A) \[{{R}^{2/3}}\]
B) R
C) \[{{R}^{1/2}}\]
D) \[{{R}^{3/2}}\]
Correct Answer: C
Solution :
Angular momentum of earth around the sun \[\operatorname{L} = l\omega = m{{R}^{2}}\omega \] \[\operatorname{L} \propto m{{R}^{2}}\omega \] ... (i) As, \[{{T}^{2}}\,\,\propto \,\,{{R}^{3}}\] (Kepler?s law) \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,T\,\,\propto \,\,{{R}^{3/2}}\] Now, \[\omega =\frac{2\pi }{T}\propto {{R}^{-3/2}}\] ... (ii) Now, putting the value of \[\omega \] from Eq. (ii),to (i) \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,L\propto {{R}^{2}}\,{{R}^{-3/2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,L\propto \sqrt{R}\]
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