A small block of super dense material has mass ` 2 xx 10^(24) kg`. It is at a height `h lt lt R `. It falls towards the earth.Find its speed when it is at a height `( h )/( 2)`

Options

A. `sqrt(( 2gh)/( 3))`
B. `sqrt(( 3gh)/( 4))`
C. `sqrt(( 3gh)/( 5))`
D. `sqrt(( gh)/( 2))`

Answer

To solve the problem of finding the speed of a small block of super dense material when it is at a height \( \frac{h}{2} \) 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
We have a block of mass \( m_b = 2 \times 10^{24} \, \text{kg} \) falling from a height \( h \) (where \( h \ll R \), with \( R \) being the radius of the Earth). We need to find its speed when it reaches a height \( \frac{h}{2} \).

### Step 2: Apply Conservation of Energy
The total mechanical energy (potential + kinetic) at the initial height \( h \) should equal the total mechanical energy at the height \( \frac{h}{2} \).

1. **Initial Potential Energy (PE_initial)** at height \( h \):
\[ PE_{\text{initial}} = -\frac{G M m_b}{R + h} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

2. **Final Potential Energy (PE_final)** at height \( \frac{h}{2} \):
\[ PE_{\text{final}} = -\frac{G M m_b}{R + \frac{h}{2}} \]

3. **Kinetic Energy (KE)** at height \( \frac{h}{2} \):
\[ KE = \frac{1}{2} m_b v^2 \]

### Step 3: Set Up the Energy Conservation Equation
Using the conservation of energy:
\[ PE_{\text{initial}} = PE_{\text{final}} + KE \]
Substituting the expressions we have:
\[ -\frac{G M m_b}{R + h} = -\frac{G M m_b}{R + \frac{h}{2}} + \frac{1}{2} m_b v^2 \]

### Step 4: Simplify the Equation
1. Rearranging the equation gives:
\[ \frac{1}{2} m_b v^2 = -\frac{G M m_b}{R + h} + \frac{G M m_b}{R + \frac{h}{2}} \]

2. Factor out \( G M m_b \):
\[ \frac{1}{2} m_b v^2 = G M m_b \left( \frac{1}{R + \frac{h}{2}} - \frac{1}{R + h} \right) \]

3. Cancel \( m_b \) from both sides:
\[ \frac{1}{2} v^2 = G M \left( \frac{1}{R + \frac{h}{2}} - \frac{1}{R + h} \right) \]

### Step 5: Calculate the Difference in Potential
1. Finding a common denominator:
\[ \frac{1}{R + \frac{h}{2}} - \frac{1}{R + h} = \frac{(R + h) - (R + \frac{h}{2})}{(R + h)(R + \frac{h}{2})} = \frac{\frac{h}{2}}{(R + h)(R + \frac{h}{2})} \]

2. Substitute back into the equation:
\[ \frac{1}{2} v^2 = G M \cdot \frac{\frac{h}{2}}{(R + h)(R + \frac{h}{2})} \]

### Step 6: Solve for \( v \)
1. Multiply both sides by 2:
\[ v^2 = \frac{G M h}{(R + h)(R + \frac{h}{2})} \]

2. Take the square root:
\[ v = \sqrt{\frac{G M h}{(R + h)(R + \frac{h}{2})}} \]

### Step 7: Final Expression
Since \( h \ll R \), we can approximate:
\[ v \approx \sqrt{\frac{G M h}{R^2}} = \sqrt{g h} \]
where \( g = \frac{G M}{R^2} \) is the acceleration due to gravity.

### Conclusion
Thus, the speed of the block when it reaches a height \( \frac{h}{2} \) is approximately:
\[ v = \sqrt{g h} \]