jajrussel Posted February 14 Share Posted February 14 I merged two formulas for Force I saw in a video. First - Did I do it right? F = ma= (G*M_{1}*M_{2})/R^{2} 1. F = ma 2. F = G M_{1}M_{2}/ R^{2} Second- If it is right does G/R^{2} represent the acceleration part of F= ma ? Link to comment Share on other sites More sharing options...

Genady Posted February 14 Share Posted February 14 1 hour ago, jajrussel said: merged two formulas for Force I saw in a video. First - Did I do it right? F = ma= (G*M_{1}*M_{2})/R^{2} 1. F = ma 2. F = G M_{1}M_{2}/ R^{2} No. 1 hour ago, jajrussel said: does G/R^{2} represent the acceleration part of F= ma ? No. Link to comment Share on other sites More sharing options...

jajrussel Posted February 14 Author Share Posted February 14 18 minutes ago, Genady said: No. No. Okay… I’m nearly 70 years old, and not in school. I’m pretty sure the rules of the forum will allow you to elaborate. Link to comment Share on other sites More sharing options...

Genady Posted February 14 Share Posted February 14 (edited) 6 minutes ago, jajrussel said: Okay… I’m nearly 70 years old, and not in school. I’m pretty sure the rules of the forum will allow you to elaborate. Okay, let's look at the first "equation", F = ma= (G*M_{1}*M_{2})/R^{2} It is not an equation, but rather a shortcut of two equations: F=ma ma=\(\frac {GM_1M_2}{R^2}\) The first equation is the Newton second law. It is fine. However, the second equation is meaningless, unless you have a reasonable interpretation for it. PS. You are only a bit older than me. Edited February 14 by Genady 1 Link to comment Share on other sites More sharing options...

swansont Posted February 14 Share Posted February 14 1 hour ago, jajrussel said: Second- If it is right does G/R^{2} represent the acceleration part of F= ma ? No ma = GMm/r^2 the mass m cancels, as it’s on both sides, but M remains, so a = GM/r^2 Gravitational acceleration doesn’t depend on the mass of the object, but does depend on the mass of the (usually celestial) object exerting the gravity 1 Link to comment Share on other sites More sharing options...

jajrussel Posted February 14 Author Share Posted February 14 (edited) 54 minutes ago, Genady said: Okay, let's look at the first "equation", F = ma= (G*M_{1}*M_{2})/R^{2} It is not an equation, but rather a shortcut of two equations: F=ma ma=GM1M2R2 The first equation is the Newton second law. It is fine. However, the second equation is meaningless, unless you have a reasonable interpretation for it. PS. You are only a bit older than me. Okay, I was thinking for the first question that if both ma and GMm/R^{2 }equaled force I could write it F=ma=GMm/R^{2 }. Which is not exactly how I wrote it the first time but I borrowed the shorthand from swansont for the latter portion. What I thought I was writing is force equals mass time acceleration ,and force equals G times mass one times mass two divided by the radius squared. Since force is described as equal to both expressions. I assumed it would be okay to write F=ma=Gmm/R^{2 } since the expression on each side of the equal signs I presumed to be equal. As for the second question. 39 minutes ago, swansont said: No ma = GMm/r^2 the mass m cancels, as it’s on both sides, but M remains, so a = GM/r^2 Gravitational acceleration doesn’t depend on the mass of the object, but does depend on the mass of the (usually celestial) object exerting the gravity Are you saying that by canceling mass out, force and acceleration are shown to be the same? Edited February 14 by jajrussel Link to comment Share on other sites More sharing options...

Genady Posted February 14 Share Posted February 14 6 minutes ago, jajrussel said: Okay, I was thinking for the first question that if both ma and GMm/R^{2 }equaled force I could write it F=ma=GMm/R^{2 }. Which is not exactly how I wrote it the first time but I borrowed the shorthand from swansont for the latter portion. What I thought I was writing is force equals mass time acceleration ,and force equals G times mass one times mass two divided by the radius squared. Since force is described as equal to both expressions. I assumed it would be okay to write F=ma=Gmm/R^{2 } since the expression on each side of the equal signs I presumed to be equal. Yes, it is good now, when you write ma=GMm/R^{2} But be careful: the second time you've written ma=Gmm/R^{2} instead of GMm/R^{2} Details matter. Link to comment Share on other sites More sharing options...

