A ball of mass m is fired vertically upwards from the surface of the earth with velocity `nv_(e)`, where `v_(e)` is the escape velocity and `nlt1`. Neglecting air resistance, to what height will the ball rise? (Take radius of the earth=R):-

Options

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

Answer

To solve the problem of how high a ball will rise when fired vertically upwards with a velocity less than the escape velocity, we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the given parameters
- Mass of the ball: \( m \)
- Initial velocity of the ball: \( v = n v_e \) (where \( n < 1 \) and \( v_e \) is the escape velocity)
- Radius of the Earth: \( R \)
- Escape velocity formula: \( v_e = \sqrt{2gR} \)

### Step 2: Write the expression for initial kinetic energy (KE) and gravitational potential energy (PE)
- Initial kinetic energy (at the surface of the Earth):
\[ KE_i = \frac{1}{2} m (n v_e)^2 = \frac{1}{2} m n^2 v_e^2 \]
- Initial gravitational potential energy (at the surface of the Earth):
\[ PE_i = -\frac{G M m}{R} \]

### Step 3: Write the expression for final kinetic energy and gravitational potential energy at maximum height \( h \)
- At maximum height \( h \), the kinetic energy will be zero (the ball momentarily stops):
\[ KE_f = 0 \]
- Final gravitational potential energy (at height \( h \)):
\[ PE_f = -\frac{G M m}{R + h} \]

### Step 4: Apply the conservation of mechanical energy
According to the conservation of energy:
\[ KE_i + PE_i = KE_f + PE_f \]
Substituting the expressions we have:
\[ \frac{1}{2} m n^2 v_e^2 - \frac{G M m}{R} = 0 - \frac{G M m}{R + h} \]

### Step 5: Simplify the equation
Cancelling \( m \) from both sides (since \( m \neq 0 \)):
\[ \frac{1}{2} n^2 v_e^2 - \frac{G M}{R} = -\frac{G M}{R + h} \]
Rearranging gives:
\[ \frac{1}{2} n^2 v_e^2 = \frac{G M}{R} - \frac{G M}{R + h} \]

### Step 6: Substitute \( v_e \) and simplify
Substituting \( v_e = \sqrt{2gR} \):
\[ \frac{1}{2} n^2 (2gR) = \frac{G M}{R} - \frac{G M}{R + h} \]
This simplifies to:
\[ n^2 g R = \frac{G M}{R} - \frac{G M}{R + h} \]

### Step 7: Factor out \( G M \)
Factoring out \( G M \):
\[ n^2 g R = G M \left( \frac{1}{R} - \frac{1}{R + h} \right) \]
This can be rewritten as:
\[ n^2 g R = G M \left( \frac{(R + h) - R}{R(R + h)} \right) = \frac{G M h}{R(R + h)} \]

### Step 8: Cross-multiply and solve for \( h \)
Cross-multiplying gives:
\[ n^2 g R^2 (R + h) = G M h \]
Rearranging terms leads to:
\[ n^2 g R^2 + n^2 g R^2 h = G M h \]
\[ n^2 g R^2 = h \left( G M - n^2 g R^2 \right) \]
Thus, solving for \( h \):
\[ h = \frac{n^2 g R^2}{G M - n^2 g R^2} \]

### Step 9: Substitute \( g = \frac{G M}{R^2} \)
Substituting \( g \) into the equation:
\[ h = \frac{n^2 \frac{G M}{R^2} R^2}{G M - n^2 \frac{G M}{R^2} R^2} \]
This simplifies to:
\[ h = \frac{n^2 R}{1 - n^2} \]

### Final Result
The maximum height \( h \) to which the ball will rise is:
\[ h = \frac{n^2 R}{1 - n^2} \]