A) \[\sqrt{r}\]
B) r3/2
C) r
D) None of these
Correct Answer: A
Solution :
According to Keplers third law \[{{T}^{2}}\propto {{r}^{3}}\Rightarrow T\propto {{r}^{3/2}}\] We have \[\omega =\frac{2\pi }{T}\] \[\therefore \]\[\omega \propto {{r}^{-3/2}}\] Now, \[L=m{{r}^{2}}\omega \Rightarrow L\propto {{r}^{2}}\times {{r}^{-3/2}}\] \[\therefore \]\[L\alpha {{r}^{1/2}}\]
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