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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Gravitation

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 $r^{\large\frac{-5}{2}}$

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

1 Answer

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.

 

answered Aug 10, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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