The velocity with which a body should be projected from the surface of the earth such that it reaches a maximum height equal to 5 times radius R fo the earth is
(M is mass of the earth )

Options

A. `sqrt((3GM)/(5R))`
B. `sqrt((4GM)/(3R))`
C. `sqrt((4GM)/(5R))`
D. `sqrt((5GM)/(3R))`

Explanation

To solve the problem of finding the velocity with which a body should be projected from the surface of the Earth to reach a maximum height equal to 5 times the radius of the Earth (5R), we can use the principle of conservation of mechanical energy. Here’s the step-by-step solution:

### Step 1: Understand the Problem
We need to find the initial velocity (V) required for a body to reach a height (h) of 5R, where R is the radius of the Earth.

### Step 2: Write the Conservation of Energy Equation
The total mechanical energy at the surface of the Earth (initial) must equal the total mechanical energy at the maximum height (final). The equation can be written as:
\[ \text{K.E.}_\text{initial} + \text{P.E.}_\text{initial} = \text{K.E.}_\text{final} + \text{P.E.}_\text{final} \]

### Step 3: Define Kinetic and Potential Energies
- At the surface of the Earth (initial):
- Kinetic Energy (K.E.) = \(\frac{1}{2} m V^2\)
- Potential Energy (P.E.) = \(-\frac{GMm}{R}\)

- At the maximum height (final):
- Kinetic Energy (K.E.) = 0 (at maximum height, the velocity is zero)
- Potential Energy (P.E.) = \(-\frac{GMm}{R + h}\), where \(h = 5R\)

### Step 4: Substitute Values into the Energy Equation
Substituting the values into the conservation of energy equation:
\[ \frac{1}{2} m V^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R + 5R} \]
This simplifies to:
\[ \frac{1}{2} m V^2 - \frac{GMm}{R} = -\frac{GMm}{6R} \]

### Step 5: Simplify the Equation
Rearranging gives:
\[ \frac{1}{2} m V^2 = \frac{GMm}{R} - \frac{GMm}{6R} \]
Factoring out \(\frac{GMm}{R}\):
\[ \frac{1}{2} m V^2 = GMm \left(\frac{1}{R} - \frac{1}{6R}\right) \]
\[ \frac{1}{2} m V^2 = GMm \left(\frac{6 - 1}{6R}\right) = \frac{5GMm}{6R} \]

### Step 6: Cancel Mass and Solve for V
Cancelling \(m\) from both sides (assuming \(m \neq 0\)):
\[ \frac{1}{2} V^2 = \frac{5GM}{6R} \]
Multiplying both sides by 2:
\[ V^2 = \frac{10GM}{6R} = \frac{5GM}{3R} \]
Taking the square root:
\[ V = \sqrt{\frac{5GM}{3R}} \]

### Final Answer
The velocity with which a body should be projected from the surface of the Earth to reach a maximum height equal to 5 times the radius of the Earth is:
\[ V = \sqrt{\frac{5GM}{3R}} \]