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Newton's law of gravity

  1. Oct 12, 2004 #1
    Two particles having masses of m and M respectively, attract each other according to Newton's law of gravity. Initially they are at rest at an infinite distance apart. Find their relative acceleration and show that their relative velocity of approach is

    v(rel) = square root(2G(M+m)/a)

    I do not even know where to start on this problem, if any could give me help that would be great.

    Thank you
  2. jcsd
  3. Oct 12, 2004 #2


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    Momentum and energy are conserved!
  4. Oct 13, 2004 #3


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    The particles are initially at rest so their angular momentum is zero, they will move towards each other along the straight line connecting them.
    As there are no external forces acting on this system of particles, their centre of mass stays steady during the motion. You can place the origin of the frame of reference there. That is

    [tex]mv+MV=0\mbox{ } \rightarrow \mbox{ }V=-\frac{M}{m}v[/tex]

    The energy of the system is conserved, and it is zero, as the particles started from rest at infinite distance apart.
    At distance 'a' the mutual potential energy is

    [tex]U_{pot}= - G\frac{mM}{a^2}[/tex].

    [tex] \frac{1}{2}mv^2+\frac{1}{2}MV^2-G\frac{mM}{a^2}=0[/tex]

    You have two equations for v and V. Solve, and determine their relative speed [tex]v_{rel}=|v-V|[/tex].

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