Two particles of masses `m_(1)` and `m_(2)` initially at rest a infinite distance from each other, move under the action of mutual gravitational pull. Show that at any instant therir relative velocity of approach is `sqrt(2G(m_(1)+m_(2))//R)` where `R` is their separation at that instant.
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The gravitatioinal force of attraction on `m_(1)` due to `m_(2)` at a separation `r` is `F_(1)=(Gm_(1)m_(2))/(r^(2))` Therefore, the acceleration of `m_(1)` is `a_(1)=(F_(1))/(m_(1))=(Gm_(2))/(r^(2))` Similarly the acceleration of `m_(2)` due to `m_(1)` is `a_(2)=(Gm_(1))/(r^(2))` The negative sign is put as `a_(2)` is directed opposite to `a_(1)`. The relative acceleration of approach is `a=a_(1)-a_(2)=(G(m_(1)+m_(2)))/(r^(2))`.........i If `v` is the relative velocity, then `a=(dv)/(dt)=(dv)/(dr) (dr)/(dt)` But `-dr//dt=v` (negative sign shows that `r` decreases with increasing `t`). `:. a=-(dv)/(dr)v`...........i From eqn i and ii we have `v dv=-(G(m_(1)+m_(2)))/(r^(2)) dr` Integrating we get `(v^(2))/w=(G(m_(1)+m_(2)))/r` At `r=oo, v=0` (given) and So `C=0`. therfore `v^(2)=(2G(m_(1)+m_(2)))/r` Let `v=v_(R)` then `r=R.` Then `v_(R)=sqrt((2(Gm_(1)+m_(2)))/R)`
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