A particle is projected vertically upward with with velocity `sqrt((2)/(3)(GM)/(R ))` from the surface of Earth. The height attained by it is (G, M, R have usual meanings)

Options

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

Answer

To solve the problem of finding the height attained by a particle projected vertically upward with a given initial velocity from the surface of the Earth, we can use the principle of conservation of energy.

### Given:
- Initial velocity \( v_0 = \sqrt{\frac{2}{3} \frac{GM}{R}} \)
- \( G \) = gravitational constant
- \( M \) = mass of the Earth
- \( R \) = radius of the Earth

### Step 1: Calculate the initial kinetic energy (KE) at the surface
The initial kinetic energy of the particle when it is projected is given by:
\[ KE = \frac{1}{2} m v_0^2 \]
Substituting the value of \( v_0 \):
\[ KE = \frac{1}{2} m \left(\sqrt{\frac{2}{3} \frac{GM}{R}}\right)^2 = \frac{1}{2} m \cdot \frac{2}{3} \cdot \frac{GM}{R} = \frac{mGM}{3R} \]

### Step 2: Calculate the gravitational potential energy (PE) at the surface
The gravitational potential energy at the surface of the Earth is given by:
\[ PE = -\frac{GMm}{R} \]

### Step 3: Calculate the total mechanical energy (E) at the surface
The total mechanical energy at the surface is the sum of kinetic and potential energy:
\[ E = KE + PE = \frac{mGM}{3R} - \frac{GMm}{R} \]
Combining the terms:
\[ E = \frac{mGM}{3R} - \frac{3mGM}{3R} = -\frac{2mGM}{3R} \]

### Step 4: Calculate the total mechanical energy at height \( h \)
At the height \( h \), the potential energy becomes:
\[ PE_h = -\frac{GMm}{R+h} \]
The kinetic energy at the maximum height is zero (as the particle comes to rest at the highest point). Thus, the total mechanical energy at height \( h \) is:
\[ E_h = 0 - \frac{GMm}{R+h} = -\frac{GMm}{R+h} \]

### Step 5: Set the total mechanical energy at the surface equal to that at height \( h \)
Using conservation of energy:
\[ -\frac{2mGM}{3R} = -\frac{GMm}{R+h} \]
Cancelling \( -GMm \) from both sides:
\[ \frac{2}{3R} = \frac{1}{R+h} \]

### Step 6: Solve for \( h \)
Cross-multiplying gives:
\[ 2(R+h) = 3R \]
Expanding and rearranging:
\[ 2R + 2h = 3R \implies 2h = 3R - 2R \implies 2h = R \implies h = \frac{R}{2} \]

### Final Answer:
The height attained by the particle is:
\[ h = \frac{R}{2} \]

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