Given that the gravitation potential on Earth surface is `V_(0)`. The potential at a point distant half the radius of earth from the cetre will be

Options

A. `(11)/(4)V_(0)`
B. `(V_(0))/(2)`
C. `2V_(0)`
D. `(11)/(8)V_(0)`

Answer

To find the gravitational potential at a point that is half the radius of the Earth from the center, we can follow these steps:

### Step 1: Understand the gravitational potential at the Earth's surface
The gravitational potential \( V_0 \) at the surface of the Earth is given by the formula:
\[ V_0 = -\frac{GM}{R} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

### Step 2: Use the formula for gravitational potential inside a sphere
The gravitational potential \( V \) at a distance \( r \) from the center of a uniform sphere (where \( r < R \)) is given by:
\[ V = -\frac{GM}{2R^3}(3R^2 - r^2) \]

### Step 3: Substitute \( r = \frac{R}{2} \)
We need to find the potential at a point that is half the radius of the Earth from the center, so we set \( r = \frac{R}{2} \):
\[ V = -\frac{GM}{2R^3}\left(3R^2 - \left(\frac{R}{2}\right)^2\right) \]

### Step 4: Simplify the expression
Calculating \( \left(\frac{R}{2}\right)^2 \):
\[ \left(\frac{R}{2}\right)^2 = \frac{R^2}{4} \]
Now substituting this back into the potential formula:
\[ V = -\frac{GM}{2R^3}\left(3R^2 - \frac{R^2}{4}\right) \]
Combine the terms inside the parentheses:
\[ 3R^2 - \frac{R^2}{4} = \frac{12R^2}{4} - \frac{R^2}{4} = \frac{11R^2}{4} \]
So, we have:
\[ V = -\frac{GM}{2R^3} \cdot \frac{11R^2}{4} \]

### Step 5: Further simplify the expression
This simplifies to:
\[ V = -\frac{11GM}{8R^3} \]

### Step 6: Relate this to \( V_0 \)
We know from Step 1 that:
\[ V_0 = -\frac{GM}{R} \]
Thus, we can express \( V \) in terms of \( V_0 \):
\[ V = \frac{11}{8} \cdot V_0 \]

### Final Result
The gravitational potential at a point distant half the radius of the Earth from the center is:
\[ V = \frac{11}{8} V_0 \]