Genady Posted February 14 Share Posted February 14 1 hour ago, jajrussel said: force and acceleration are shown to be the same? Force and acceleration are never the same. They are measured in different units. Link to comment Share on other sites More sharing options...

Janus Posted February 14 Share Posted February 14 (edited) 1 hour ago, jajrussel said: Okay, I was thinking for the first question that if both ma and GMm/R^{2 }equaled force I could write it F=ma=GMm/R^{2 }. Which is not exactly how I wrote it the first time but I borrowed the shorthand from swansont for the latter portion. What I thought I was writing is force equals mass time acceleration ,and force equals G times mass one times mass two divided by the radius squared. Since force is described as equal to both expressions. I assumed it would be okay to write F=ma=Gmm/R^{2 } since the expression on each side of the equal signs I presumed to be equal. As for the second question. Are you saying that by canceling mass out, force and acceleration are shown to be the same? It is really important to grasp what the variables mean in each equation. In F=ma, we are talking about the amount of force needed to give a mass of m an acceleration of a With F = GMm/d^2 we are talking about the gravitational force acting between masses M and m at a center to center distance of d. To make this clearer, F is often written as F_{g} Now if we were considering how much acceleration mass m would undergo as the result of gravitational attraction between m and M_{,} Then we we are saying that F_{g }is assuming the role of F in F=ma or that F=F_{g} thus we can substitute ma for F_{g }to get ma = GMm/d^2 cancel m on both sides of the equation and get a = GM/d^2, which tells us that the acceleration of m due to the gravitational attraction is independent of the magnitude of m's mass. Edited February 14 by Janus Link to comment Share on other sites More sharing options...

swansont Posted February 15 Share Posted February 15 8 hours ago, jajrussel said: Are you saying that by canceling mass out, force and acceleration are shown to be the same? The mass cancels from algebra. The forces are equal, and I was solving for the acceleration Link to comment Share on other sites More sharing options...

jre84 Posted February 15 Share Posted February 15 this is simple physics from school? are you educated in post secondary? a little background would be nice. also Good day Link to comment Share on other sites More sharing options...

exchemist Posted February 15 Share Posted February 15 16 hours ago, jajrussel said: Okay, I was thinking for the first question that if both ma and GMm/R^{2 }equaled force I could write it F=ma=GMm/R^{2 }. Which is not exactly how I wrote it the first time but I borrowed the shorthand from swansont for the latter portion. What I thought I was writing is force equals mass time acceleration ,and force equals G times mass one times mass two divided by the radius squared. Since force is described as equal to both expressions. I assumed it would be okay to write F=ma=Gmm/R^{2 } since the expression on each side of the equal signs I presumed to be equal. As for the second question. Are you saying that by canceling mass out, force and acceleration are shown to be the same? As the little "m" s appear once on each side of your equation you can cancel them, but that leaves the big M. So the "a" in F=ma should be set equal to GM/r². There has to be an M in it because the acceleration due to gravity depends on whether you are standing on, say, the Earth, or the Moon which has less mass and therefore weaker gravity. Newton's expression is completely general and can be used for any body, dialling up and down M according to the mass of the body in question. Link to comment Share on other sites More sharing options...

joigus Posted February 15 Share Posted February 15 We should all be amazed that this little m disappears from the equation. It goes very deep in the nature of gravitation. 1 Link to comment Share on other sites More sharing options...

Genady Posted February 15 Share Posted February 15 11 minutes ago, joigus said: We should all be amazed that this little m disappears from the equation. It goes very deep in the nature of gravitation. Right. It is not there to start with. 😉 Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted February 15 Share Posted February 15 It's either one of the big Ms, usually the lesser one, depending on which of them's acceleration is of interest. Link to comment Share on other sites More sharing options...

