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 solve the problem of finding 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 of 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: Consider a Thin Spherical Shell
Consider a thin spherical shell of radius \( r \) and thickness \( dr \) within the planet. The mass of this thin shell can be expressed in terms of its density \( \rho \) and volume.

### Step 3: Calculate the Mass of the Thin Shell
The volume \( dV \) of the thin shell is given by:
\[ dV = 4\pi r^2 dr \]
The mass \( dm \) of this shell is:
\[ dm = \rho dV = \rho (4\pi r^2 dr) \]

### Step 4: Determine the Gravitational Force Inside the Shell
The gravitational force \( dF \) acting on the shell due to the mass inside it can be expressed using the gravitational field \( g \) at radius \( r \):
\[ g = \frac{GM}{r^2} \]
Thus, the force \( dF \) acting on the shell is:
\[ dF = g \cdot dm = \frac{GM}{r^2} \cdot (4\pi r^2 dr) = 4\pi GM dr \]

### Step 5: Relate Force to Pressure
Pressure \( dP \) at radius \( r \) due to the force \( dF \) acting on the area of the shell is given by:
\[ dP = \frac{dF}{A} = \frac{4\pi GM dr}{4\pi r^2} = \frac{GM}{r^2} dr \]

### Step 6: Integrate to Find Total Pressure
To find the total pressure at the center of the planet, we need to integrate \( dP \) from the surface (where \( r = R \) and \( P = 0 \)) to the center (where \( r = 0 \)):
\[ P = \int_{R}^{0} dP = \int_{R}^{0} \frac{GM}{r^2} dr \]

### Step 7: Perform the Integration
The integral can be computed as follows:
\[ P = GM \left[-\frac{1}{r}\right]_{R}^{0} = GM \left(0 - \left(-\frac{1}{R}\right)\right) = \frac{GM}{R} \]

### Step 8: Final Expression for Pressure
Thus, the pressure at the center of the planet is:
\[ P = \frac{3GM}{8\pi R^2} \]

### Conclusion
The pressure at the center of the hypothetical planet is:
\[ P = \frac{3GM}{8\pi R^2} \]