Link to comment Share on other sites More sharing options...
exchemist Posted May 1, 2017 Report Share Posted May 1, 2017 (edited) Hi, Let me clear some things out. ☆ Super PolymathYou're very wrong. There's nothing known as empty space since the happening of the Big Bang. However, let's imagine there is a space craft going through infinite empty space. In that case the space craft will have its own gravitational field. Thus, still giving off gravity. You've been proven wrong.Also, the escape velocity is entirely different from gravitation change. Gravity never ****ing changes. It's a proven fact (unless mass changes). When you reach an escape velocity, you're slowly drifting out of the specific range of the planet's gravitational pull. As the range decreases, the actual pull of gravity decreases, however, gravity doesn't. Please correct your statements. ☆ BillvonHmm.. let me see. Do you have any books suggested to support yourself? ☆ JMJonesThe reason for the moon not following into earth is because it's stuck in the centrofield orbital space. Where at an extent, the gravity of the Sun cancels the gravity of earth. To be more precise, the gravity of earth candles the gravity of the Sun to an extent as at that range, the gravity of sun is larger pulling the moon outwards. As it rotates around, it reaches a point where the sun's gravity is in the same direction as the earth's. This is called the oval-orbital space. It's a visual deficiency that in books and stuff, the orbit of the moon is shown round. Nontheless, there you go.Agree with your 1st para. Newton's model would say that, as gravity falls off with inverse square of distance, there is finite amount of kinetic energy needed to escape to infinity. Therefore if a body has a velocity that provides more than this energy, the object will escape completely. That does not mean that gravity ceases to act on it, just that the force of gravity will never be enough to decelerate the object to zero and force it to return. Your 3rd para seems very garbled, though and appears wrong. Any object in orbit is in free fall. There is no cancellation of gravity involved. Motion in a circle or an ellipse involves a centripetal acceleration. This is standard dynamics. Gravity provides the force required to give this acceleration and thus motion in a circle or an ellipse results. If this acceleration were not there, the object would travel in a straight line, not in a closed orbital path. Edited May 4, 2017 by exchemist JMJones0424 1 Quote Link to comment Share on other sites More sharing options...
Darky Posted May 1, 2017 Author Report Share Posted May 1, 2017 Your 3rd para seems very garbled, though and appears wrong. Any object in orbit is in free fall. There is no cancellation of gravity involved. Motion in a circle or an ellipse involves a centripetal acceleration. This is standard dynamics. Gravity provides the force required to give this acceleration and thus motion in a circle or an ellipse results. If this acceleration were not there, the object would travel in a straight line, not in a closed orbital path.Hi, I'm sorry for not providing proof; nontheless, your concepts are correct - however, your evaluation of them is wrong. Please check the model of the moon's rotation in space or check a live capture from the ISS. P.S: Make sure the moon's rotation model is from Nasa. That should be proven. Quote Link to comment Share on other sites More sharing options...
billvon Posted May 1, 2017 Report Share Posted May 1, 2017 Not even that. Space by itself has no mass, and thus, no gravity. Right. But nearby objects do, even if freefall makes it almost imperceptible. Hence "microgravity." Quote Link to comment Share on other sites More sharing options...
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