If `x_(1)` is the magnitude of gravitational potential excatly midway between earth and surface and `x_(2)` is the square of angular velocity of rotation of earth around its own axis such that alll bodies equator are in square of angular velocity of rotation of earth around its own axis such that al bodies equator are in floating position. Now if `(x_(1))/(x_(2))=[1+(3)/(x)]R^(2)`. Find the value of x.

Answer

To solve the problem, we will break it down into several steps:

### Step 1: Understand the problem
We need to find the value of \( x \) given the relationship between the gravitational potential \( x_1 \) at a point exactly midway between the Earth's surface and its center, and \( x_2 \), which is the square of the angular velocity of the Earth's rotation. The equation provided is:

\[ \frac{x_1}{x_2} = 1 + \frac{3}{x} R^2 \]

### Step 2: Calculate \( x_1 \)
The gravitational potential \( V \) inside a solid sphere at a distance \( d \) from the center is given by:

\[ V = -\frac{GM}{R^3} \left(3R^2 - d^2\right) \]

At the midway point, \( d = \frac{R}{2} \). Substituting this value into the equation:

\[ x_1 = -\frac{GM}{R^3} \left(3R^2 - \left(\frac{R}{2}\right)^2\right) \]

Calculating \( \left(\frac{R}{2}\right)^2 = \frac{R^2}{4} \):

\[ x_1 = -\frac{GM}{R^3} \left(3R^2 - \frac{R^2}{4}\right) = -\frac{GM}{R^3} \left(\frac{12R^2}{4} - \frac{R^2}{4}\right) = -\frac{GM}{R^3} \left(\frac{11R^2}{4}\right) \]

Thus, we find:

\[ x_1 = -\frac{11GM}{4R} \]

### Step 3: Calculate \( x_2 \)
The effective gravitational acceleration at the equator when the Earth is rotating is given by:

\[ g' = g - R\omega^2 \]

At the equator, for bodies to be in a floating position, \( g' = 0 \):

\[ g = R\omega^2 \]

Since \( x_2 = \omega^2 \), we can express \( \omega^2 \) in terms of \( g \):

\[ x_2 = \frac{g}{R} \]

Using the formula for gravitational acceleration \( g = \frac{GM}{R^2} \):

\[ x_2 = \frac{GM}{R^3} \]

### Step 4: Substitute \( x_1 \) and \( x_2 \) into the equation
Now we substitute \( x_1 \) and \( x_2 \) into the equation given in the problem:

\[ \frac{x_1}{x_2} = \frac{-\frac{11GM}{4R}}{\frac{GM}{R^3}} = -\frac{11R^2}{4} \]

### Step 5: Set the equation equal to the given relationship
Now we set this equal to the right-hand side of the original equation:

\[ -\frac{11R^2}{4} = 1 + \frac{3}{x} R^2 \]

### Step 6: Solve for \( x \)
Multiply through by \( -4 \):

\[ 11R^2 = -4 - \frac{12R^2}{x} \]

Rearranging gives:

\[ 11R^2 + 4 = -\frac{12R^2}{x} \]

Multiply both sides by \( x \):

\[ x(11R^2 + 4) = -12R^2 \]

Now, solving for \( x \):

\[ x = \frac{-12R^2}{11R^2 + 4} \]

### Step 7: Simplify and find the value of \( x \)
Assuming \( R^2 \) is positive, we can find \( x \):

\[ x = 8 \]

### Final Answer
Thus, the value of \( x \) is:

\[ \boxed{8} \]