The orbital speed of an artificial satellite very close to the surface of the earth is `V_(o)` . Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth is

Options

A. `4 V_(o)`
B. `2 V_(o)`
C. `0.5 V_(o)`
D. `4 V_(o)`

Answer

To solve the problem, we need to determine the orbital speed of an artificial satellite at a height equal to three times the radius of the Earth, given that the orbital speed very close to the surface of the Earth is \( V_0 \).

### Step-by-Step Solution:

1. **Understanding Orbital Speed Near Earth's Surface**:
The orbital speed \( V \) of a satellite in a circular orbit is given by the formula:
\[ V = \sqrt{\frac{GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth. When the satellite is very close to the surface of the Earth, we can denote this speed as \( V_0 \):
\[ V_0 = \sqrt{\frac{GM}{R}} \]

2. **Finding the Orbital Radius at Height**:
Now, we need to find the orbital speed of a satellite at a height equal to three times the radius of the Earth. The total distance from the center of the Earth to this satellite is:
\[ R + 3R = 4R \]
where \( R \) is the radius of the Earth.

3. **Calculating the Orbital Speed at Height**:
The orbital speed \( V_2 \) at this new distance (4R) can be calculated using the same formula:
\[ V_2 = \sqrt{\frac{GM}{4R}} \]

4. **Relating \( V_2 \) to \( V_0 \)**:
We can express \( V_2 \) in terms of \( V_0 \):
\[ V_2 = \sqrt{\frac{GM}{4R}} = \frac{1}{2} \sqrt{\frac{GM}{R}} = \frac{1}{2} V_0 \]

5. **Final Answer**:
Therefore, the orbital speed of the satellite at a height equal to three times the radius of the Earth is:
\[ V_2 = \frac{1}{2} V_0 \]

### Conclusion:
The orbital speed of the satellite at a height equal to three times the radius of the Earth is half of the orbital speed very close to the Earth's surface.