An artificial satellite is orbiting at a height of 1800 km from the earth's surface. The earth's radius is 6300 km and `g=10m/s^(2)` on its surface. What is the radial acceleration of the satellite?

Options

A. `6m//s^(2)`
B. `7m//s^(2)`
C. `8m//s^(2)`
D. `9m//s^(2)`

Explanation

To find the radial acceleration of the satellite orbiting the Earth at a height of 1800 km, we can follow these steps:

### Step 1: Identify the given data
- Height of the satellite (H) = 1800 km = 1800 × 10^3 m = 1.8 × 10^6 m
- Radius of the Earth (R) = 6300 km = 6300 × 10^3 m = 6.3 × 10^6 m
- Acceleration due to gravity at the Earth's surface (g) = 10 m/s²

### Step 2: Calculate the total distance from the center of the Earth to the satellite
The total distance (r) from the center of the Earth to the satellite is the sum of the Earth's radius and the height of the satellite:
\[ r = R + H = 6.3 \times 10^6 \, \text{m} + 1.8 \times 10^6 \, \text{m} = 8.1 \times 10^6 \, \text{m} \]

### Step 3: Use the formula for gravitational acceleration at a distance r from the center of the Earth
The gravitational acceleration (g') at distance r from the center of the Earth can be calculated using the formula:
\[ g' = \frac{g \cdot R^2}{r^2} \]
Substituting the known values:
\[ g' = \frac{10 \, \text{m/s}^2 \cdot (6.3 \times 10^6 \, \text{m})^2}{(8.1 \times 10^6 \, \text{m})^2} \]

### Step 4: Calculate the values
Calculating \( R^2 \) and \( r^2 \):
- \( R^2 = (6.3 \times 10^6)^2 = 39.69 \times 10^{12} \, \text{m}^2 \)
- \( r^2 = (8.1 \times 10^6)^2 = 65.61 \times 10^{12} \, \text{m}^2 \)

Now substituting these values into the equation for g':
\[ g' = \frac{10 \cdot 39.69 \times 10^{12}}{65.61 \times 10^{12}} \]
\[ g' = \frac{396.9}{65.61} \approx 6.05 \, \text{m/s}^2 \]

### Step 5: Conclusion
The radial (centripetal) acceleration of the satellite is approximately:
\[ a = g' \approx 6.05 \, \text{m/s}^2 \]