If earth has uniform density, and radius 'R'. The value of acceleration due to gravity at distance d above the surface is same as the acceleration due to gravity at distance d below the surface. If `d=((sqrtx-1)/(2))R`, then find x.

Answer

To solve the problem, we need to find the value of \( x \) given that the acceleration due to gravity at a distance \( d \) above the surface of the Earth is equal to the acceleration due to gravity at a distance \( d \) below the surface. The distance \( d \) is given as:

\[ d = \frac{\sqrt{x} - 1}{2} R \]

where \( R \) is the radius of the Earth.

### Step-by-step Solution:

1. **Understanding the Acceleration due to Gravity:**
- The acceleration due to gravity at a height \( h \) above the surface of the Earth is given by:
\[ g_h = \frac{GM}{(R + h)^2} \]
- The acceleration due to gravity at a depth \( d \) below the surface is given by:
\[ g_d = \frac{GM'}{(R - d)^2} \]
- Here, \( M' \) is the mass of the Earth enclosed within the radius \( R - d \).

2. **Using Uniform Density:**
- Since the Earth has uniform density, the mass \( M' \) at depth \( d \) can be expressed as:
\[ M' = \rho \cdot \frac{4}{3} \pi (R - d)^3 \]
- The total mass \( M \) of the Earth can be expressed as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

3. **Setting Up the Equations:**
- The condition states that \( g_h = g_d \):
\[ \frac{GM}{(R + d)^2} = \frac{GM'}{(R - d)^2} \]
- Cancelling \( G \) from both sides:
\[ \frac{M}{(R + d)^2} = \frac{M'}{(R - d)^2} \]

4. **Substituting for \( M' \):**
- Substitute \( M' \) in the equation:
\[ \frac{M}{(R + d)^2} = \frac{\rho \cdot \frac{4}{3} \pi (R - d)^3}{(R - d)^2} \]
- Simplifying gives:
\[ \frac{M}{(R + d)^2} = \rho \cdot \frac{4}{3} \pi (R - d) \]

5. **Expressing \( M \) in terms of \( \rho \):**
- From the total mass of the Earth:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]
- Substitute this into the equation:
\[ \frac{\rho \cdot \frac{4}{3} \pi R^3}{(R + d)^2} = \rho \cdot \frac{4}{3} \pi (R - d) \]
- Cancelling \( \rho \cdot \frac{4}{3} \pi \) gives:
\[ \frac{R^3}{(R + d)^2} = (R - d) \]

6. **Cross Multiplying:**
- Cross multiplying gives:
\[ R^3 = (R - d)(R + d)^2 \]

7. **Expanding the Right Side:**
- Expanding the right side:
\[ R^3 = (R - d)(R^2 + 2Rd + d^2) \]
- This expands to:
\[ R^3 = R^3 + 2R^2d + Rd^2 - dR^2 - 2d^2R - d^3 \]
- Simplifying gives:
\[ 0 = 2R^2d + Rd^2 - dR^2 - 2d^2R - d^3 \]

8. **Rearranging the Terms:**
- Rearranging leads to:
\[ d^3 + (R - 2R^2)d + 2R^2d = 0 \]

9. **Substituting for \( d \):**
- Substitute \( d = \frac{\sqrt{x} - 1}{2} R \):
\[ \left(\frac{\sqrt{x} - 1}{2} R\right)^3 + (R - 2R^2)\left(\frac{\sqrt{x} - 1}{2} R\right) + 2R^2\left(\frac{\sqrt{x} - 1}{2} R\right) = 0 \]

10. **Solving for \( x \):**
- After simplifying and solving, we find that \( x = 5 \).

### Final Answer:
\[ x = 5 \]