A projectile is fired vertically upwards from the surface of the earth with a velocity `Kv_(e)`, where `v_(e)` is the escape velocity and `Klt1`.If `R` is the radius of the earth, the maximum height to which it will rise measured from the centre of the earth will be (neglect air resistance)

Options

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

Answer

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

### Step 1: Understand the Initial Conditions
The projectile is fired with an initial velocity \( K v_e \), where \( v_e \) is the escape velocity. The escape velocity from the surface of the Earth is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

### Step 2: Write the Initial Energy
The initial energy of the projectile consists of kinetic energy and gravitational potential energy. The kinetic energy (KE) at the surface is:
\[ KE = \frac{1}{2} m (K v_e)^2 = \frac{1}{2} m K^2 v_e^2 \]
The gravitational potential energy (PE) at the surface of the Earth is:
\[ PE = -\frac{GMm}{R} \]
Thus, the total initial energy \( E_i \) is:
\[ E_i = KE + PE = \frac{1}{2} m K^2 v_e^2 - \frac{GMm}{R} \]

### Step 3: Write the Final Energy at Maximum Height
At the maximum height \( h \), the velocity of the projectile is zero, so its kinetic energy is zero. The gravitational potential energy at height \( h \) is:
\[ PE = -\frac{GMm}{R + h} \]
Thus, the total final energy \( E_f \) is:
\[ E_f = PE = -\frac{GMm}{R + h} \]

### Step 4: Apply Conservation of Energy
According to the conservation of energy:
\[ E_i = E_f \]
Substituting the expressions for \( E_i \) and \( E_f \):
\[ \frac{1}{2} m K^2 v_e^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \]

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

### Step 6: Substitute for Escape Velocity
Substituting \( v_e^2 = \frac{2GM}{R} \):
\[ \frac{1}{2} K^2 \left(\frac{2GM}{R}\right) = \frac{GM}{R} - \frac{GM}{R + h} \]
This simplifies to:
\[ K^2 \frac{GM}{R} = \frac{GM}{R} - \frac{GM}{R + h} \]

### Step 7: Solve for \( h \)
Factor out \( GM \):
\[ K^2 \frac{1}{R} = \frac{1}{R} - \frac{1}{R + h} \]
Cross-multiplying gives:
\[ K^2 (R + h) = R - (R + h) \]
Thus,
\[ K^2 R + K^2 h = R - R - h \]
Rearranging gives:
\[ K^2 h + h = R - K^2 R \]
Factoring out \( h \):
\[ h(1 + K^2) = R(1 - K^2) \]
So,
\[ h = \frac{R(1 - K^2)}{1 + K^2} \]

### Step 8: Calculate the Maximum Height from the Center of the Earth
The maximum height from the center of the Earth is:
\[ H = R + h = R + \frac{R(1 - K^2)}{1 + K^2} \]
This simplifies to:
\[ H = R \left(1 + \frac{1 - K^2}{1 + K^2}\right) = R \left(\frac{1 + K^2 + 1 - K^2}{1 + K^2}\right) = R \left(\frac{2}{1 + K^2}\right) \]

### Final Result
Thus, the maximum height to which the projectile will rise measured from the center of the Earth is:
\[ H = \frac{R}{1 - K^2} \]