A particle is fired vertically upward with a speed of `15 kms^-1`. With what speed will it move in interstellar space. Assume only earth's gravitational field.

Explanation

To solve the problem of finding the speed of a particle in interstellar space after being fired vertically upward with an initial speed of 15 km/s, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the Initial Speed to SI Units**:
The initial speed \( v \) is given as \( 15 \, \text{km/s} \). We need to convert this to meters per second:
\[ v = 15 \, \text{km/s} = 15,000 \, \text{m/s} \]

2. **Write the Expressions for Kinetic and Potential Energy**:
The kinetic energy (KE) at the Earth's surface is given by:
\[ KE = \frac{1}{2} m v^2 \]
The potential energy (PE) at the Earth's surface is given by:
\[ PE = -\frac{G M m}{r} \]
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.

3. **Total Energy at Earth's Surface**:
The total energy (E) at the Earth's surface is the sum of kinetic and potential energy:
\[ E = KE + PE = \frac{1}{2} m v^2 - \frac{G M m}{r} \]

4. **Energy in Interstellar Space**:
In interstellar space, the kinetic energy is:
\[ KE' = \frac{1}{2} m v'^2 \]
The potential energy is zero (since we are considering only Earth's gravitational field):
\[ PE' = 0 \]
Thus, the total energy in interstellar space is:
\[ E' = KE' + PE' = \frac{1}{2} m v'^2 + 0 = \frac{1}{2} m v'^2 \]

5. **Conservation of Energy**:
Since energy is conserved, we can set the total energy at the Earth's surface equal to the total energy in interstellar space:
\[ \frac{1}{2} m v^2 - \frac{G M m}{r} = \frac{1}{2} m v'^2 \]

6. **Canceling Mass**:
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 - \frac{G M}{r} = \frac{1}{2} v'^2 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ v'^2 = v^2 - \frac{2 G M}{r} \]

8. **Substituting Known Values**:
We know:
- \( v = 15,000 \, \text{m/s} \)
- \( G = 6.67 \times 10^{-11} \, \text{m}^3/\text{kg s}^2 \)
- \( M = 6 \times 10^{24} \, \text{kg} \)
- \( r = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)

Now substituting these values into the equation:
\[ v'^2 = (15,000)^2 - 2 \cdot \frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24})}{6.4 \times 10^6} \]

9. **Calculating the Gravitational Potential Energy Term**:
Calculate \( \frac{G M}{r} \):
\[ \frac{G M}{r} = \frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24})}{6.4 \times 10^6} \approx 6.25 \times 10^7 \, \text{m}^2/\text{s}^2 \]

10. **Final Calculation**:
Now plug this back into the equation:
\[ v'^2 = (15,000)^2 - 2 \cdot (6.25 \times 10^7) \]
\[ v'^2 = 225,000,000 - 125,000,000 = 100,000,000 \]
\[ v' = \sqrt{100,000,000} = 10,000 \, \text{m/s} = 10 \, \text{km/s} \]

### Final Answer:
The speed of the particle in interstellar space will be \( 10 \, \text{km/s} \).