A) \[-\frac{k}{2r}\]
B) \[-\frac{k}{r}\]
C) \[-\frac{2\,k}{r}\]
D) \[-\frac{4\,k}{r}\]
Correct Answer: A
Solution :
\[\frac{m{{\nu }^{2}}}{r}=\frac{k}{{{r}^{2}}}\,\,\Rightarrow \,\,m{{\nu }^{2}}=\frac{k}{r}\,\,\,\,\therefore \,\,\,KE=\frac{1}{2}m{{\nu }^{2}}=\frac{k}{2r}\] \[PE=\int{Fdr}=\int{\frac{k}{{{r}^{2}}}dr=-\frac{k}{r}}\] \[\therefore \,\,\,\,Total\,\,energy\,\,=\,\,KE+PE=\frac{k}{2r}-\frac{k}{r}=-\frac{k}{2r}\]
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