A particle is fired upword with a speed of `20km//s` . The speed with which it will move in interstellar space is

Options

A. `8.8km//s`
B. `16.5km//s`
C. `4.6km//s`
D. `10km//s`

Answer

To solve the problem of a particle fired upward with a speed of 20 km/s and to find the speed with which it will move in interstellar space, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
The total mechanical energy (kinetic + potential) at the starting point must equal the total mechanical energy at the point in interstellar space.

### Step 2: Write the Initial Energy Equation
The initial kinetic energy (KE_initial) when the particle is fired is given by:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 \]
where \( v = 20 \, \text{km/s} = 20000 \, \text{m/s} \).

The initial potential energy (PE_initial) at the surface of the Earth is:
\[ PE_{\text{initial}} = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the particle, and \( R \) is the radius of the Earth.

### Step 3: Write the Final Energy Equation
In interstellar space, the final potential energy (PE_final) is zero, and the final kinetic energy (KE_final) is:
\[ KE_{\text{final}} = \frac{1}{2} mv'^2 \]
where \( v' \) is the speed in interstellar space.

### Step 4: Set Up the Energy Conservation Equation
According to the conservation of energy:
\[ KE_{\text{initial}} + PE_{\text{initial}} = KE_{\text{final}} + PE_{\text{final}} \]
Substituting the expressions we have:
\[ \frac{1}{2} mv^2 - \frac{GMm}{R} = \frac{1}{2} mv'^2 + 0 \]

### Step 5: Simplify the Equation
Canceling \( m \) from all terms (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 - \frac{GM}{R} = \frac{1}{2} v'^2 \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ v'^2 = v^2 - \frac{2GM}{R} \]

### Step 7: Calculate Escape Velocity
The escape velocity \( v_e \) from the Earth is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
Using the known value of escape velocity \( v_e \approx 11.2 \, \text{km/s} \):
\[ v_e^2 = (11.2 \, \text{km/s})^2 = 125.44 \, \text{km}^2/\text{s}^2 \]

### Step 8: Substitute Values
Now substitute \( v = 20 \, \text{km/s} \) and \( v_e^2 \) into the equation:
\[ v'^2 = (20 \, \text{km/s})^2 - 2 \times 125.44 \, \text{km}^2/\text{s}^2 \]
\[ v'^2 = 400 - 250.88 \]
\[ v'^2 = 149.12 \]

### Step 9: Calculate Final Velocity
Taking the square root gives:
\[ v' = \sqrt{149.12} \approx 12.2 \, \text{km/s} \]

### Conclusion
The speed with which the particle will move in interstellar space is approximately **12.2 km/s**.