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Integrating Newton's law

  1. Aug 18, 2013 #1
    Hello everyone,
    What does it mean if I integrate Newton's law of universal gravitation with respect to r.
    F= GMm/r^2 become 3GMm/r^3 . Is this the work needed to escape a gravitational pull ?

    Thank you
     
  2. jcsd
  3. Aug 18, 2013 #2
    Dear Mr-R. Welcome to Physics Forums.

    You integrated incorrectly. Please try again.
     
  4. Aug 18, 2013 #3
    This is the work-energy theorem.

    $$\int ^b _a \vec{F} \cdot d \vec{r} = \Delta KE = W $$

    If you integrate ##\frac{GMm}{r^2}## you get the potential energy.

    Try integrating it again correctly.
     
  5. Aug 18, 2013 #4
    Oh thats quiet embarrassing. It should be = GMm/r which is the work. Thank you very much.
     
  6. Aug 18, 2013 #5
    No, that's potential energy. In this case, ## W = \Delta U = \frac{GMm}{r_2}- \frac{GMm}{r_1}##
     
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