A rocket is launched vertically from the surface of earth with an initial velocity `v`. How far above the surface of earth it will go? Neglect the air resistance.

Options

A. `(R_(E)v^(2))/(gR_(E)-v^(2))`
B. `(R_(E)v^(2))/(gR_(E)+v^(2))`
C. `(R_(E)v^(2))/(2gR_(E)-v^(2))`
D. `(R_(E)v^(2))/(2gR_(E)+v^(2))`

Explanation

To solve the problem of how far above the surface of the Earth a rocket will go when launched vertically with an initial velocity \( v \), 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 of the rocket at the launch point (initial) will be equal to the total mechanical energy at the maximum height (final). The initial energy consists of kinetic energy and gravitational potential energy, while the final energy consists only of gravitational potential energy.

### Step 2: Write the Initial Energy Equation
The initial energy \( E_i \) when the rocket is at the surface of the Earth is given by:
\[ E_i = KE + PE = \frac{1}{2} m v^2 - \frac{GMm}{R} \]
where:
- \( KE \) is the kinetic energy,
- \( PE \) is the gravitational potential energy,
- \( m \) is the mass of the rocket,
- \( v \) is the initial velocity,
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

### Step 3: Write the Final Energy Equation
At the maximum height \( h \), the energy \( E_f \) is given by:
\[ E_f = PE = -\frac{GMm}{R+h} \]

### Step 4: Set Initial Energy Equal to Final Energy
Using the conservation of energy, we have:
\[ \frac{1}{2} m v^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

### Step 5: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 - \frac{GM}{R} = -\frac{GM}{R+h} \]
Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{GM}{R} - \frac{GM}{R+h} \]

### Step 6: Combine the Terms
To combine the terms on the right side, we find a common denominator:
\[ \frac{1}{2} v^2 = GM \left( \frac{(R+h) - R}{R(R+h)} \right) = \frac{GMh}{R(R+h)} \]

### Step 7: Rearrange to Solve for \( h \)
Now, we can rearrange the equation to isolate \( h \):
\[ \frac{1}{2} v^2 \cdot R(R+h) = GMh \]
Expanding and rearranging gives:
\[ \frac{1}{2} v^2 R + \frac{1}{2} v^2 h = GMh \]
\[ \frac{1}{2} v^2 R = GMh - \frac{1}{2} v^2 h \]
\[ \frac{1}{2} v^2 R = h \left( GM - \frac{1}{2} v^2 \right) \]
Thus,
\[ h = \frac{\frac{1}{2} v^2 R}{GM - \frac{1}{2} v^2} \]

### Step 8: Substitute \( GM \)
Using \( GM = gR^2 \) (where \( g \) is the acceleration due to gravity at the surface of the Earth):
\[ h = \frac{\frac{1}{2} v^2 R}{gR^2 - \frac{1}{2} v^2} \]

### Final Result
This gives us the height \( h \) that the rocket will reach above the surface of the Earth.