Two hypothetical planets of masses m1 and m2 are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed when their separation is ?d?? (Speed of \[{{m}_{1}}\] is \[{{v}_{1}}\] and that of \[{{m}_{2}}\] is \[{{v}_{2}}\])
[JEE Main Online Paper ( Held On 12 Apirl 2014 )
A) \[{{v}_{1}}={{v}_{2}}\]
B) \[{{v}_{1}}={{m}_{2}}\sqrt{\frac{2G}{d\left( {{m}_{1}}+{{m}_{2}} \right)}}\] \[{{v}_{2}}={{m}_{1}}\sqrt{\frac{2G}{d\left( {{m}_{1}}+{{m}_{2}} \right)}}\]
C) \[{{v}_{1}}={{m}_{1}}\sqrt{\frac{2G}{d\left( {{m}_{1}}+{{m}_{2}} \right)}}\]\[{{v}_{2}}={{m}_{2}}\sqrt{\frac{2G}{d\left( {{m}_{1}}+{{m}_{2}} \right)}}\]
D) \[{{v}_{2}}={{m}_{2}}\sqrt{\frac{2G}{{{m}_{1}}}}\] \[{{v}_{2}}={{m}_{2}}\sqrt{\frac{2G}{{{m}_{2}}}}\]
Correct Answer: B
Solution :
We choose reference point, infinity, where total energy of the system is zero. So, initial energy of the system = 0 Final energy \[=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}-\frac{G{{m}_{1}}{{m}_{2}}}{d}\] From conservation of energy, Initial energy = Final energy \[\therefore \]\[0=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}-\frac{G{{m}_{1}}{{m}_{2}}}{d}\] or\[\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{1}}v_{2}^{2}=\frac{G{{m}_{1}}{{m}_{2}}}{d}\] ?.(1) By conservation of linear momentum \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=0\]or\[\frac{{{v}_{1}}}{{{v}_{2}}}=-\frac{{{m}_{2}}}{{{m}_{1}}}\Rightarrow {{v}_{2}}=-\frac{{{m}_{1}}}{{{m}_{2}}}{{v}_{1}}\] Putting value of \[{{v}_{2}}\]in equation (1), we get \[{{m}_{1}}v_{1}^{2}+{{m}_{2}}{{\left( -\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{2}}} \right)}^{2}}=\frac{2G{{m}_{1}}{{m}_{2}}}{d}\] \[\frac{{{m}_{1}}{{m}_{2}}v_{1}^{2}+m_{1}^{2}v_{1}^{2}}{{{m}_{2}}}=\frac{2G{{m}_{1}}{{m}_{2}}}{d}\] \[{{v}_{1}}=\sqrt{=\frac{2Gm_{2}^{2}}{d({{m}_{1}}+{{m}_{2}})}}={{m}_{2}}\sqrt{=\frac{2G}{d({{m}_{1}}+{{m}_{2}})}}\] Similarly\[={{v}_{2}}=-{{m}_{1}}\sqrt{\frac{2G}{d({{m}_{1}}+{{m}_{2}})}}\]
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