Calculate the escape speed of an atmospheric particle which is 1000 km above the earth's surface . (Radius of the = 6400 km nd acceleration due to gravity =` 9.8 ms ^(-2)`

Options

A. `6.5 `km/s
B. 8 km/s
C. 10 km/s
D. `11.2 ` km/s

Answer

To calculate the escape speed of an atmospheric particle located 1000 km above the Earth's surface, we can follow these steps:

### Step 1: Understand the Escape Speed Formula
The escape speed \( V_e \) from a height \( h \) above the Earth's surface is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R + h}} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth,
- \( h \) is the height above the Earth's surface.

### Step 2: Calculate the Effective Radius
Given:
- Radius of the Earth \( R = 6400 \) km \( = 6400 \times 10^3 \) m,
- Height \( h = 1000 \) km \( = 1000 \times 10^3 \) m.

The effective radius \( R + h \) is:
\[ R + h = 6400 \times 10^3 + 1000 \times 10^3 = 7400 \times 10^3 \text{ m} \]

### Step 3: Calculate the Gravitational Constant \( GM \)
We know that the acceleration due to gravity at the Earth's surface \( g = 9.8 \, \text{m/s}^2 \) is given by:
\[ g = \frac{GM}{R^2} \]
From this, we can express \( GM \) as:
\[ GM = g \cdot R^2 \]
Substituting the values:
\[ GM = 9.8 \, \text{m/s}^2 \cdot (6400 \times 10^3 \, \text{m})^2 \]

### Step 4: Substitute into the Escape Speed Formula
Now, substituting \( GM \) and \( R + h \) into the escape speed formula:
\[ V_e = \sqrt{\frac{2 \cdot (9.8 \cdot (6400 \times 10^3)^2)}{7400 \times 10^3}} \]

### Step 5: Calculate the Escape Speed
Calculating the above expression step-by-step:
1. Calculate \( (6400 \times 10^3)^2 \):
\[ (6400 \times 10^3)^2 = 40960000 \times 10^{6} = 4.096 \times 10^{13} \, \text{m}^2 \]
2. Calculate \( GM \):
\[ GM = 9.8 \cdot 4.096 \times 10^{13} = 4.01808 \times 10^{14} \, \text{m}^3/\text{s}^2 \]
3. Substitute \( GM \) back into the escape speed formula:
\[ V_e = \sqrt{\frac{2 \cdot 4.01808 \times 10^{14}}{7400 \times 10^3}} = \sqrt{\frac{8.03616 \times 10^{14}}{7400 \times 10^3}} \]
4. Calculate \( 7400 \times 10^3 = 7.4 \times 10^6 \):
\[ V_e = \sqrt{\frac{8.03616 \times 10^{14}}{7.4 \times 10^6}} = \sqrt{1.085 \times 10^{8}} \approx 1041.57 \, \text{m/s} \]

### Step 6: Convert to km/s
Finally, converting \( V_e \) from m/s to km/s:
\[ V_e \approx 1.04157 \, \text{km/s} \approx 10.42 \, \text{km/s} \]

### Final Answer
Thus, the escape speed of the atmospheric particle at a height of 1000 km above the Earth's surface is approximately:
\[ \boxed{10.42 \, \text{km/s}} \]