isaac Posted April 12, 2020 Report Share Posted April 12, 2020 "Escape velocity" was invented by "physics professors" to explain to themselves why when they "calculate" the equation for the velocity of motion of a body in the Universe via energy balance, they get one velocity (escape velocity), and when they do the same via the force balance then they get another velocity (orbital velocity). Now in "physics" for the same phenomenon there are two definitions (equations) !!! Even in his "Principia", Newton "drew" a cannon to our professors and explained to them that "escape velocity" and "orbital velocity" are one and the same! The same thing, to our physics professors, was "drawn and calculated" by Kepler, even before Newton!! But our "physics professors" are "smarter" than both Kepler and Newton!!! That's why they don't understand neither Kepler nor Newton, but perform some "their physics". In this "physics" Newton's gravitational constant is derived from the "orbital velocity" derived by the professors, but it can also be derived from the "escape velocity" as defined by Newton and Kepler. By definition of our professors: G=rv^{2}/M, while it is by definition of Newton and Kepler: G=rv^{2}/(2M). So which one is the right one? Well Kepler's and Newton's of course !!! Due to the complete misunderstanding of Newton's Law of Force, as well as the three principles of operation of that Force, our "physics professors" have "wrongly" determined Newton's gravitational constant G, and thus the masses of all bodies in the Universe have been wrongly determined! To hide this shame, "physics professors" had to invent "dark mass and energy," to explain why the measured velocities of stars are 2^{0.5} higher than those "theoretically derived" by our "physics professors". They, now, that shame of their own, like a cuckoo's egg, are laying it to Newton! Like, there is no Newton "clue" what force is, what gravity is, let alone what is dark mass and energy! (The whole "physics" after Newton is wrong!) The orbit is a place in space where the potential and kinetic energies of the body are in equilibrium (E_{p}=E_{k}). Therefore, there is no "Escape Velocity" but there is only "Orbital Velocity" and it is the same as "Escape Velocity"! Proof: If a body of unit mass has a potential energy GM/r in the gravitational field, how much more energy do we need to take that body to infinity? So once again so much (GM/r)! With 2GM/r the body will end up at the end of the Universe! If we now convert this energy into kinetic energy, it will be 2E_{p}=2E_{k}, that is, E_{p}=E_{k} , so again this "final" velocity will be equal to the orbital velocity! In fact, no body can ever escape either its or anyone's gravity - the Law! The gravitational field of each mass covers the entire Universe, and their interaction is called "action at a distance" and is the cause of the Law of Conservation of Energy! Long live Newton's Law of Force! May the Force be with you! Quote Link to comment Share on other sites More sharing options...

exchemist Posted April 12, 2020 Report Share Posted April 12, 2020 (edited) Just for any non-crank readers there may be (?) , the forgoing post is total ballocks of course. If one calculates the gravitational potential energy gained by a body being separated progressively from another one, e.g. a rocket leaving the Earth, it turns out that, even at infinite separation, the gain in GPE has a finite value. So, if the body possesses a velocity such that its kinetic energy is greater than this GPE value, the body will follow a path that is no longer a closed orbit. It has thus "escaped", even though it continues to feel a force at any finite separation. It turns out, when one does the maths, that this velocity is independent of the mass of the body in question and thus has the same value for a large object or a small one. This velocity value is what is meant by the term "escape velocity". Edited April 12, 2020 by exchemist Quote Link to comment Share on other sites More sharing options...

OceanBreeze Posted April 14, 2020 Report Share Posted April 14, 2020 The orbit is a place in space where the potential and kinetic energies of the body are in equilibrium (E_{p}=E_{k}). Wrong. When you start out wrong, you end up even more wrong.Orbit is a place where the kinetic energy of a satellite is equal to half its gravitational potential energy. Therefore, there is no "Escape Velocity" but there is only "Orbital Velocity" and it is the same as "Escape Velocity"! Proof: If a body of unit mass has a potential energy GM/r in the gravitational field, how much more energy do we need to take that body to infinity? So once again so much (GM/r)! With 2GM/r the body will end up at the end of the Universe! If we now convert this energy into kinetic energy, it will be 2Ep=2Ek, that is, Ep=Ek , so again this "final" velocity will be equal to the orbital velocity! You’re reasoning is so wrong it would give aspirin a headache. It would be a waste of time to correct you. Quote Link to comment Share on other sites More sharing options...

isaac Posted April 16, 2020 Author Report Share Posted April 16, 2020 Gentlemen, So what kind of nonsense is it that my post, from pure theoretical physics, is placed under the Silli Clames? Well I didn't bring up any Silli Clames, I just quoted Newton and Kepler! So is it my fault that my "interlocutors" do not even have an elementary knowledge of mathematics and physics, and of decency not to speak ?! But if there is such a level of knowledge, in this forum, that no one understood what I wrote and what I asked, I would have to explain it again to you as in elementary school. So, what is Kepler's law then? What is Kepler's law talking about? What is the amount of Newton's gravitational constant G when we derive it from Kepler's law? Kepler's law reads: T^{2}/r^{3} = 4(pi)^{2}/(GM), and it speaks of the "orbital velocity" of orbiting the planet around the Sun! So how much is "orbital velocity" when we derive it from Kepler's Law? If we introduce velocity v over the circular frequency T = 2r(pi)/v into Kepler's law, we obtain: GM = rv^{2}/2 from where we derive the "orbital velocity" v^{2} = 2GM/r. Our "physics professors" claim it to be - "escape velocity"! So whose Silli Clame is that? So is my question What is "escape velocity" not justified? If, now, from Kepler's law GM = rv^{2}/2 we derive Newton's gravitational constant G = rv^{2}/(2M) we see that it is different from Kevendish's that reads G = rv^{2}/M, by which he "determined" G! So whose Silli Clame is that? Well, isn't my question "What is actually the gravitational constant G" justified? If we write Kepler's law as: GM/r = v^{2}/2 we can see that Kepler claims that the orbit is a place where Potential energy (GM/r) equals Kinetic energy (v^{2}/2) per unit mass! So is Kepler's law a Silli Claim? So is it my fault that those who spit on my post actually spit on Newton and Kepler? Shame! So I ask you to get my post back to where I published it, in theoretical physics. Thank you! Quote Link to comment Share on other sites More sharing options...

