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Question

Two masses $$M_1$$ and $$M_2$$ at an infinite distance from each other and initially at rest, start interacting gravitationally. Find their velocity of approach when they are distances apart.


Solution

Since they move under mutual attraction and no external force acts on them, their momentum and energy are conserved. Therefore,
$$\therefore 0=\dfrac{1}{2}M_1v^2_1+\dfrac{1}{2}M_2v^2_2-\dfrac{GM_1M_2}{s}$$
It is zero because in the beginning, both kinetic energy and potential energy are zero.
$$0=M_1v_1+M_2v_2$$
Solving the equations,
$$v^2_1=\dfrac{2GM^2_2}{s(M_1+M_2)}$$
and $$v^2_2=\dfrac{2GM^2_1}{s(M_1+M_2)}$$
V(velocity of approach)$$=v_1-(-v_2)=v_1+v_2$$

$$=\sqrt{\dfrac{2G(M_1+M_2)}{s}}$$

Physics

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