Question

The mass density of a planet of radius$R$ varies with the distance$r$ from its center as$\rho \left(r\right)={\rho }_0(1-(r^2/R^2\left)\right)$. Then the gravitational field is maximum at:

• A. $r=1/\surd 3R$
• B. $r=(\surd 3/\surd 4)R$
• C. $r=R$
• D. $r=(\surd 5/\surd 9)R$

Answer 1

The correct option is D

$r=(\surd 5/\surd 9)R$

Step 1. Given data:

The radius of the planet$=R$

The distance from the center$=r$

The mass density$\rho \left(r\right)={\rho }_0\left(1-\left(r^2/R^2\right)\right)$

Step 2. Formula used:

$dm=\rho dv$ (for continuous mass distribution)

Step 3. Calculating the distance where the gravitational field is maximum

The total mass of the planet is,

$$\begin{gathered} M=\int \rho \left(r\right)dv \\ =\int {\rho }_0\left(1-\left(\frac{r^2}{R^2}\right)\right).4{\mathrm{\pi r}}^2\mathrm{dr} \\ =4{\mathrm{\pi \rho }}_0\left(\frac{{\mathrm{r}}^3}{3}-\frac{{\mathrm{r}}^5}{5{\mathrm{R}}^2}\right) \end{gathered}$$

The gravitational field is,

$E_g=\frac{GM}{R^2}$

$\begin{array}{rcl}{E}_g& =& \frac{G4{\mathrm{\pi \rho }}_0\left(\frac{{\mathrm{r}}^3}{3}-\frac{{\mathrm{r}}^5}{5{\mathrm{R}}^2}\right)}{r^2}\\ & =& G4{\mathrm{\pi \rho }}_0\left(\frac{\mathrm{r}}{3}-\frac{{\mathrm{r}}^3}{5{\mathrm{R}}^2}\right)\\ \frac{d{E}_g}{dr}& =& \left(\frac{1}{3}-\frac{3{\mathrm{r}}^2}{5{\mathrm{R}}^2}\right)=0\\ & \Rightarrow & \frac{1}{3}=\frac{3{\mathrm{r}}^2}{5{\mathrm{R}}^2}\\ & \Rightarrow & r=\frac{\sqrt{5}}{\sqrt{9}}R\end{array}$

Thus, the gravitational field is maximum$r=\left(\frac{\sqrt{5}}{\sqrt{9}}\right)R$.

Hence, option D is the correct answer.