Question

The pressure caused by the gravitational pull inside the earth at $a$ distance a measure from its centre, when its mass and radius are $m$ and $R$ respectively is given by

• A. $\frac{8}{3}\frac{G{m}^2}{\pi R^4}(1+\frac{a^2}{R^2})$
• B. $\frac{8}{3}\frac{G{m}^2}{\pi R^2}(1-\frac{a^2}{R^2})$
• C. $\frac{3}{8}\frac{G{m}^2}{\pi R^4}(1-{\left(\frac{a}{R}\right)}^2)$
• D. $\frac{8}{3}\frac{G{m}^2}{\pi R^4}(1-\frac{a}{R})$

Answer 1

The correct option is C $\frac{3}{8}\frac{G{m}^2}{\pi R^4}(1-{\left(\frac{a}{R}\right)}^2)$

As the pressure opposing the weight, so $dp=\rho g_{r}dr$
where $\rho =\frac{m}{4/3\pi R^3},g_r=\frac{Gmr}{R^3},$earth's internal gravity, $r$ indicates the distance from earth's center.
now, $p={\int }_a^R\frac{m}{4/3\pi R^3}.\frac{Gmr}{R^3}dr=\frac{3G{m}^2}{4\pi R^6}{\int }_a^{R}rdr$
or,$p=\frac{3G{m}^2}{4\pi R^6}\frac{1}{2}(R^2-a^2)=\frac{3G{m}^2}{8\pi R^4}[1-{\left(\frac{a}{R}\right)}^2]$
ans:(C)