The mass density of a spherical galaxy varies as \frac{K}{r} over a large distance 'r' from its center. In that region, a small star is in a circular orbit of radius R. Then the period of revolution, T depends on R as:
Option: 1 T^{2}\propto R
 
Option: 2 T^{2}\propto R^{3}
Option: 3 T^{2}\propto \frac{1}{R^{3}}  
Option: 4 T\propto R

Answers (1)

\begin{array}{l} \mathrm{dm}=\rho \mathrm{dv} \\ \\ \mathrm{dm}=\left(\frac{\mathrm{k}}{\mathrm{r}}\right)\left(4 \pi \mathrm{r}^{2} \mathrm{dr}\right) \end{array}

\begin{array}{l} \mathrm{dm}=4 \pi \mathrm{krdr} \\ \mathrm{M}=\int_{0}^{\mathrm{R}} \mathrm{dm}=\int_{0}^{\mathrm{R}} 4 \pi \mathrm{krdr} \\\\ \mathrm{M}=4 \pi \mathrm{k} [\frac{\mathrm{r}^{2}}{2} ]_{0}^{\mathrm{R}} \\ \\ \mathrm{M}=2 \pi \mathrm{k}\left(\mathrm{R}^{2}-0\right) \\ \mathrm{M}=2 \pi \mathrm{k} \mathrm{R}^{2} \end{array}

for circular motion, gravitational force will provide the required centripetal force 

i.e 

\dpi{120} \begin{array}{l} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}} \\ \\ \frac{\mathrm{G}\left(2 \pi \mathrm{k} \mathrm{R}^{2}\right) \mathrm{m}}{\mathrm{R}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{R}} \\ \Rightarrow \mathrm{v}=\sqrt{2 \pi \mathrm{GkR}} \\ \text {And Time period }= \mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}} \\ \\ \mathrm{T}=\frac{2 \pi \mathrm{R}}{(\sqrt{2 \pi \mathrm{GkR}})} \propto \sqrt{\mathrm{R}} \\ \\ \text {So } \mathrm{T}^{2} \propto \mathrm{R} \end{array}

 

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