© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the given density function
The mass density of the planet is given as
$ \rho(r) = \rho_0 \left(1 - \frac{r^2}{R^2}\right). $
Step 2: Set up the expression for mass enclosed within radius r
The mass enclosed within a sphere of radius $ r $ is found by integrating the density over the volume from the center up to radius $ r $:
$ M(r) = \int_{0}^{r} 4\pi r'^2 \, \rho(r') \, dr'. $
Substituting the given density,
$ M(r) = \int_{0}^{r} 4\pi r'^2 \, \rho_0 \left(1 - \frac{{r'}^2}{R^2}\right) dr'. $
Step 3: Perform the integration
Take out the constants and integrate term-by-term:
$ M(r) = 4\pi \rho_0 \int_{0}^{r} \left(r'^2 - \frac{r'^4}{R^2}\right) dr'.$
We split this integral:
$ \int_{0}^{r} r'^2 \, dr' = \frac{r^3}{3},
\quad \int_{0}^{r} r'^4 \, dr' = \frac{r^5}{5}. $
Hence,
$ M(r) = 4\pi \rho_0 \left[\frac{r^3}{3} - \frac{r^5}{5 R^2}\right]. $
Step 4: Write the gravitational field inside the planet
The gravitational field at a distance $ r $ from the center is given by
$ g(r) = \frac{G \, M(r)}{r^2}. $
Substituting $ M(r) $ from above,
$
g(r)
= \frac{G}{r^2} \left(
4\pi \rho_0 \left[\frac{r^3}{3} - \frac{r^5}{5 R^2}\right]
\right)
= 4\pi G \rho_0 \left[
\frac{r}{3} - \frac{r^3}{5 R^2}
\right].
$
Factor out $ r $:
$
g(r) = \frac{4\pi G \rho_0}{3} \, r \left(1 - \frac{3 r^2}{5 R^2}\right).
$
Step 5: Find the value of r for which g(r) is maximum
To find the maximum of $ g(r) $, differentiate $ g(r) $ with respect to $ r $ and set it to zero:
Let
$
g(r) = A \, r \left(1 - \alpha r^2\right),
\quad
\text{where }
A = \frac{4\pi G \rho_0}{3}
\text{ and }
\alpha = \frac{3}{5 R^2}.
$
Differentiate:
$
\frac{dg}{dr}
= A \left[
\left(1 - \alpha r^2\right)
+ r \left(-2 \alpha r\right)
\right]
= A \left[
1 - \alpha r^2 - 3 \alpha r^2
\right]
= A \left[
1 - 3\alpha r^2
\right].
$
Set $ \frac{dg}{dr} = 0 $:
$
1 - 3 \alpha r^2 = 0
\quad \Longrightarrow \quad
3 \alpha r^2 = 1
\quad \Longrightarrow \quad
r^2 = \frac{1}{3 \alpha}.
$
Since $ \alpha = \frac{3}{5 R^2}, $
$
r^2
= \frac{1}{3 \times \frac{3}{5 R^2}}
= \frac{1}{\frac{9}{5 R^2}}
= \frac{5 R^2}{9}
\quad \Longrightarrow \quad
r = \sqrt{\frac{5}{9}} \, R.
$
Step 6: Conclude the value of r for maximum gravitational field
Thus, the gravitational field is maximum at
$
r = \sqrt{\frac{5}{9}} R.
$
This matches the given correct answer.
