What is the maximum height attained by a body projected with a velocity equal to one- third of the escape velocity from the surface of the earth? (Radius of the earth=R)

Options

A. `R//2`
B. `R//3`
C. `R//5`
D. `R//8`

Answer

To find the maximum height attained by a body projected with a velocity equal to one-third of the escape velocity from the surface of the Earth, we can follow these steps:

### Step 1: Calculate the Escape Velocity
The escape velocity (\(V_e\)) from the surface of the Earth is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.

### Step 2: Determine the Initial Velocity
Since the body is projected with a velocity equal to one-third of the escape velocity, we can express this as:
\[ V_h = \frac{1}{3} V_e = \frac{1}{3} \sqrt{\frac{2GM}{R}} \]

### Step 3: Calculate the Kinetic Energy at Launch
The kinetic energy (\(KE\)) of the body at the moment of projection can be calculated using the formula:
\[ KE = \frac{1}{2} m V_h^2 \]
Substituting \(V_h\):
\[ KE = \frac{1}{2} m \left(\frac{1}{3} \sqrt{\frac{2GM}{R}}\right)^2 = \frac{1}{2} m \cdot \frac{1}{9} \cdot \frac{2GM}{R} = \frac{mGM}{9R} \]

### Step 4: Calculate the Potential Energy at Maximum Height
At the maximum height (\(h\)), the body will momentarily come to rest, and all its kinetic energy will have been converted into gravitational potential energy (\(PE\)):
\[ PE = \frac{mGM}{R + h} \]
At maximum height, we can set the kinetic energy equal to the potential energy:
\[ \frac{mGM}{9R} = \frac{mGM}{R + h} \]

### Step 5: Solve for Maximum Height
We can cancel \(mGM\) from both sides (assuming \(m \neq 0\) and \(GM \neq 0\)):
\[ \frac{1}{9R} = \frac{1}{R + h} \]
Cross-multiplying gives:
\[ R + h = 9R \]
Thus:
\[ h = 9R - R = 8R \]

### Conclusion
The maximum height attained by the body is:
\[ h = 8R \]