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

Options

A. `sqrt((GM)/(R))`
B. `sqrt((2GM)/(R))`
C. `sqrt((1)/(4)(GM)/(R))`
D. `sqrt((GM)/(2R))`

Explanation

To solve the problem of determining 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 (v) required for a body to reach a height (h) equal to the radius of the Earth (R).

### Step 2: Apply Conservation of Mechanical Energy
The total mechanical energy at the point of projection (initial) should equal the total mechanical energy at the maximum height (final).

The initial mechanical energy consists of kinetic energy (KE) and gravitational potential energy (PE):
- Initial KE = (1/2)mv²
- Initial PE = -GMm/R (where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth)

At the maximum height (h = R), the potential energy becomes:
- Final PE = -GMm/(R + R) = -GMm/(2R)
- Final KE = 0 (as the body comes to rest at the maximum height)

### Step 3: Set Up the Energy Conservation Equation
Using the conservation of energy:
\[ \text{Initial Energy} = \text{Final Energy} \]
\[ \frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{2R} \]

### Step 4: Simplify the Equation
Canceling mass (m) from both sides (assuming m ≠ 0):
\[ \frac{1}{2}v^2 - \frac{GM}{R} = -\frac{GM}{2R} \]

Rearranging gives:
\[ \frac{1}{2}v^2 = \frac{GM}{R} - \frac{GM}{2R} \]
\[ \frac{1}{2}v^2 = \frac{GM}{2R} \]

### Step 5: Solve for v
Multiplying both sides by 2:
\[ v^2 = \frac{GM}{R} \]
Taking the square root:
\[ v = \sqrt{\frac{GM}{R}} \]

### Step 6: Relate to g
We know that \( g = \frac{GM}{R^2} \), so we can express \( GM \) as \( gR \):
\[ v = \sqrt{gR} \]

### Final Answer
The velocity of the body with which it was projected is \( v = \sqrt{gR} \).

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