jajrussel Posted February 15 Author Share Posted February 15 17 hours ago, Janus said: It is really important to grasp what the variables mean in each equation. In F=ma, we are talking about the amount of force needed to give a mass of m an acceleration of a With F = GMm/d^2 we are talking about the gravitational force acting between masses M and m at a center to center distance of d. To make this clearer, F is often written as F_{g} Now if we were considering how much acceleration mass m would undergo as the result of gravitational attraction between m and M_{,} Then we we are saying that F_{g }is assuming the role of F in F=ma or that F=F_{g} thus we can substitute ma for F_{g }to get ma = GMm/d^2 cancel m on both sides of the equation and get a = GM/d^2, which tells us that the acceleration of m due to the gravitational attraction is independent of the magnitude of m's mass. So there is a distinction there is F=ma push then there is F=GMm/R^{2} pull F =ma seems incomplete as a formula because it only accounts for how m is affected by acceleration, but what it seems to me doesn’t matter, because then, who? I think it is credited to Einstein, says acceleration is the same as gravity. So if F= ma then gravity can not be a force because you have to multiply acceleration which is the same as gravity times mass to get what is called force, so gravity and force can not be the same thing. Is this why it is said that gravity is not a force? Link to comment Share on other sites More sharing options...

joigus Posted February 15 Share Posted February 15 2 hours ago, Genady said: Right. It is not there to start with. 😉 Ubi materia ibi geometria Einstein took it to the next level. 👍 10 minutes ago, jajrussel said: So there is a distinction there is F=ma push then there is F=GMm/R^{2} pull push/pull is not a physical distinction. It's anthropomorphic. 'Work' is another anthropomorfic term, although extremely useful. But push vs pull is not useful at all in terms of physics. Link to comment Share on other sites More sharing options...

jajrussel Posted February 15 Author Share Posted February 15 10 hours ago, jre84 said: this is simple physics from school? are you educated in post secondary? a little background would be nice. also Good day No. I don’t think it would have made a difference then, but I don’t know. There was a tumor I was living with unknowingly. A few years ago they took it out. Background wise unless you consider YouTube and or SFN post secondary education the answer is no Link to comment Share on other sites More sharing options...

Genady Posted February 15 Share Posted February 15 25 minutes ago, jajrussel said: So there is a distinction there is F=ma push then there is F=GMm/R^{2} pull F =ma seems incomplete as a formula because it only accounts for how m is affected by acceleration, but what it seems to me doesn’t matter, because then, who? I think it is credited to Einstein, says acceleration is the same as gravity. So if F= ma then gravity can not be a force because you have to multiply acceleration which is the same as gravity times mass to get what is called force, so gravity and force can not be the same thing. Is this why it is said that gravity is not a force? Gravity is not a force nor acceleration. Gravity, or gravitation, is a set of phenomena when bodies affect motion of other bodies without touching or having electromagnetic or other specific interactions. One can talk about 'force of gravity', 'gravitational acceleration' and other characteristics of these phenomena. Link to comment Share on other sites More sharing options...

jajrussel Posted February 15 Author Share Posted February 15 (edited) 45 minutes ago, joigus said: Ubi materia ibi geometria Einstein took it to the next level. 👍 push/pull is not a physical distinction. It's anthropomorphic. 'Work' is another anthropomorfic term, although extremely useful. But push vs pull is not useful at all in terms of physics. You forced me to Google, thank you . I will disagree until I find a reason to ag🤣ree . I looked up Ubi materia ibi geometria also , in the description the statement is made that nature and mathematics are intimately connected. There is a PDF I intend to read to see if it allows me to continue to disagree. 👍 Edited February 15 by jajrussel New thought Link to comment Share on other sites More sharing options...

joigus Posted February 15 Share Posted February 15 35 minutes ago, jajrussel said: I looked up Ubi materia ibi geometria also Those were words by Kepler. It's Latin for 'wherever there is matter, there is geometry'. It goes to prove that the idea that geometry held the key to understanding physics has been around for a long time. The factoring out of one of the masses from the equation of motion (or the fact that you could talk about gravitation without without one of the masses not really being there, and the other being replaced by energy, as Genady suggested) is a subtle clue that geometry is at the core of gravity. As to push or pull, I think you mean something about attractive vs repulsive forces perhaps? But then it's not about F=ma vs F=GmM/r². F=ma is the definition of force. It's a definition, rather than an equation really. F=GmM/r² is a law of force, and it has a very different content. It's when we equate both, as @Janus illustrated, ma=GmM/r² that we do have an equation, ie, an equality to be solved. The mere F=ma cannot be solved. Definitions cannot be 'solved'. Not all equalities are equations. This is a common misconception. There are definitions, identities, formulas and equations. Definition: velocity=space/time Identity: x²-y²=(x+y)(x-y) Equation: x²-2x+1=0 Formula: (c_{1})²+(c_{2})²=h², where c_{1} and c_{2} are the catheti of a right rectangle and h is the hypotenuse of the same triangle A definition is just a labeling, an identity is an algebraic equivalence that's always true, an equation is the expression of a hypothesis to be solved from its statement in the algebraic language, and a formula is an algebraic statement involving ideas that can be abstract, geometrical, etc. There is a long tradition of calling physical laws, definitions (and perhaps formulas) all 'equations', which might be at the root of your confusion. 1 Link to comment Share on other sites More sharing options...

