A) \[-G\]
B) \[-\frac{8}{3}G\]
C) \[-\frac{4}{3}G\]
D) \[-4G\]
Correct Answer: D
Solution :
The resulting gravitational potential, \[V=-2G\left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... \right]\] \[\Rightarrow \] \[V=-2G\left[ 1+\frac{1}{2}+\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{3}}}... \right]\] \[\Rightarrow \] \[V=-2G{{\left( 1+\frac{1}{2} \right)}^{-1}}\] \[\Rightarrow \] \[V=-\frac{2G}{\left( 1-\frac{1}{2} \right)}=-\frac{2G}{\left( \frac{1}{2} \right)}=-4G\]
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