question_answer 72)

                  An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler?s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that \[T=\,\frac{k}{R}\,\sqrt{\frac{{{r}^{3}}}{g}},\] where k is a dimensionless constant and g is acceleration due to gravity.                                

Answer:

                  \[T\propto {{r}^{a}}\,{{g}^{b}}\,{{R}^{C}}\] or\[T=K\,{{r}^{a}}\,{{g}^{b}}\,{{R}^{C}}\]                        ?.. (i)                 Writing dimensional formula of quantity                 \[[{{M}^{0}}{{L}^{0}}T]\,=\,{{[L]}^{a}}{{[L{{T}^{-2}}]}^{b}}{{[L]}^{z}}\]                 \[=[{{M}^{0}}{{L}^{a+b+c}}{{T}^{-2b}}]\]                 Using Principle of homogeneity                 \[a+b+c=0\]                                       ?...(ii)                 \[-2b=1\]                 or \[b=-\frac{1}{2}\]                                       ??(iii)                 Since \[{{T}^{2}}\propto \,{{r}^{3}}\] or \[\,T\,\,=\,K\,{{r}^{3/2}}\] \[\therefore \]\[a\,\,=3/2\]                 From eqn. (ii) \[c\,=\,\frac{1}{2}-\frac{3}{2}=-1\]                 Put values of \[a,\,b\] and \[z\] in eqn. (i)                 \[T\,=\,k\,{{r}^{3/2\,}}\,{{g}^{-1/2}}{{R}^{-1}}\,=\,\frac{K}{R}\sqrt{\frac{{{r}^{3}}}{g}}\]