A) \[2\sqrt{\frac{G}{d}({{M}_{1}}+{{M}_{2}})}\]
B) \[2\sqrt{\frac{2G}{d}({{M}_{1}}+{{M}_{2}})}\]
C) \[2\sqrt{\frac{Gm}{d}({{M}_{1}}+{{M}_{2}})}\]
D) \[2\sqrt{\frac{Gm({{M}_{1}}+{{M}_{2}})}{d({{R}_{1}}+{{R}_{2}})}}\]
Correct Answer: A
Solution :
Gravitational potential at mid point \[V=\frac{-G{{M}_{1}}}{d/2}+\frac{-G{{M}_{2}}}{d/2}\] Now, \[PE=m\times V=\frac{-2Gm}{d}({{M}_{1}}+{{M}_{2}})\] [m = mass of particle] So, for projecting particle from mid point to infinity \[KE\,=\,|\,PE\,|\] \[\Rightarrow \,\frac{1}{2}m{{v}^{2}}=\frac{2\,Gm}{d}({{M}_{1}}+{{M}_{2}})\] \[\Rightarrow \,v=2\sqrt{\frac{G\,({{M}_{1}}+{{M}_{2}})}{d}}\]You need to login to perform this action.
You will be redirected in
3 sec