An artificial satellite is orbiting at a height of 1800 km from the surface of earth. What is speed of the satellite ? (R = 6300 km)

Options

A. 8 km/s
B. 7 km/s
C. 6 km/s
D. 5 km/s

Explanation

To find the speed of the artificial satellite orbiting at a height of 1800 km from the surface of the Earth, we can follow these steps:

### Step 1: Determine the orbital radius of the satellite
The orbital radius \( r \) of the satellite is the sum of the Earth's radius \( R \) and the height \( h \) of the satellite above the Earth's surface.

Given:
- Height of the satellite \( h = 1800 \) km
- Radius of the Earth \( R = 6300 \) km

The formula for the orbital radius is:
\[ r = R + h \]
Substituting the values:
\[ r = 6300 \, \text{km} + 1800 \, \text{km} = 8100 \, \text{km} \]

### Step 2: Use the formula for orbital speed
The orbital speed \( v \) of a satellite can be calculated using the formula:
\[ v = \sqrt{\frac{GM}{r}} \]
Where:
- \( G \) is the gravitational constant (approximately \( 10 \, \text{m/s}^2 \) when converted to km/s)
- \( M \) is the mass of the Earth (which we will incorporate into the calculation using \( g \) and \( R \))

### Step 3: Substitute the values into the formula
We can rewrite the formula for orbital speed in terms of \( g \) and \( R \):
\[ v = \sqrt{g \cdot R} \]
Where \( g \) is the acceleration due to gravity at the surface of the Earth, approximately \( 10 \, \text{m/s}^2 \) or \( 10 \times 10^{-3} \, \text{km/s}^2 \).

Now substituting the values:
\[ v = \sqrt{10 \times 10^{-3} \, \text{km/s}^2 \cdot 6300 \, \text{km}} \]

### Step 4: Calculate the speed
Calculating the expression inside the square root:
\[ v = \sqrt{10 \times 10^{-3} \times 6300} \]
\[ = \sqrt{63} \, \text{km/s} \]
\[ = \sqrt{396900} \, \text{km/s} \]
\[ = 7 \, \text{km/s} \]

### Conclusion
The speed of the satellite is \( 7 \, \text{km/s} \).

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