A hypothetical planet in the shape of a sphere is completely made of an incompressible fluid and has a mass M and radius R. If the pressure at the surface of the planet is zero, then the pressure at the centre of the planet is [G = universal constant of gravitation]

Options

A. `P=(3GM^(2))/(8piR^(4))`
B. `P=(3GM^(2))/(4piR^(4))`
C. `P=(3GM^(2))/(8pi^(2)R^(4))`
D. `P=(3GM^(2))/(4pi^(2)R^(4))`

Answer

To find the pressure at the center of a hypothetical planet made of an incompressible fluid, we can follow these steps:

### Step 1: Understand the Problem
We have a spherical planet with mass \( M \) and radius \( R \). The pressure at the surface of the planet is given to be zero. We need to find the pressure at the center of the planet.

### Step 2: Concept of Pressure
Pressure is defined as the force per unit area. In this case, the force contributing to the pressure at the center is due to the gravitational attraction of the mass of the fluid above it.

### Step 3: Gravitational Field Inside the Sphere
For a solid sphere, the gravitational field \( g \) at a distance \( r \) from the center is given by:
\[ g = -\frac{GM}{R^3} r \]
where \( G \) is the universal gravitational constant, \( M \) is the total mass of the planet, and \( R \) is the radius of the planet.

### Step 4: Consider a Spherical Shell
To find the pressure at the center, we consider a thin spherical shell of thickness \( dr \) at a distance \( r \) from the center. The mass \( dm \) of this shell can be expressed in terms of its volume and density.

### Step 5: Density Calculation
The density \( \rho \) of the planet is given by:
\[ \rho = \frac{M}{\frac{4}{3} \pi R^3} \]
The mass \( dm \) of the thin shell is:
\[ dm = \rho \cdot dV = \rho \cdot 4\pi r^2 dr = \frac{M}{\frac{4}{3} \pi R^3} \cdot 4\pi r^2 dr = \frac{3M}{R^3} r^2 dr \]

### Step 6: Force on the Shell
The gravitational force \( df \) acting on this shell is:
\[ df = g \cdot dm = -\frac{GM}{R^3} r \cdot \frac{3M}{R^3} r^2 dr = -\frac{3GM^2}{R^6} r^3 dr \]

### Step 7: Pressure Differential
The pressure differential \( dp \) at the center due to this force is:
\[ dp = \frac{df}{\text{Area}} = \frac{-\frac{3GM^2}{R^6} r^3 dr}{4\pi r^2} = -\frac{3GM^2}{4\pi R^6} r dr \]

### Step 8: Integrate to Find Total Pressure
To find the total pressure at the center \( P_c \), we integrate \( dp \) from \( r = 0 \) to \( r = R \):
\[ P_c = \int_0^R -\frac{3GM^2}{4\pi R^6} r dr \]
Calculating the integral:
\[ P_c = -\frac{3GM^2}{4\pi R^6} \cdot \left[\frac{r^2}{2}\right]_0^R = -\frac{3GM^2}{4\pi R^6} \cdot \frac{R^2}{2} \]
\[ P_c = \frac{3GM^2}{8\pi R^4} \]

### Final Answer
Thus, the pressure at the center of the planet is:
\[ P_c = \frac{3GM^2}{8\pi R^4} \]