Calculating Relative Acceleration and Velocity with Newton's Law of Gravity

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Two particles with masses m and M attract each other according to Newton's law of gravity, starting from rest at an infinite distance. The conservation of momentum and energy is crucial, as the center of mass remains stationary during their approach. The potential energy at distance 'a' is given by U_pot = -G(mM/a^2), and the total energy is zero. By applying the conservation equations, the relative velocity of approach is derived as v(rel) = √(2G(M+m)/a). This analysis effectively demonstrates the relationship between gravitational attraction and relative motion.
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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
 
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Momentum and energy are conserved!
 
starbaj12 said:
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

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

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

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

U_{pot}= - G\frac{mM}{a^2}.

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

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

ehild
 
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