First slide

Gravitational field intensity

Question

The mass density of a planet of radius R varies with the distance r from its center as $ρ (r) = ρ_{0} (1 - \frac{r^{2}}{R^{2}})$ . Then the gravitational field is maximum at :

Difficult

A

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

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

$r = R$
D

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

Solution

For points inside the planet, at distance r from the center $(r < R)$ , g is given by
$\begin{array}{l} E = \frac{G M_{e n c l o s e d}}{r^{2}} \\ h e r e M_{e n c l o s e d} = \int_{0}^{r} ρ d V = \int_{0}^{r} ρ_{0} (1 - \frac{r^{2}}{R^{2}}) 4 π r^{2} d r \\ \Rightarrow E = \frac{G \int_{0}^{r} ρ_{0} (1 - \frac{r^{2}}{R^{2}}) 4 π r^{2} d r}{r^{2}} \\ \Rightarrow E = G ρ_{0} 4 π (\frac{r}{3} - \frac{r^{3}}{5 R^{2}}) \\ For \max i m a, \frac{d E}{d r} = 0 \\ \Rightarrow \frac{d}{d r} G ρ_{0} 4 π (\frac{r}{3} - \frac{r^{3}}{5 R^{2}}) = 0 \\ \Rightarrow \frac{d}{d r} (\frac{r}{3} - \frac{r^{3}}{5 R^{2}}) = 0 \\ ∴ \frac{1}{3} - \frac{3 r^{2}}{5 R^{2}} = 0 \\ \Rightarrow r = \sqrt{\frac{5}{9}} R \end{array}$

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