A satellite is revolving around the earth. Ratio of its orbital speed and escape speed will be.

Options

A. `1/(sqrt2)`
B. `sqrt2`
C. `sqrt3`
D. `2 sqrt2`

Answer

To find the ratio of the orbital speed (V₀) of a satellite revolving around the Earth to the escape speed (Vₑ) from the Earth's gravitational field, we can follow these steps:

### Step 1: Define the orbital speed (V₀)
The orbital speed of a satellite in a circular orbit around the Earth is given by the formula:
\[ V₀ = \sqrt{\frac{GM}{r}} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( r \) is the distance from the center of the Earth to the satellite.

### Step 2: Define the escape speed (Vₑ)
The escape speed from the gravitational field of the Earth is given by the formula:
\[ Vₑ = \sqrt{\frac{2GM}{r}} \]

### Step 3: Calculate the ratio of orbital speed to escape speed
We need to find the ratio \( \frac{V₀}{Vₑ} \):
\[ \frac{V₀}{Vₑ} = \frac{\sqrt{\frac{GM}{r}}}{\sqrt{\frac{2GM}{r}}} \]

### Step 4: Simplify the ratio
This can be simplified as follows:
\[ \frac{V₀}{Vₑ} = \frac{\sqrt{GM/r}}{\sqrt{2GM/r}} = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Conclusion
Thus, the ratio of the orbital speed to the escape speed is:
\[ \frac{V₀}{Vₑ} = \frac{1}{\sqrt{2}} \]

### Final Answer
The correct option is \( \frac{1}{\sqrt{2}} \).

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