A body is projected vertically upwards from the earth's surface. The body reaches a height equal to one half of radius of earth. Calculate the initial speed with which the body was projected initially.
Radius of earth = 6,400 km
Mass of earth `= 6 xx 10^(24) kg`

Answer

To solve the problem of calculating the initial speed with which a body is projected vertically upwards from the Earth's surface, we can use the principle of conservation of energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
A body is projected upwards and reaches a height equal to half the radius of the Earth. We need to find the initial speed (V) of the body.

### Step 2: Define the Variables
- Let \( R \) be the radius of the Earth, which is given as \( 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \).
- Let \( M \) be the mass of the Earth, given as \( 6 \times 10^{24} \, \text{kg} \).
- Let \( g \) be the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

### Step 3: Apply Conservation of Energy
The total mechanical energy at the surface of the Earth (initial state) must equal the total mechanical energy at the maximum height (final state).

1. **Initial Energy (at the surface)**:
- Kinetic Energy (KE) = \( \frac{1}{2} m V^2 \)
- Potential Energy (PE) = \( -\frac{G M m}{R} \)

Therefore, the total initial energy \( E_1 \) is:
\[ E_1 = \frac{1}{2} m V^2 - \frac{G M m}{R} \]

2. **Final Energy (at height \( h = \frac{R}{2} \))**:
- At maximum height, the kinetic energy is zero (the body momentarily stops).
- The potential energy at height \( h \) is:
\[ PE = -\frac{G M m}{R + h} = -\frac{G M m}{R + \frac{R}{2}} = -\frac{G M m}{\frac{3R}{2}} = -\frac{2 G M m}{3R} \]

Therefore, the total final energy \( E_2 \) is:
\[ E_2 = 0 - \frac{2 G M m}{3R} \]

### Step 4: Set Up the Energy Conservation Equation
Since energy is conserved:
\[ E_1 = E_2 \]
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \frac{1}{2} m V^2 - \frac{G M m}{R} = -\frac{2 G M m}{3R} \]

### Step 5: Simplify the Equation
1. Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} V^2 - \frac{G M}{R} = -\frac{2 G M}{3R} \]

2. Rearranging gives:
\[ \frac{1}{2} V^2 = \frac{G M}{R} - \frac{2 G M}{3R} \]

3. Combine the terms on the right:
\[ \frac{1}{2} V^2 = \frac{3 G M}{3R} - \frac{2 G M}{3R} = \frac{G M}{3R} \]

### Step 6: Solve for V
Multiply both sides by 2:
\[ V^2 = \frac{2 G M}{3R} \]
Taking the square root:
\[ V = \sqrt{\frac{2 G M}{3R}} \]

### Step 7: Substitute Known Values
Substituting \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \), \( M = 6 \times 10^{24} \, \text{kg} \), and \( R = 6400 \times 10^3 \, \text{m} \):
\[ V = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (6 \times 10^{24})}{3 \times (6400 \times 10^3)}} \]

Calculating this gives:
\[ V \approx 6456.6 \, \text{m/s} \approx 6.45 \, \text{km/s} \]

### Final Answer
The initial speed with which the body was projected is approximately **6.45 km/s**.

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