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Law of universal gravitation

  1. Aug 11, 2006 #1
    hello all

    i was wondering if we could use newton's F=GMn/r^2 to apply it to non-point masses, and specifically, if they two masses come completely within each other, would it still hold?
     
  2. jcsd
  3. Aug 11, 2006 #2

    jtbell

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    Not directly. In general, you have to use integral calculus. Conceptually, you divide both M and m into lots of tiny pieces dM and dm, all at different locations, of course; then find the force of attraction between each combination of dM and dm (each combination has a different r); then add all those forces as vectors.

    If each object's mass is distributed in a spherically-symmetric way, and they don't overlap, then you can pretend that each object is a point object located at the object's center. I've read that Newton invented integral calculus in order to prove this.

    But of the objects overlap, you can't do this.
     
  4. Aug 12, 2006 #3

    quasar987

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    Absolutely![itex]^1[/itex] It is the Glory of what one might call the center of mass theorem. Given a body of density [itex]\rho[/itex] (or more generally even, given any system of particles whatsoever), you can calculate the "center of mass" of this body, defined by

    [tex]\vec{R} = \frac{\int_{\mbox{body}}\vec{r}\rho dV}{\int_{\mbox{body}}\rho dV} = \frac{\int_{\mbox{body}}\vec{r}\rho dV}{M}[/tex]

    and the equation of motion for this point is

    [tex]\vec{F} = M \frac{d^2}{dt^2}\vec{R}[/tex]

    where [itex]\vec{F} [/itex] is the sum of the forces acting on the body.


    Qualitatively, what this is saying is that for any body, its center of mass moves according to Newton's second law of motion as if it were a point-particle of mass M!


    [itex]^1[/itex] but not directly, as jtbell said. It can be applied, but we must pass by the concept of the center of mass.
     
  5. Aug 12, 2006 #4
    i see. so, if i drop something through earth, assuming that there is a hole through the other side, the gravity is the strongest at the center of earth? since r^2 approaches 0.
     
  6. Aug 12, 2006 #5

    HallsofIvy

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    jtbelland quasar987 answered you correctly and you misinterpreted their answers completely! If a small mass is inside the earth then you have to use [itex]\frac{GmM}{r^2}[/itex] at each point within the earth, integrating over the volume of the earth. One thing that shows is that the mass of the earth outside the radius of the object from the center of the earth cancels out. Only the mass inside that radius is important and you can use [itex]\frac{GmM}{r^2}[/itex] for that. But the mass falls off as r3 so in fact the total force on an object within the earth is proportional to r. The net force on an object at the center of the earth is 0. That should be obvious: all the forces are outwardly directed and are symmetric.
     
  7. Aug 12, 2006 #6

    Doc Al

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    Center of mass

    Careful! This does not mean that you can directly apply F=GMm/r^2 to calculate the gravitational attraction between two extended objects by treating them as point masses located at their center of mass. This only works for special geometries, such as spherically symmetric mass distributions (see Newton's sphere theorem). In general you must use integral calculus as jtbell explained.

    It is certainly true that the net force on an extended object can be used to calculate the acceleration of the object's center of mass according to F=ma, but that is a different problem. (That is what I would call the "center of mass theorem".)
     
  8. Aug 12, 2006 #7
    oh. i get it. thanks a million.

    but may i ask, does those formulae apply to the quantum levels as well?
     
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