The escape velocity of a body from earth's surface is `v_e`. The escape velocity of the same body from a height equal to `7R` from earth's surface will be

Options

A. `v_(e)//sqrt(2)`
B. `v_(e)//2`
C. `v_(e)//2sqrt(2)`
D. `v_(e)//4`

Explanation

To find the escape velocity of a body from a height equal to \(7R\) above the Earth's surface, we can follow these steps:

### Step 1: Understand Escape Velocity
Escape velocity is the minimum velocity required for an object to break free from the gravitational pull of a celestial body without any further propulsion. The formula for escape velocity \(v_e\) from a distance \(r\) from the center of the Earth is given by:

\[ v_e = \sqrt{\frac{2GM}{r}} \]

where \(G\) is the gravitational constant and \(M\) is the mass of the Earth.

### Step 2: Determine the Distance from the Center of the Earth
1. The radius of the Earth is denoted as \(R\).
2. When the body is at the Earth's surface, the distance from the center of the Earth is \(R\).
3. When the body is at a height of \(7R\) above the Earth's surface, the total distance from the center of the Earth becomes:

\[ r = R + 7R = 8R \]

### Step 3: Calculate Escape Velocity at Height \(7R\)
Using the escape velocity formula at the height \(7R\):

\[ v_{e}' = \sqrt{\frac{2GM}{8R}} \]

### Step 4: Relate it to the Escape Velocity at the Surface
We already know the escape velocity at the surface:

\[ v_e = \sqrt{\frac{2GM}{R}} \]

Now, we can express \(v_{e}'\) in terms of \(v_e\):

\[ v_{e}' = \sqrt{\frac{2GM}{8R}} = \sqrt{\frac{1}{8}} \cdot \sqrt{\frac{2GM}{R}} = \frac{1}{\sqrt{8}} v_e = \frac{1}{2\sqrt{2}} v_e \]

### Step 5: Final Result
Thus, the escape velocity from a height equal to \(7R\) from the Earth's surface is:

\[ v_{e}' = \frac{1}{2\sqrt{2}} v_e \]

### Conclusion
The escape velocity of the body from a height equal to \(7R\) from the Earth's surface is \(\frac{1}{2\sqrt{2}} v_e\).

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