A rocket is launched vertical from the surface of the earth of radius R with an initial speed `v`. If atmospheric resistance is neglected, then maximum height attained by the rocket is

Options

A. `h=(R)/(((2gR)/(v^(2))-1))`
B. `h=(R)/(((2gR)/(v^(2))+1))`
C. `h=(R^(2))/(((2gR)/(v^(2))-1))`
D. `h=(R^(2))/(((2gR)/(v^(2))+1))`

Explanation

To find the maximum height attained by a rocket launched vertically from the surface of the Earth with an initial speed \( v \), we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
The total mechanical energy (kinetic + potential) at the surface of the Earth will be equal to the total mechanical energy at the maximum height.

### Step 2: Write the Energy Equations
1. **At the surface (point 1)**:
- Kinetic Energy (KE) = \( \frac{1}{2} m v^2 \)
- Potential Energy (PE) = \( -\frac{GMm}{R} \)

Total Energy at point 1:
\[ E_1 = KE + PE = \frac{1}{2} m v^2 - \frac{GMm}{R} \]

2. **At the maximum height (point 2)**:
- Kinetic Energy (KE) = 0 (at maximum height, the velocity is zero)
- Potential Energy (PE) = \( -\frac{GMm}{R + h} \)

Total Energy at point 2:
\[ E_2 = 0 - \frac{GMm}{R + h} \]

### Step 3: Set the Total Energies Equal
Using the conservation of energy:
\[ E_1 = E_2 \]
\[ \frac{1}{2} m v^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \]

### Step 4: Simplify the Equation
1. Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 - \frac{GM}{R} = -\frac{GM}{R + h} \]

2. Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{GM}{R} - \frac{GM}{R + h} \]

### Step 5: Combine the Right Side
1. Find a common denominator:
\[ \frac{1}{2} v^2 = GM \left( \frac{(R + h) - R}{R(R + h)} \right) \]
\[ \frac{1}{2} v^2 = \frac{GMh}{R(R + h)} \]

### Step 6: Solve for \( h \)
1. Cross-multiply to isolate \( h \):
\[ \frac{1}{2} v^2 R(R + h) = GMh \]
\[ \frac{1}{2} v^2 R^2 + \frac{1}{2} v^2 Rh = GMh \]

2. Rearranging gives:
\[ \frac{1}{2} v^2 R^2 = GMh - \frac{1}{2} v^2 Rh \]
\[ \frac{1}{2} v^2 R^2 = h \left( GM - \frac{1}{2} v^2 R \right) \]

3. Finally, solve for \( h \):
\[ h = \frac{\frac{1}{2} v^2 R^2}{GM - \frac{1}{2} v^2 R} \]

### Step 7: Substitute \( GM \) with \( gR^2 \)
Using the relation \( g = \frac{GM}{R^2} \):
\[ h = \frac{\frac{1}{2} v^2 R^2}{gR - \frac{1}{2} v^2} \]

### Final Expression for Maximum Height
Thus, the maximum height \( h \) attained by the rocket is:
\[ h = \frac{v^2 R}{2g - \frac{v^2}{R}} \]