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)`

Answer

To calculate the escape speed of an atmospheric particle that is 1000 km above the Earth's surface, we will use the formula for escape speed:

\[ v = \sqrt{\frac{2GM}{r + h}} \]

Where:
- \( G \) is the universal gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) is the mass of the Earth, approximately \( 5.97 \times 10^{24} \, \text{kg} \)
- \( r \) is the radius of the Earth
- \( h \) is the height above the Earth's surface

### Step 1: Convert all units to meters
- Radius of the Earth, \( r = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} = 6.4 \times 10^6 \, \text{m} \)
- Height above the Earth's surface, \( h = 1000 \, \text{km} = 1000 \times 10^3 \, \text{m} = 1 \times 10^6 \, \text{m} \)

### Step 2: Calculate the total distance from the center of the Earth
\[ r + h = 6.4 \times 10^6 \, \text{m} + 1 \times 10^6 \, \text{m} = 7.4 \times 10^6 \, \text{m} \]

### Step 3: Substitute values into the escape speed formula
Now we substitute \( G \), \( M \), and \( r + h \) into the escape speed formula:

\[ v = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{7.4 \times 10^6}} \]

### Step 4: Calculate the numerator
First, calculate \( 2GM \):

\[ 2GM = 2 \times (6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \approx 7.94 \times 10^{14} \, \text{m}^3/\text{s}^2 \]

### Step 5: Calculate the escape speed
Now, we can calculate \( v \):

\[ v = \sqrt{\frac{7.94 \times 10^{14}}{7.4 \times 10^6}} \approx \sqrt{1.07 \times 10^8} \approx 10340 \, \text{m/s} \]

### Final Answer
The escape speed of an atmospheric particle which is 1000 km above the Earth's surface is approximately \( 10,340 \, \text{m/s} \).

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