Two particles of masses `m_(1)` & `m_(2)` separated by a distance d are at rest initially. If they move towards each other under mutual gravitational interaction, where will they meet.

Explanation

Initially particles are at rest, so `V_(cm)=0`.
`V_(cm)=(m_(1) times 0+m_(2) times 0)/(m_(1)+m_(2))=0`
`F_(ext)=0`, so `V_(cm)` remain unchanged
`(m_(1)+v_(1)+m_(2)v_(2))/(m_(1)+m_(2))=0," ie, "m_(1)v_(1)+m_(2)v_(2)=0`
But `bar(v)=(Deltavec(r))/(Deltat)`
`therefore m_(1)(Deltavec(r)_(1))/(Deltat)+m_(2)(Deltavec(r)_(2))/(Deltat)=0`
`m_(1)Deltar_(1)+m_(2)Deltar_(2)=0`
`m_(1)d_(1)+m_(2)d_(2)=0" (since" Deltavec(r)=vec(d)`)
`d=d_(1)+d_(2)`,
`" "m_(1)d_(1)-m_(2)d_(2)=0" "` Since direction of `d_(2)` is opposite to that of `d_(1)`.
`m_(1)d_(1)=m_(2)d_(2)`
`therefore d_(1)=(m_(2)d)/(m_(1)+m_(2))" "d_(2)=(m_(1)d)/(m_(1)+m_(2))`
`d_(1)" & "d_(2)` represent the position of centre of mass from masses `m_(1) and m_(2)` respectively, so the two bodies centre of mass of the system.