A) \[{{\left[ 2G\frac{({{m}_{1}}-{{m}_{2}})}{r} \right]}^{1/2}}\]
B) \[{{\left[ \frac{2G}{r}({{m}_{1}}+{{m}_{2}}) \right]}^{1/2}}\]
C) \[{{\left[ \frac{r}{2G({{m}_{1}}{{m}_{2}})} \right]}^{1/2}}\]
D) \[{{\left[ \frac{2G}{r}{{m}_{1}}{{m}_{2}} \right]}^{1/2}}\]
Correct Answer: B
Solution :
[b] Let Velocities of these masses at r distance from each other be \[{{v}_{1}}\] and \[{{v}_{2}}\] respectively. By conservation of momentum \[{{m}_{1}}{{v}_{1}}-{{m}_{2}}{{v}_{2}}=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}}\] ...(i) By conservation of energy change in P.E.= change in K.E. \[\frac{G{{m}_{1}}{{m}_{2}}}{r}=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\frac{m_{1}^{2}v_{1}^{2}}{{{m}_{1}}}+\frac{m_{2}^{2}v_{2}^{2}}{{{m}_{2}}}=\frac{2G{{m}_{1}}{{m}_{2}}}{r}\] ...(ii) On solving equation (i) and (ii) \[{{v}_{1}}=\sqrt{\frac{2Gm_{2}^{2}}{r({{m}_{1}}+{{m}_{2}})}}\] and \[{{v}_{2}}=\sqrt{\frac{2Gm_{1}^{2}}{r({{m}_{1}}+{{m}_{2}})}}\] \[\therefore \,\,\,\,\,\,\,\,\,{{v}_{app}}=|{{v}_{1}}|+|{{v}_{2}}|=\sqrt{\frac{2G}{r}({{m}_{1}}+{{m}_{2}})}\]
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