A spring balance extends by 2 cm due to a mass on surface of Earth. What will be the extension at a height `2R_(e )` from earth’s surface ?

Options

A. `(1)/(2)cm`
B. `(1)/(4)cm`
C. `(2)/(9)cm`
D. `(1)/(9)cm`

Explanation

To solve the problem, we need to find the extension of a spring balance when a mass is placed on it at a height of \(2R_e\) from the Earth's surface, given that the extension at the surface is \(2 \, \text{cm}\).

### Step-by-step Solution:

1. **Understanding the Extension at the Surface**:
- At the surface of the Earth, the extension \(x\) of the spring balance due to a mass \(m\) is given by:
\[ x = \frac{mg}{k} \]
- Here, \(g\) is the acceleration due to gravity at the surface of the Earth, and \(k\) is the spring constant.

2. **Given Information**:
- The extension at the surface is \(x = 2 \, \text{cm}\).
- Therefore, we can write:
\[ 2 = \frac{mg}{k} \quad \text{(1)} \]

3. **Finding the Acceleration Due to Gravity at Height \(2R_e\)**:
- The formula for the acceleration due to gravity at a height \(h\) from the surface of the Earth is:
\[ g' = \frac{GM}{(R_e + h)^2} \]
- For \(h = 2R_e\), we have:
\[ g' = \frac{GM}{(R_e + 2R_e)^2} = \frac{GM}{(3R_e)^2} = \frac{GM}{9R_e^2} \]
- Since \(g = \frac{GM}{R_e^2}\), we can relate \(g'\) to \(g\):
\[ g' = \frac{g}{9} \quad \text{(2)} \]

4. **Finding the Extension at Height \(2R_e\)**:
- At height \(2R_e\), the new extension \(x'\) can be expressed as:
\[ x' = \frac{mg'}{k} \]
- Substituting \(g'\) from equation (2):
\[ x' = \frac{m \cdot \frac{g}{9}}{k} = \frac{mg}{9k} \]

5. **Relating \(x'\) to the Original Extension**:
- From equation (1), we know that \(x = \frac{mg}{k}\). Therefore:
\[ x' = \frac{1}{9} \cdot \frac{mg}{k} = \frac{1}{9} \cdot 2 \, \text{cm} = \frac{2}{9} \, \text{cm} \]

6. **Final Answer**:
- The extension of the spring balance at a height of \(2R_e\) from the Earth's surface is:
\[ x' = \frac{2}{9} \, \text{cm} \]