The escape velocity of a body from the surface of the earth is `V_(1)` and from an altitude equal to twice the radius of the earth, is, `V_(2)`. Then

Options

A. `V_(1)=V_(2)`
B. `V_(1)=7V_(2)`
C. `V_(1)=sqrt(3)V_(2)`
D. `V_(1)=sqrt(2)V_(2)`

Explanation

To find the relationship between the escape velocities \( V_1 \) (from the surface of the Earth) and \( V_2 \) (from an altitude equal to twice the radius of the Earth), we can follow these steps:

### Step 1: Write the formula for escape velocity
The escape velocity \( V \) from a distance \( r \) from the center of the Earth is given by the formula:
\[ V = \sqrt{\frac{2GM}{r}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

### Step 2: Calculate \( V_1 \)
At the surface of the Earth, the distance \( r \) is equal to the radius of the Earth \( R \). Therefore, the escape velocity \( V_1 \) is:
\[ V_1 = \sqrt{\frac{2GM}{R}} \]

### Step 3: Calculate \( V_2 \)
When the body is at an altitude equal to twice the radius of the Earth, the total distance from the center of the Earth becomes:
\[ r = R + 2R = 3R \]
Now, we can calculate the escape velocity \( V_2 \) at this distance:
\[ V_2 = \sqrt{\frac{2GM}{3R}} \]

### Step 4: Relate \( V_2 \) to \( V_1 \)
Now, we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = \sqrt{\frac{2GM}{3R}} = \frac{1}{\sqrt{3}} \sqrt{\frac{2GM}{R}} = \frac{1}{\sqrt{3}} V_1 \]

### Step 5: Conclusion
Thus, we find the relationship between \( V_1 \) and \( V_2 \):
\[ V_2 = \frac{1}{\sqrt{3}} V_1 \quad \text{or} \quad V_1 = \sqrt{3} V_2 \]

### Final Answer
The relationship between the escape velocities is:
\[ V_1 = \sqrt{3} V_2 \]

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