A spherical planet far out in space has a mass `M_(0)` and diameter `D_(0)`. A particle of mass m falling freely near the surface of this planet will experience an accelertion due to gravity which is equal to

Options

A. `4 GM_(p)//D_(p)^(2)`
B. `GM_(p)m//D_(p)^(2)`
C. `GM_(p)m//D_(p)^(2)`
D. `4GM_(p)m//D_(p)^(2)`

Explanation

To find the acceleration due to gravity experienced by a particle of mass \( m \) falling freely near the surface of a spherical planet with mass \( M_0 \) and diameter \( D_0 \), we can follow these steps:

### Step 1: Understand the relationship between gravitational force and acceleration
The gravitational force \( F \) acting on the particle of mass \( m \) due to the planet is given by Newton's law of gravitation:
\[ F = \frac{G M_0 m}{r^2} \]
where \( G \) is the universal gravitational constant and \( r \) is the radius of the planet.

### Step 2: Express the radius in terms of diameter
The radius \( r \) of the planet can be expressed in terms of its diameter \( D_0 \):
\[ r = \frac{D_0}{2} \]

### Step 3: Substitute the radius into the gravitational force equation
Substituting \( r \) into the gravitational force equation gives:
\[ F = \frac{G M_0 m}{\left(\frac{D_0}{2}\right)^2} \]
This simplifies to:
\[ F = \frac{G M_0 m}{\frac{D_0^2}{4}} = \frac{4 G M_0 m}{D_0^2} \]

### Step 4: Relate force to acceleration
According to Newton's second law, the force acting on the particle is also equal to the mass of the particle multiplied by its acceleration \( a \):
\[ F = m a \]

### Step 5: Set the two expressions for force equal to each other
Equating the two expressions for force gives:
\[ m a = \frac{4 G M_0 m}{D_0^2} \]

### Step 6: Solve for acceleration \( a \)
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a = \frac{4 G M_0}{D_0^2} \]

### Conclusion
The acceleration due to gravity experienced by the particle near the surface of the planet is:
\[ a = \frac{4 G M_0}{D_0^2} \]

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