A man of mass m starts falling towards a planet of mass M and radius R. As he reaches near to the surface realizes that he will pass through a small hole in the planet. As he enters the hole he seen that the planet really made of two places a spherical shell of negligible thickness of mass `(2M)/(3)` and a point mass `(M)/(3)` of centre. Change in the force of gravity experienced by the man is

Options

A. `(2)/(3)(GM)/(R^(2))`
B. `0`
C. `(1)/(3)(GMm)/(R^(2))`
D. `(4)/(3)(GMm)/(R^(2))`

Explanation

To solve the problem of the change in gravitational force experienced by a man falling towards a planet with a specific structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure of the Planet**:
The planet consists of two parts:
- A spherical shell with mass \( \frac{2M}{3} \)
- A point mass at the center with mass \( \frac{M}{3} \)

2. **Gravitational Field Inside the Shell**:
According to the shell theorem, the gravitational field inside a spherical shell of uniform density is zero. Therefore, when the man is inside the shell, he experiences no gravitational force from the shell.

\[ E_{\text{shell}} = 0 \]

3. **Gravitational Force Due to the Point Mass**:
When the man is at a distance \( r \) from the center of the planet, the only gravitational force acting on him is due to the point mass at the center. The gravitational force \( F_{\text{new}} \) can be calculated using Newton's law of gravitation:

\[ F_{\text{new}} = \frac{G \cdot \left(\frac{M}{3}\right) \cdot m}{r^2} \]

4. **Gravitational Force Near the Surface**:
When the man is outside the shell, the gravitational force \( F_{\text{old}} \) he experiences is due to the total mass of the planet:

\[ F_{\text{old}} = \frac{G \cdot M \cdot m}{R^2} \]

5. **Change in Gravitational Force**:
The change in gravitational force experienced by the man as he passes from outside the shell to inside the shell can be calculated as:

\[ \Delta F = F_{\text{old}} - F_{\text{new}} \]

Substituting the expressions for \( F_{\text{old}} \) and \( F_{\text{new}} \):

\[ \Delta F = \frac{G \cdot M \cdot m}{R^2} - \frac{G \cdot \left(\frac{M}{3}\right) \cdot m}{r^2} \]

6. **Simplifying the Expression**:
To simplify, we can factor out \( G \cdot m \):

\[ \Delta F = G \cdot m \left( \frac{M}{R^2} - \frac{M}{3r^2} \right) \]

Assuming \( r \) approaches \( R \) as the man reaches the surface:

\[ \Delta F = G \cdot m \left( \frac{M}{R^2} - \frac{M}{3R^2} \right) \]

This simplifies to:

\[ \Delta F = G \cdot m \cdot \frac{2M}{3R^2} \]

7. **Final Result**:
Thus, the change in gravitational force experienced by the man is:

\[ \Delta F = \frac{2GmM}{3R^2} \]