If an object is projected vertically upwards with speed , half the escape speed of earth , then the aximum height attained by it is [ R is radius of earth ]

Options

A. R
B. `R/2`
C. `2 R`
D. `R/3`

Answer

To solve the problem of finding the maximum height attained by an object projected vertically upwards with a speed equal to half the escape speed of Earth, we can use the principle of conservation of energy.

### Step-by-Step Solution:

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

2. **Determine Initial Velocity**:
The initial velocity \( V \) of the object is half the escape velocity:
\[ V = \frac{V_E}{2} = \frac{1}{2} \sqrt{2gR} = \frac{\sqrt{2gR}}{2} \]

3. **Calculate Initial Kinetic Energy**:
The initial kinetic energy \( KE_i \) of the object when it is projected is given by:
\[ KE_i = \frac{1}{2} m V^2 = \frac{1}{2} m \left(\frac{\sqrt{2gR}}{2}\right)^2 = \frac{1}{2} m \left(\frac{2gR}{4}\right) = \frac{mgR}{4} \]

4. **Calculate Initial Potential Energy**:
The initial potential energy \( PE_i \) of the object at the surface of the Earth is:
\[ PE_i = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

5. **Set Up Conservation of Energy Equation**:
At the maximum height \( H \), the kinetic energy will be zero, and the potential energy will be:
\[ PE_f = -\frac{GMm}{R + H} \]
By conservation of energy, we have:
\[ KE_i + PE_i = KE_f + PE_f \]
Substituting the values we calculated:
\[ \frac{mgR}{4} - \frac{GMm}{R} = 0 - \frac{GMm}{R + H} \]

6. **Simplify the Equation**:
Rearranging the equation gives:
\[ \frac{mgR}{4} - \frac{GMm}{R} = -\frac{GMm}{R + H} \]
Multiplying through by \( R(R + H) \) to eliminate the denominators:
\[ \frac{mgR^2(R + H)}{4} - GMm(R + H) = -GMmR \]

7. **Substituting \( GM \)**:
We know from the definition of \( g \) that \( g = \frac{GM}{R^2} \), thus \( GM = gR^2 \). Substituting this into the equation:
\[ \frac{mgR^2(R + H)}{4} - gR^2(R + H) = -gR^2 \]

8. **Solving for Height \( H \)**:
Rearranging gives:
\[ \frac{mgR^2(R + H)}{4} = gR^2(R + H) - gR^2 \]
Simplifying further, we can isolate \( H \):
\[ \frac{mgR^2}{4} = gR^2 - gR^2\frac{R}{R + H} \]
Solving this will yield:
\[ H = \frac{R}{3} \]

### Final Answer:
The maximum height attained by the object is:
\[ H = \frac{R}{3} \]