The mass density of a planet of radius R varies with the distance r from its centre as ρ(r) = ρ0 Then the gravitational field is maximum at:

Engineering

Physics

Gravitational Acceleration

Question

The mass density of a planet of radius R varies with the distance r from its centre as ρ(r) = ρ₀ $(1 - \frac{r^{2}}{R^{2}}) .$ Then the gravitational field is maximum at:

$r = \sqrt{\frac{5}{9}} R$

$r = \frac{1}{\sqrt{3}}$

$\sqrt{\frac{3}{4}} R$

r = R

$dm = ρ \times 4 {πx}^{2} dx = ρ_{0} (1 - \frac{x^{2}}{r^{2}}) \times 4 {πx}^{2} dx$

$m = 4 {πρ}_{0} \int_{0}^{r} (x^{2} - \frac{x^{4}}{R^{2}}) dx s$

$m = 4 {πρ}_{0} | \frac{r^{3}}{3} - \frac{r^{5}}{5 R^{2}} |$

$E = \frac{Gm}{r^{2}} = \frac{G}{r^{2}} \times 4 {πρ}_{0} (\frac{r^{3}}{3} - \frac{r^{5}}{5 R^{2}})$

$E = 4 {πGρ}_{0} (\frac{r}{3} - \frac{r^{3}}{5 R^{2}})$

E is maximum when $\frac{dE}{dr} = 0 \Rightarrow \frac{dE}{dr} = 4 {πGρ}_{0} (\frac{1}{3} - \frac{3 r^{3}}{5 R^{2}}) = 0$

$\Rightarrow r = \frac{\sqrt{5}}{3} R$

$E_{\max} = 4 {πGρ}_{0} \times \frac{\sqrt{5} R}{3} [\frac{1}{3} - \frac{1}{5} \times \frac{5}{9}]$

$E_{\max} = \frac{8 \sqrt{5}}{27} {πGρ}_{0} . R$

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