Question

A star of mass $M$ and radius $R$ is made up of gases. The average gravitational pressure compressing the star due to gravitational pull of the gases making up the star depends on $R$ as

• A. $1/R^4$
• B. $1/R$
• C. $1/R^2$
• D. $1/R^6$

Answer 1

Answer: (A)

For a spherical shell of radius $r$ and thickness $dr$, weight of layer is balanced by pressure on layer.

$\therefore$ Weight $=\left\{p-\left(p+\Delta p\right)\right\}4\pi r^2$
$\Rightarrow mg=-\Delta p.4\pi r^2$
$\Rightarrow 4\pi r^2.dr.p.g.=-\Delta p.4\pi r^2$
$\Rightarrow -\Delta p=pgdr$
$\Rightarrow -\Delta p=p^2.\frac{4}{3}\pi Rdr$
So, $P_{av}=\frac{1}{R}\underset{0}{\overset{R}{\int }}\Delta p=\frac{4}{3}\pi p^{2}R\underset{0}{\overset{R}{\int }}dr$
$=\frac{4}{3}\frac{\pi MR.R}{{\left(V\right)}^2}$
$=\frac{4}{3}\pi .\frac{M{R}^2}{{\left(\frac{4}{3}\pi R^3\right)}^2}$
$\Rightarrow P_{av}=\frac{3}{4\pi }.\frac{M}{R^4}$
$\Rightarrow P_{av}\alpha \frac{1}{R^4}$