A roket of mass M is launched vertically upwards from trhe surface of the earth with an intial speed v . Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the roket above the surface of the earth is

Options

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

Explanation

To solve the problem of finding the maximum height attained by a rocket launched from the surface of the Earth with an initial speed \( v \), we will use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify the Initial Energy:**
The initial energy of the rocket when it is launched consists of its kinetic energy and gravitational potential energy. The kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} M v^2 \]
The gravitational potential energy (PE) at the surface of the Earth is:
\[ PE = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the rocket, and \( R \) is the radius of the Earth.

2. **Total Initial Energy:**
The total initial energy \( E_i \) at the surface of the Earth is:
\[ E_i = KE + PE = \frac{1}{2} M v^2 - \frac{GMm}{R} \]

3. **Energy at Maximum Height:**
At the maximum height \( h \), the rocket will momentarily come to rest, so its kinetic energy will be zero. The gravitational potential energy at height \( h \) is:
\[ PE_h = -\frac{GMm}{R+h} \]
Therefore, the total energy \( E_f \) at maximum height is:
\[ E_f = PE_h = -\frac{GMm}{R+h} \]

4. **Apply Conservation of Energy:**
According to the conservation of energy, the total initial energy must equal the total energy at maximum height:
\[ E_i = E_f \]
Substituting the expressions we have:
\[ \frac{1}{2} M v^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

5. **Rearranging the Equation:**
Rearranging the equation to isolate \( h \):
\[ \frac{1}{2} M v^2 = -\frac{GMm}{R+h} + \frac{GMm}{R} \]
\[ \frac{1}{2} M v^2 = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) \]

6. **Finding a Common Denominator:**
The term \( \frac{1}{R} - \frac{1}{R+h} \) can be simplified:
\[ \frac{1}{R} - \frac{1}{R+h} = \frac{(R+h) - R}{R(R+h)} = \frac{h}{R(R+h)} \]
Thus, we can rewrite the energy equation as:
\[ \frac{1}{2} M v^2 = GMm \cdot \frac{h}{R(R+h)} \]

7. **Solving for \( h \):**
Rearranging gives:
\[ h = \frac{R(R+h)}{2M v^2 / GMm} \]
This leads to a quadratic equation in \( h \). Solving this quadratic equation will yield the maximum height \( h \).

8. **Final Expression:**
After solving the quadratic equation, we can express the maximum height \( h \) in terms of \( R \) and \( v \).