Question

A large fluid oscillates in shape under the influence of its own gravitational field. Using dimensional analysis find the expression for period of oscillation $\left(T\right)$ in terms of radius of star $\left(R\right)$, mean density of fluid $\left(\rho \right)$ and universal gravitational constant $\left(G\right)$.

Answer 1

We have the following relation

$T\propto R^a{\rho }^b{G}^c$

or $T=k{R}^a{\rho }^b{G}^c$

here

$\left[T\right]=\left[M^0{L}^0{T}^1\right]$

$\left[R\right]=\left[M^0{L}^1{T}^0\right]$

$\left[\rho \right]=\left[M^1{L}^{-3}{T}^0\right]$

$\left[G\right]=\left[M^{-1}{L}^3{T}^{-2}\right]$

So, we have

$\left[M^0{L}^0{T}^1\right]=[M^0{L}^1{T}^0{]}^a.[M^1{L}^{-3}{T}^0{]}^b.[M^{-1}{L}^3{T}^{-2}{]}^c$

$\left[M^0{L}^0{T}^1\right]=\left[M^{b-c}{L}^{a-3b+3c}{T}^{-2c}\right]$

Now by comparing the two sides we get

$b-c=0\Rightarrow b=c$

$a-3b+3c=0\Rightarrow a=3(b-c)=0$

$-2c=1\Rightarrow c=-\frac{1}{2}$

So,

$b=-\frac{1}{2},a=0$

Thus, from the first relation, we get

$T=\frac{k}{(\rho .G{)}^{1/2}}$