A particle of mass m is projected upwards with velocity `v=(v_(e))/(2)`, where `v_(e)` is the escape velocity then at the maximum height the potential energy of the particle is : (R is radius of earth and M is mass of earth)

Options

A. `(-GMm)/(2R)`
B. `(-GMm)/(4R)`
C. `(-3GMm)/(4R)`
D. `(-2GMm)/(3R)`

Answer

To solve the problem step by step, we will use the principle of conservation of energy.

### Step 1: Understand the given information
We have a particle of mass \( m \) projected upwards with a velocity \( v = \frac{v_e}{2} \), where \( v_e \) is the escape velocity. We need to find the potential energy at the maximum height.

### Step 2: Write down the expression for escape velocity
The escape velocity \( v_e \) is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 3: Calculate the initial kinetic energy (KE) and potential energy (PE) at the surface of the Earth
The initial kinetic energy at the surface (point A) when the particle is projected is:
\[ KE_A = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{v_e}{2}\right)^2 = \frac{1}{2} m \left(\frac{v_e^2}{4}\right) = \frac{m v_e^2}{8} \]

The initial potential energy at the surface (point A) is given by:
\[ PE_A = -\frac{GMm}{R} \]

### Step 4: Write the total energy at point A
The total energy at point A is:
\[ E_A = KE_A + PE_A = \frac{m v_e^2}{8} - \frac{GMm}{R} \]

### Step 5: Calculate the potential energy at the maximum height (point P)
At the maximum height (point P), the kinetic energy is zero (since the particle stops momentarily). Therefore, the total energy at point P is just the potential energy:
\[ E_P = PE_P \]

### Step 6: Apply the conservation of energy principle
According to the conservation of energy:
\[ E_A = E_P \]
Thus,
\[ \frac{m v_e^2}{8} - \frac{GMm}{R} = PE_P \]

### Step 7: Substitute the expression for escape velocity
Substituting \( v_e^2 = \frac{2GM}{R} \) into the equation:
\[ \frac{m \left(\frac{2GM}{R}\right)}{8} - \frac{GMm}{R} = PE_P \]
This simplifies to:
\[ \frac{mGM}{4R} - \frac{GMm}{R} = PE_P \]

### Step 8: Simplify the equation
Taking a common factor of \( \frac{GMm}{R} \):
\[ PE_P = \frac{GMm}{R} \left(\frac{1}{4} - 1\right) = \frac{GMm}{R} \left(-\frac{3}{4}\right) \]
Thus, we have:
\[ PE_P = -\frac{3GMm}{4R} \]

### Final Answer
The potential energy of the particle at the maximum height is:
\[ PE_P = -\frac{3GMm}{4R} \]