XPost: sci.math   
   From: PointedEars@web.de   
      
   Ross Finlayson wrote:   
   > When you think about Newton's laws, or when I do,   
   > or when one does, or as one may, think about Newton's   
   > laws, as we think about Newton's laws, one might make   
   > for a "deconstructive account" of them.   
   >   
   > rest/rest motion/motion equal/opposite   
   > <- isn't that three laws?   
      
   Once again your writing is so confused that one cannot discern what you mean.   
      
   > F = ma   
      
   This is a formulation of Newton's Second Law of Motion.   
      
   Originally (in 1686/1687) Newton wrote this law in Latin words that as an   
   equation he would have later written (or has written?)   
      
     F = ṗ,   
      
   (F = dot p, where by a dot above a symbol he defined the derivative of the   
   corresponding quantity with respect to time, a notation that we are still   
   using today), where he had previously defined the equivalent of   
      
     p = m v.   
      
   He called this, what we call "momentum" today, "the quantity of motion", and   
   defined that it "arises from Velocity [v] and the quantity of Matter [mass   
   m] together" ("Philosophiæ Naturalis Principia Mathematica": "Definitiones").   
      
       [The notion of vectors did not exist yet -- they were introduced only in   
        the 19th century by Hamilton --, so Newton would not have written this   
        as a vector equation.]   
      
      
      
   (Cambridge University, Cambridge Digital Library. High resolution digitised   
   version of Newton's own copy of the first edition, interleaved with blank   
   pages for his annotations and corrections.  Cited in:   
   )   
      
   One can see that this is equivalent to   
      
     F = m a,   
      
   considering that the mass is assumed to be constant¹, and that acceleration   
   is the derivative of velocity with respect to time:   
      
     F = ṗ = m v̇ = m a.   
      
   I have heard that the law was first formulated as "F = m a" by Leonhard   
   Euler ca. 100 years later, but so far I have not found corroborating evidence.   
      
   ___   
   ¹ This assumption must be challenged in special cases.  For example, the   
     mass of a rocket decreases due to pushing out fuel for propulsion.  As   
     a result, the *rocket equation* features a time derivative of mass that   
     arises from the product rule of derivation: F = ṗ = m v̇ + ṁ v = m a +   
   ṁ v.   
      
      
      
   > <- that's also F(t) = m d^2 x /dt^2, differential d   
      
   Correct.   
      
   > F = g Mm / d^2   
      
   That is a *wrong* notation of Newton's Law of Universal Gravitation.   
      
   AISB, one must use "G" for the gravitational *constant* here; "g" is the   
   symbol for the gravitational *acceleration* instead, whose magnitude is given   
   by   
      
     g(d) = G M/d^2,   
      
   which is therefore NOT constant in general.   
      
   However, this form of the magnitude of the gravitational acceleration only   
   holds for spherically-symmetric objects with mass M when d means the   
   distance from the center of mass, i. e. the _radial_ distance.   
      
   In general, the *field of* gravitational acceleration g⃗ (the gravitational   
   field) is given by Gauss' Law for gravitation:   
      
     ∇ ⋅ g⃗ = -4π G ρ,   
      
   where ρ is the mass density field.   
      
   AISB (multiple times) for a *uniform* *spherically-symmetric* mass density,   
   using Gauss' Theorem, one finds Newton's Law of Universal Gravitation:   
      
     ∰_V dV (∇ ⋅ g⃗ ) = ∯_A dA⃗ ⋅ g⃗   
                      = -4π r² g(r) = -4π G ∰_V dV ρ = -4π G M   
     <==>                      g(r) =                       G M/r²   
     ===>                      g⃗(r⃗) = -(G M/r²)   r⃗/r   
     <==>              F⃗_g = m g⃗(r⃗) = -(G M m/r²) r⃗/r,   
      
   where V is a spherical volume that fully contains the region where the mass   
   density is non-zero (i.e. the entire, assumed spherically-symmetric,   
   celestial object where we would like to determine the gravitational field).   
      
   > <- F(t) = g Mm / d(t)^2 proportional inverse-square distance d   
   >   
   > Force is a function of time.   
      
   In general, yes; but not necessarily.   
      
   Because the law in this form only applies to spherically-symmetric mass   
   distributions, one usually writes "r" (for "radius") instead of "d".   
   [Another reason is to avoid confusion with the operator for the total   
   derivative which also is "d".]   
      
   Newton's Law of Gravitation gives rise to the differential equation   
      
     r̈ ≔ (d²/dt²) r = G M/r²   
      
   from   
      
     F = m a = m (d²/dt²) r = G M m/r².   
      
   Solving this differential equation, one can calculate, for example, the time   
   that it takes for an object to fall in vacuum when the fact that the   
   gravitational acceleration varies with altitude (to begin with) cannot be   
   neglected.  This is important, for example, to land space probes safely on   
   comets and asteroids (which has been done).   
      
   There is at best a tenuous relation of this to special and general   
   relativity, so F'up2 sci.physics instead.   
      
   --   
   PointedEars   
      
   Twitter: @PointedEars2   
   Please do not cc me. / Bitte keine Kopien per E-Mail.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)