OceanBreeze Posted April 17, 2020 Report Share Posted April 17, 2020 Gentlemen,So what kind of nonsense is it that my post, from pure theoretical physics, is placed under the Silli Clames? Well I didn't bring up any Silli Clames, I just quoted Newton and Kepler! So is it my fault that my "interlocutors" do not even have an elementary knowledge of mathematics and physics, and of decency not to speak ?! But if there is such a level of knowledge, in this forum, that no one understood what I wrote and what I asked, I would have to explain it again to you as in elementary school. So, what is Kepler's law then? What is Kepler's law talking about? What is the amount of Newton's gravitational constant G when we derive it from Kepler's law? Kepler's law reads: T^{2}/r^{3} = 4(pi)^{2}/(GM), and it speaks of the "orbital velocity" of orbiting the planet around the Sun! So how much is "orbital velocity" when we derive it from Kepler's Law? If we introduce velocity v over the circular frequency T = 2r(pi)/v into Kepler's law, we obtain: GM = rv^{2}/2 from where we derive the "orbital velocity" v^{2} = 2GM/r. We obtain no such thing! Substituting [math] T=\frac { 2\pi r }{ v } [/math] into [math] \frac { { T }^{ 2 } }{ { r }^{ 3 } } [/math] Yields [math] \frac { 4{ \pi }^{ 2 }{ r }^{ 2 } }{ { v }^{ 2 }{ r }^{ 3 } } =\frac { 4{ \pi }^{ 2 } }{ GM } [/math] Which reduces to [math] { v }^{ 2 }=\frac { GM }{ r } [/math] So, your claim is shown to be silly and the rest of your post is just more of the same nonsense Our "physics professors" claim it to be - "escape velocity"! So whose Silli Clame is that? So is my question What is "escape velocity" not justified? If, now, from Kepler's law GM = rv2/2 we derive Newton's gravitational constant G = rv2/(2M) we see that it is different from Kevendish's that reads G = rv2/M, by which he "determined" G! So whose Silli Clame is that? Well, isn't my question "What is actually the gravitational constant G" justified? If we write Kepler's law as: GM/r = v2/2 we can see that Kepler claims that the orbit is a place where Potential energy (GM/r) equals Kinetic energy (v2/2) per unit mass! So is Kepler's law a Silli Claim? So is it my fault that those who spit on my post actually spit on Newton and Kepler? Shame! So I ask you to get my post back to where I published it, in theoretical physics. Thank you! It is becoming increasing clear from reading your posts that you are only here to disrupt this forum with nonsensical “math” and “physics”. I will continue to move your nonsense to silly claims or maybe just delete it but I won't be bothering to correct it. Quote Link to comment Share on other sites More sharing options...

isaac Posted April 22, 2020 Author Report Share Posted April 22, 2020 Sorry, OceanBreeze, I was wrong, I admit, but not because my claims were wrong, but because I based my statement on the wrong version of Kepler's law! Instead of using "Kepler's law of orbit", I "mistakenly" cited "Kepler's law" about the motion of the pendulum: T = 2(pi)(r/g)^{0.5}, respectively, T^{2} = 4(pi)^{2}r/g, with g = GM/r^{2}, which gives T^{2}/r^{3} = 4(pi)^{2}/(GM), but that is why (in fact, by my mistake) you have presented your proof, for your claim, that E_{k} = E_{p}/2 is in orbit, namely: GM/r = v^{2}, which is the same as GM/2r = v^{2}/2, that is, E_{p}/2 = E_{k}, which I think is wrong, because, as I stated above, it is derived from the Law of motion of pendulum! When the mass moves in orbit then it is constantly in equilibrium, and when the mass is moving like a pendulum it is constantly out of balance! The average velocity of oscillation is "twice as low" as the same mass that would orbit! That is why it is not the time, nor the place, for you to exult and just spit on "my views". It would be better if you could help me answer "my questions" together: What is escape velocity? Why in today's "physics" are there "escape velocity" and "orbital velocity"? So how can the poor mass "decide" at what velocity to move? So how does a poor physicist "decide" from which equation to determine "how much is actually" Newton's gravitational constant G? Obviously, we're talking about a fundamental area of physics, where it can be very easy to make mistakes! Therefore, it is not the point of my post that you correct me or that I correct you, but rather to correct "physics" together. I will prove what I claim, and you prove what you claim! Fair enough!? Quote Link to comment Share on other sites More sharing options...

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.