A rocket is launched with velocity 10 km / s . If radius of earth is R , then maximum height attained by it will be

Options

A. 2 R
B. 3 R
C. 4 R
D. 5 R

Answer

To find the maximum height attained by a rocket launched with a velocity of 10 km/s from the surface of the Earth, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
The rocket is launched from the surface of the Earth with an initial velocity \( v_0 = 10 \, \text{km/s} \). The radius of the Earth is denoted as \( R \).

### Step 2: Write the Conservation of Energy Equation
At the surface of the Earth, the total mechanical energy (kinetic + potential) can be expressed as:
\[ E_{\text{initial}} = KE + PE = \frac{1}{2} m v_0^2 - \frac{GMm}{R} \]
Where:
- \( KE = \frac{1}{2} m v_0^2 \) is the kinetic energy,
- \( PE = -\frac{GMm}{R} \) is the gravitational potential energy (negative because it is defined relative to infinity).

At the maximum height \( h \), the velocity of the rocket becomes zero, so the total mechanical energy is:
\[ E_{\text{final}} = PE = -\frac{GMm}{R + h} \]

### Step 3: Set the Initial Energy Equal to the Final Energy
Using conservation of energy:
\[ \frac{1}{2} m v_0^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \]

### Step 4: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_0^2 - \frac{GM}{R} = -\frac{GM}{R + h} \]

### Step 5: Rearrange the Equation
Rearranging gives us:
\[ \frac{1}{2} v_0^2 = \frac{GM}{R} - \frac{GM}{R + h} \]
\[ \frac{1}{2} v_0^2 = GM \left( \frac{1}{R} - \frac{1}{R + h} \right) \]

### Step 6: Find a Common Denominator
The right-hand side can be simplified:
\[ \frac{1}{R} - \frac{1}{R + h} = \frac{(R + h) - R}{R(R + h)} = \frac{h}{R(R + h)} \]
Thus, we have:
\[ \frac{1}{2} v_0^2 = GM \cdot \frac{h}{R(R + h)} \]

### Step 7: Solve for \( h \)
Rearranging gives:
\[ h = \frac{R(R + h)}{2GM} v_0^2 \]
This can be further simplified, but we can also substitute \( g = \frac{GM}{R^2} \) (acceleration due to gravity at the surface of the Earth):
\[ h = \frac{R^2}{2g} v_0^2 - \frac{Rh}{2g} \]
Solving for \( h \) leads to:
\[ h \left(1 + \frac{1}{2g} v_0^2 \right) = \frac{R^2 v_0^2}{2g} \]
\[ h = \frac{R^2 v_0^2}{2g \left(1 + \frac{1}{2g} v_0^2 \right)} \]

### Step 8: Substitute Values
Substituting \( v_0 = 10 \times 10^3 \, \text{m/s} \) and \( g \approx 9.8 \, \text{m/s}^2 \):
\[ h \approx \frac{R^2 (10^4)^2}{2 \times 9.8 \left(1 + \frac{(10^4)^2}{2 \times 9.8}\right)} \]

### Step 9: Approximate the Result
After solving, we find that the maximum height \( h \) is approximately \( 4R \).

### Conclusion
Thus, the maximum height attained by the rocket is approximately \( 4R \).

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