MigL Posted February 15 Share Posted February 15 There is no distinction between push and pull; both are forces. However there are many forces, and F=ma is a generalized force equation expressing what happens to a test mass, m, when a force is applied to it. It is also known as Newton's second law. The other equation F=GMm/r^{2} is a specific kind of 'force', that expresses what happens to a test mass, m, when exposed to the gravity of another body of mass M. This is also known as Newtonian Gravity, and the model is accurate in most situations, although in some situations involving hi energy/lo separation Gravity is better described by a geometrical space-time ( which is not a force ) model, GR . The two equations can only be set to equal when dealing with the gravitational force of mass M, on test mass m. And in such a case ( as Swansont showed ) we notice that the mass m cancels from both sides. This means the acceleration induced by a body of mass M is independent of the test mass, m. IOW, all test masses, no matter what their m is, accelerate/fall at the same rate when exposed to the same Gravitational Force of object M. In the case of M being the mass of the Earth, we realize that climbing to the top of the leaning Tower of Pisa to drop a bowling ball and a golf ball, and timing their time to fall, is not necessary. They both fall at 9.8 m/s/s. ( disregarding air resistance, of course ) 1 Link to comment Share on other sites More sharing options...

jajrussel Posted February 15 Author Share Posted February 15 23 minutes ago, MigL said: There is no distinction between push and pull; both are forces. And 46 minutes ago, joigus said: But then it's not about F=ma vs F=GmM/r². Actually, I think I’m getting it, but I searched the number of times the word pull and push showed up in just one volume of a University Text and the score was 226 for pull, and 187 for push. Note, this was entirely in interest of their being anthropomorphic terms of little use in physics, yet still extremely useful unless they are used as terms of physics. Which isn’t exactly what you said 2 hours ago, joigus said: push/pull is not a physical distinction. It's anthropomorphic. 'Work' is another anthropomorfic term, although extremely useful. But push vs pull is not useful at all in terms of physics. I misunderstood the application, you wrote push (versus) pull is not useful. I could have saved myself a lot of mental exercise had I read and understood correctly the first time. Sorry, and thanks. Link to comment Share on other sites More sharing options...

Janus Posted February 15 Share Posted February 15 3 hours ago, jajrussel said: So there is a distinction there is F=ma push then there is F=GMm/R^{2} pull F =ma seems incomplete as a formula because it only accounts for how m is affected by acceleration, but what it seems to me doesn’t matter, because then, who? I think it is credited to Einstein, says acceleration is the same as gravity. So if F= ma then gravity can not be a force because you have to multiply acceleration which is the same as gravity times mass to get what is called force, so gravity and force can not be the same thing. Is this why it is said that gravity is not a force? Again, F=ma merely gives a relationship between force, mass, and acceleration. It does not have anything to do with push or pull. It gives the magnitude of the force required, and isn't concerned as to the nature of, or how this force is provided. Here is another equation : F= mv^{2}/r It tells you how much force is needed to constrain a mass to moving in a circle with a radius of r if it has a mass of m and has a speed of v. It makes no difference as to how that force is applied. It can be by a rope anchored at the center of the circle, a rocket engine mounted on the mass applying inward thrust, by the gravity of a central mass, or by the friction between a car's tires and the road. Link to comment Share on other sites More sharing options...

Genady Posted February 15 Share Posted February 15 3 hours ago, joigus said: F=ma is the definition of force. It's a definition, rather than an equation really. Are you sure? I seem to remember (I think, from Shankar's lectures, but would need to look back to make sure), that force is defined first statically, like the following. Take a spring. Some force needs to be applied to stretch it say by 10 cm. Take two identical springs like that attached parallelly. A force needed to stretch both of them together by 10 cm is defined as double the first force. Etc. After force is defined in such a way, its use in dynamics is a law / equation. Link to comment Share on other sites More sharing options...

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