\(dm =\left(\frac{ k }{ r }\right)\left(4 \pi r ^{2} dr \right)\)
\(dm =4 \pi krdr\)
\(M =\int_{0}^{ R } d m =\int_{0}^{ R } 4 \pi krdr\)
\(M=\left.4 \pi k \frac{r^{2}}{2}\right|_{0} ^{R}\)
\(M =2 \pi k \left( R ^{2}-0\right)\)
\(M =2 \pi kR ^{2}\)
for circular motion gravitational force will provide required centripital force or
\(\frac{ GMm }{ R ^{2}}=\frac{ mv ^{2}}{ R }\)
\(\frac{G\left(2 \pi kR ^{2}\right) m }{ R ^{2}}=\frac{ mv ^{2}}{ R } \Rightarrow v =\sqrt{2 \pi GkR }\)
Time period \(\quad T =\frac{2 \pi R }{ v }\)
\(T =\frac{2 \pi R }{\sqrt{2 \pi GkR }} \propto \sqrt{ R }\)
or \(T^{2} \propto R\)
[પૃથ્વી ની ત્રિજયા $R =6400\, km , \sqrt{3}=1.732$ ]