At a depth `h_(1)=(R)/(2)` from the surface of the earth acceleration due to gravity is `g_(1).` It’s value changes by `Deltag_(1)` when one moves down further by 1 km. At a height `h_(2)` above the urface of the earth acceleration due to gravity is `g_(9).` It’s value changes by `Deltag_(2)`when one moves up further by 1 km. If `Deltag_(1)=Deltag_(2)` find `h_(2).` Assume the earth to be a uniform phere of radius R.

Answer

To solve the problem, we need to find the height \( h_2 \) above the surface of the Earth where the change in acceleration due to gravity \( \Delta g_2 \) is equal to the change in acceleration due to gravity \( \Delta g_1 \) at a depth \( h_1 = \frac{R}{2} \).

### Step-by-step Solution:

1. **Understanding the Depth and Acceleration Due to Gravity**:
At a depth \( h_1 = \frac{R}{2} \) from the surface of the Earth, the acceleration due to gravity \( g_1 \) is given by the formula:
\[ g_1 = g \left(1 - \frac{h_1}{R}\right) = g \left(1 - \frac{R/2}{R}\right) = g \left(1 - \frac{1}{2}\right) = \frac{g}{2} \]

2. **Change in Gravity at Depth**:
The change in acceleration due to gravity \( \Delta g_1 \) when moving down further by 1 km (or \( 1000 \, \text{m} \)) can be derived from:
\[ \Delta g_1 = \frac{dg_1}{dh_1} \cdot \Delta h_1 \]
The derivative \( \frac{dg_1}{dh_1} \) can be calculated as:
\[ \frac{dg_1}{dh_1} = -\frac{g}{R} \]
Therefore,
\[ \Delta g_1 = -\frac{g}{R} \cdot 1000 \]

3. **Understanding the Height and Acceleration Due to Gravity**:
At a height \( h_2 \) above the surface of the Earth, the acceleration due to gravity \( g_2 \) is given by:
\[ g_2 = \frac{g R^2}{(R + h_2)^2} \]

4. **Change in Gravity at Height**:
The change in acceleration due to gravity \( \Delta g_2 \) when moving up further by 1 km can be expressed as:
\[ \Delta g_2 = \frac{dg_2}{dh_2} \cdot \Delta h_2 \]
The derivative \( \frac{dg_2}{dh_2} \) can be calculated as:
\[ \frac{dg_2}{dh_2} = -\frac{2gR^2}{(R + h_2)^3} \]
Therefore,
\[ \Delta g_2 = -\frac{2gR^2}{(R + h_2)^3} \cdot 1000 \]

5. **Setting the Changes Equal**:
Since \( \Delta g_1 = \Delta g_2 \), we can set the two expressions equal to each other:
\[ -\frac{g}{R} \cdot 1000 = -\frac{2gR^2}{(R + h_2)^3} \cdot 1000 \]
Canceling \( -1000g \) from both sides gives:
\[ \frac{1}{R} = \frac{2R^2}{(R + h_2)^3} \]

6. **Cross Multiplying and Solving for \( h_2 \)**:
Cross multiplying yields:
\[ (R + h_2)^3 = 2R^3 \]
Taking the cube root of both sides:
\[ R + h_2 = \sqrt[3]{2} R \]
Thus,
\[ h_2 = \sqrt[3]{2} R - R = R(\sqrt[3]{2} - 1) \]

### Final Answer:
\[ h_2 = R(\sqrt[3]{2} - 1) \]