A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at r (r < R), then the correct option(s) is (are)

Options

A. P(r=0)=0
B. `(P(r=3R//4))/(P(r=2R//3))=63/80`
C. `(P(r=3R//5))/(P(r=2R//5))=16/21`
D. `(P(r=R//2))/(P(r=R//3))=20/27`

Explanation

To solve the problem, we need to find the pressure \( P(r) \) at a distance \( r \) from the center of a spherical body of radius \( R \) that consists of a fluid of constant density and is in equilibrium under its own gravity.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a spherical body of radius \( R \) filled with a fluid of constant density \( \rho \).
- We need to find the pressure \( P(r) \) at a point inside the sphere at a distance \( r \) from the center, where \( r < R \).

2. **Consider a Shell**:
- Consider a thin spherical shell of thickness \( dr \) at a distance \( r \) from the center. The inner pressure at this shell is \( P(r) \) and the outer pressure is \( P(r) + dP \).

3. **Hydrostatic Equilibrium**:
- According to hydrostatic equilibrium, the change in pressure across the shell is given by:
\[ P(r) - (P(r) + dP) = -\rho g \, dr \]
- This simplifies to:
\[ -dP = \rho g \, dr \]

4. **Expression for Acceleration due to Gravity \( g \)**:
- The acceleration due to gravity \( g \) at a distance \( r \) from the center of the sphere is given by:
\[ g = \frac{G M(r)}{r^2} \]
- Where \( M(r) \) is the mass enclosed within the radius \( r \):
\[ M(r) = \rho \cdot \text{Volume} = \rho \cdot \frac{4}{3} \pi r^3 \]
- Thus,
\[ g = \frac{G \rho \cdot \frac{4}{3} \pi r^3}{r^2} = \frac{4}{3} \pi G \rho r \]

5. **Substituting \( g \) into the Pressure Equation**:
- Now substituting \( g \) back into the pressure equation:
\[ -dP = \rho \left(\frac{4}{3} \pi G \rho r\right) dr \]
- This gives:
\[ dP = -\frac{4}{3} \pi G \rho^2 r \, dr \]

6. **Integrating to Find Pressure**:
- Integrate both sides from \( R \) to \( r \):
\[ \int_{P(R)}^{P(r)} dP = -\frac{4}{3} \pi G \rho^2 \int_{R}^{r} r \, dr \]
- The left side becomes \( P(r) - P(R) \) and the right side becomes:
\[ -\frac{4}{3} \pi G \rho^2 \left[\frac{r^2}{2}\right]_{R}^{r} = -\frac{4}{3} \pi G \rho^2 \left(\frac{r^2}{2} - \frac{R^2}{2}\right) \]
- Therefore, we have:
\[ P(r) - P(R) = -\frac{4}{3} \pi G \rho^2 \left(\frac{r^2 - R^2}{2}\right) \]

7. **Finding \( P(R) \)**:
- At the surface \( r = R \), the pressure \( P(R) \) is usually taken as zero (atmospheric pressure or reference pressure). Thus:
\[ P(r) = \frac{4}{3} \pi G \rho^2 \left(\frac{R^2 - r^2}{2}\right) \]

8. **Final Expression**:
- The final expression for pressure at a distance \( r \) from the center is:
\[ P(r) = \frac{2}{3} \pi G \rho (R^2 - r^2) \]

### Conclusion:
The pressure \( P(r) \) at a distance \( r \) from the center of the sphere is proportional to the difference between the square of the radius of the sphere and the square of the distance \( r \).