# Light planet is revolving around a very massive star in a circular orbit of radius 'r' with a period of revolution T. On what power of 't' will the square of period depend if the gravitational force of attraction between planet and the star is proportional to $r^{\large\frac{-5}{2}}$

$(a)\;T^2 \; \alpha \; r^{\large\frac{+7}{2}} \quad (b)\;T^2 \; \alpha \; r^{\large\frac{-7}{2}} \quad (c)\; T^2 \; \alpha \; r^{\large\frac{3}{2}} \quad (d)\;T^2 \; \alpha \; r^{\large\frac{-3}{2}}$

As the gravitation provides the centripetal force
$\large\frac{mv^2}{r}=\frac{k}{r^{\frac{5}{2}}}$
$v^2=\large\frac{k}{mr^{\Large\frac{3}{2}}}$
$T=\large\frac{2 \pi r}{v}$$=2 \pi r \sqrt {\large\frac{mr^{\Large\frac{3}{2}}}{k}} T^2=\large\frac{4 \pi^2 m}{k}$$r ^{\large\frac{7}{2}}$
$T^2 \; \alpha \; r^{\large\frac{+7}{2}}$
Hence a is the correct answer.

edited Feb 17, 2014 by meena.p