A spherical planet far out in space has mass 2M and radius a. A particle of mass m is falling freely near its surface. What will be the acceleration of that particle ?

Options

A. `(GM)/(a^(2))`
B. `(3GM)/(a^(2))`
C. `(2GM)/(a^(2))`
D. `(4GM)/(a^(2))`

Answer

To find the acceleration of a particle of mass \( m \) falling freely near the surface of a spherical planet with mass \( 2M \) and radius \( a \), we can follow these steps:

### Step 1: Identify the gravitational force acting on the particle
The gravitational force \( F \) acting on the particle of mass \( m \) due to the planet can be calculated using Newton's law of gravitation:

\[ F = \frac{G \cdot M_p \cdot m}{r^2} \]

Where:
- \( G \) is the gravitational constant,
- \( M_p \) is the mass of the planet,
- \( r \) is the distance from the center of the planet to the particle (which is approximately the radius \( a \) of the planet when the particle is near the surface).

### Step 2: Substitute the known values
Given that the mass of the planet \( M_p = 2M \) and the radius \( r = a \), we can substitute these values into the equation:

\[ F = \frac{G \cdot (2M) \cdot m}{a^2} \]

### Step 3: Calculate the gravitational field strength \( g \)
The gravitational field strength \( g \) at the surface of the planet is defined as the force per unit mass experienced by a small mass \( m \):

\[ g = \frac{F}{m} = \frac{G \cdot (2M)}{a^2} \]

### Step 4: Determine the acceleration of the particle
Since the particle is in free fall, its acceleration \( a \) is equal to the gravitational field strength \( g \):

\[ a = g = \frac{G \cdot (2M)}{a^2} \]

### Final Result
Thus, the acceleration of the particle falling freely near the surface of the planet is:

\[ a = \frac{2GM}{a^2} \]