A body attains a height equal to the radius of the earth. the velocity of the body with which is was projected is

Options

A. `sqrt((GM)/(R))`
B. `sqrt((2 GN)/(R))`
C. `sqrt((5)/(4)(GM)/(R))`
D. `sqrt((3 GM)/(R))`

Explanation

To solve the problem of finding the velocity with which a body must be projected to reach a height equal to the radius of the Earth, we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We need to find the initial velocity (u) of a body projected from the surface of the Earth that reaches a height (h) equal to the radius of the Earth (R). At this height, the body's velocity will be zero.

### Step 2: Set Up the Energy Conservation Equation
The total mechanical energy at the point of projection (point A) and at the maximum height (point B) should be equal.

At point A (on the surface of the Earth):
- Kinetic Energy (KE) = \( \frac{1}{2} mv^2 \)
- Potential Energy (PE) = \( -\frac{GMm}{R} \) (where G is the gravitational constant, M is the mass of the Earth, and m is the mass of the body)

At point B (at height R):
- Kinetic Energy (KE) = 0 (since the velocity is zero)
- Potential Energy (PE) = \( -\frac{GMm}{2R} \) (the distance from the center of the Earth is now 2R)

### Step 3: Write the Energy Conservation Equation
Using the conservation of energy:
\[ \text{Total Energy at A} = \text{Total Energy at B} \]
\[ \frac{1}{2} mv^2 - \frac{GMm}{R} = -\frac{GMm}{2R} \]

### Step 4: Simplify the Equation
Rearranging the equation gives:
\[ \frac{1}{2} mv^2 = -\frac{GMm}{2R} + \frac{GMm}{R} \]
\[ \frac{1}{2} mv^2 = \frac{GMm}{2R} \]

### Step 5: Cancel out the mass (m)
Since m appears on both sides, we can cancel it out:
\[ \frac{1}{2} v^2 = \frac{GM}{2R} \]

### Step 6: Solve for v
Multiplying both sides by 2 gives:
\[ v^2 = \frac{GM}{R} \]
Taking the square root of both sides results in:
\[ v = \sqrt{\frac{GM}{R}} \]

### Step 7: Final Expression
Thus, the velocity with which the body must be projected is:
\[ v = \sqrt{\frac{GM}{R}} \]