InterviewSolution
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InterviewSolution
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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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Answer» The gravitational 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 accleration of `m_(2)` due to `m_(1)` is `a_(2) = - (Gm_(1))/(r^(2))` the negative sign being put as `a_(2)` is directed opposite to `a_(1)`. The relative acclleration of approach is `a = a_(1) - a_(2) = (G(m_(1) + m_(2)))/(r^(2))`...(1) 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`...(2) From (1) and (2) we have `vdv = - (G(m_(1) + m_(2)))/r^(2)dr` Integrating we get `(v^(2))/(2) = (G(m_(1)+m_(2)))/(r) + C` At `r = oo`, v = 0 (given) and so C = 0 `therefore v^(2) = (2G (m_(1)+m_(2)))/(r)` Let `v = v_(R)` when `r = R` Then `v_(R)=sqrt(((2G(m_(1)+m_(2)))/(R)))` . |
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