A) \[\sqrt{r}\]
B) \[{{r}^{3/2}}\]
C) \[r\]
D) none of these
Correct Answer: A
Solution :
From Keplers third law of planetary motion, the square of the orbital period (T) of planets is directly proportional to the cube of the semi-major axis of the orbital. \[\Rightarrow \] \[{{T}^{2}}\propto {{a}^{3}}\] \[\Rightarrow \] \[T\propto {{a}^{3/2}}\] Given, \[a=r\], and \[\omega =\frac{2\,\pi }{T}=\] angular speed \[\therefore \] \[\omega \propto {{r}^{-3/2}}\] Also angular momentum \[L=m{{r}^{2}}\omega \] \[\Rightarrow \] \[L\propto {{r}^{2}}\times {{r}^{-3/2}}\] \[\Rightarrow \] \[L\propto {{r}^{1/2}}\].
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