Answer to Question #210155 in Mechanics | Relativity for Ayesha

Question

a sphere whose radius at time t is b+acosnt is surrounded by liquid extending to infinity under no force . prove that the pressure at distance r from the centre is less than the pressure at an infinite distance by pn^2/r(b+acosnt)[a(1-3sin^2 nt)+bcos nt+a/2r^3 (sin^2 nt)(b+acos nt)^3]

Answer 1

Given,

The radius of the sphere at the time t is $(r)=b+a\cos(nt)$

$dr=-an\sin(nt)$

Let the uniform density of the sphere be $\rho$

$dV = A.dr$

So, mass of the partial sphere $(dm)=\rho A.dr$

Weight of the sphere $(W)=\rho g A. dr$

$g=\frac{GM}{r^2}=\frac{4\pi Gr^3}{3r^2}=\frac{4}{3}\pi Gr$

Let the pressure at the radius r is $P$

So, pressure at $r+dr$ be $P+dP$

Now, equating the downward force and upward force

$A(P+dP)+\frac{4\pi A G\rho^2 rdr}{3}-\frac{4\pi r^3}{3}\rho\times\frac{4}{3}\pi Gr=AP$

$\Rightarrow (dP)+\frac{4\pi G\rho^2 rdr}{3}-\frac{4\pi r^3}{3A}\rho\times\frac{4}{3}\pi Gr=0$

$\Rightarrow$ $dP=\frac{4\pi r^3}{3A}\rho\times\frac{4}{3}\pi Gr-\frac{4\pi G\rho^2 rdr)}{3}$

$\Rightarrow dP=\frac{4\pi (b+a\cos(nt))^4}{3A}\rho\times\frac{4}{3}\pi G-\frac{4\pi G\rho^2 (b+a\cos(nt))(-an\sin(nt)}{3}$