Consider a spherical planet of radius R . Its density varies with the distance of its centre t as `rho=A-Br`, where A and B are positive constants. Now answer the following questions.
Acceleration due to gravity at a distance `r(rltR)` from its centre is

Options

A. `(4)/(3)piGr(A-Br)`
B. `(4)piGr(A-Br)`
C. `(4)/(3)piGr(A-(3)/(4)Br)`
D. `(4)/(3)piGr(A-(3)/(2)Br)`

Answer

To find the acceleration due to gravity at a distance \( r \) from the center of a spherical planet with a variable density given by \( \rho = A - Br \), we will follow these steps:

### Step 1: Define the density function
The density of the planet varies with the distance from its center and is given by:
\[ \rho(r) = A - Br \]
where \( A \) and \( B \) are positive constants.

### Step 2: Calculate the mass enclosed within radius \( r \)
To find the mass \( M \) enclosed within a radius \( r \), we need to integrate the density over the volume of the sphere of radius \( r \). The volume element in spherical coordinates is given by:
\[ dV = 4\pi r'^2 dr' \]
Thus, the mass \( M \) can be calculated as:
\[ M = \int_0^r \rho(r') dV = \int_0^r (A - Br') (4\pi r'^2) dr' \]

### Step 3: Evaluate the integral
Now we evaluate the integral:
\[ M = 4\pi \int_0^r (A - Br') r'^2 dr' \]
This can be split into two parts:
\[ M = 4\pi \left( A \int_0^r r'^2 dr' - B \int_0^r r'^3 dr' \right) \]
Calculating the integrals:
1. \(\int_0^r r'^2 dr' = \frac{r^3}{3}\)
2. \(\int_0^r r'^3 dr' = \frac{r^4}{4}\)

Substituting these results back:
\[ M = 4\pi \left( A \cdot \frac{r^3}{3} - B \cdot \frac{r^4}{4} \right) \]
\[ M = \frac{4\pi}{3} A r^3 - \pi B r^4 \]

### Step 4: Use the formula for acceleration due to gravity
The acceleration due to gravity \( g \) at a distance \( r \) from the center is given by:
\[ g = \frac{GM}{r^2} \]
Substituting for \( M \):
\[ g = \frac{G \left( \frac{4\pi}{3} A r^3 - \pi B r^4 \right)}{r^2} \]
Simplifying this, we get:
\[ g = \frac{4\pi G}{3} A r - \pi G B r^2 \]

### Final Expression
Thus, the acceleration due to gravity at a distance \( r \) from the center of the planet is:
\[ g = \frac{4\pi G}{3} A r - \pi G B r^2 \]