The escape velocity of a body on the earth.s surface is `V_(e )`. A body is thrown vertically up with a speed of `(kV_(e)) (k lt 1)`. The maximum height reached by the body above the earth is

Options

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

Answer

To solve the problem, we need to find the maximum height \( h \) reached by a body thrown vertically upwards with an initial speed of \( kV_e \), where \( V_e \) is the escape velocity from the Earth's surface and \( k < 1 \).

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( V_e \) from the Earth's surface is given by the formula:
\[ V_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity and \( R \) is the radius of the Earth.

2. **Initial Velocity**:
The body is thrown with an initial velocity of:
\[ v = kV_e = k\sqrt{2gR} \]

3. **Conservation of Energy**:
At the maximum height \( h \), the kinetic energy will be converted into gravitational potential energy. The total mechanical energy at the surface and at the maximum height must be equal.

The initial kinetic energy (KE) when the body is thrown is:
\[ KE = \frac{1}{2} mv^2 = \frac{1}{2} m (kV_e)^2 = \frac{1}{2} m (k^2 \cdot 2gR) = mgRk^2 \]

The potential energy (PE) at the surface is:
\[ PE_{surface} = -\frac{GMm}{R} \]
At the maximum height \( h \), the potential energy becomes:
\[ PE_{max\ height} = -\frac{GMm}{R+h} \]

4. **Setting Up the Energy Equation**:
The total energy at the surface equals the total energy at maximum height:
\[ KE + PE_{surface} = PE_{max\ height} \]
Substituting the expressions we have:
\[ mgRk^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

5. **Simplifying the Equation**:
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gRk^2 - \frac{GM}{R} = -\frac{GM}{R+h} \]
Rearranging gives:
\[ gRk^2 + \frac{GM}{R+h} = \frac{GM}{R} \]

6. **Cross Multiplying**:
Cross-multiplying to eliminate the fractions:
\[ (gRk^2)(R+h) + GM = GM \cdot R \]
Expanding and simplifying:
\[ gR^2k^2 + gRk^2h + GM = GMR \]
Rearranging gives:
\[ gRk^2h = GMR - gR^2k^2 - GM \]

7. **Solving for \( h \)**:
Now isolate \( h \):
\[ h = \frac{GMR - gR^2k^2 - GM}{gRk^2} \]
Since \( g = \frac{GM}{R^2} \), substitute \( g \):
\[ h = \frac{R - Rk^2 - R}{k^2} = \frac{Rk^2}{1 - k^2} \]

Thus, the maximum height \( h \) reached by the body above the Earth's surface is:
\[ h = \frac{Rk^2}{1 - k^2} \]

### Final Answer:
The maximum height reached by the body above the Earth's surface is:
\[ h = \frac{Rk^2}{1 - k^2} \]