A body hanging from a spring strethces it by 1cm at the earth's surface. How much will the same body stretch the spring at a place 1600km above the earth's surface? (Radius of the earth 6400km)

Options

A. 1.28 cm
B. 0.64 cm
C. 3.6 cm
D. 0.12 cm

Explanation

To solve the problem step by step, we will use the relationship between the stretching of the spring and the gravitational force acting on the body at different heights.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body stretches a spring by 1 cm at the Earth's surface.
- We need to find out how much the same body will stretch the spring when it is 1600 km above the Earth's surface.

2. **Identify Given Data**:
- Stretch at Earth's surface, \( x = 1 \, \text{cm} \)
- Height above Earth's surface, \( h = 1600 \, \text{km} \)
- Radius of the Earth, \( R_e = 6400 \, \text{km} \)

3. **Gravitational Force at Earth's Surface**:
- The weight of the body at the Earth's surface is given by \( mg \), where \( g \) is the acceleration due to gravity at the surface.

4. **Gravitational Force at Height \( h \)**:
- The acceleration due to gravity at height \( h \) can be calculated using the formula:
\[ g' = \frac{g R_e^2}{(R_e + h)^2} \]
- Here, \( g' \) is the acceleration due to gravity at the height \( h \).

5. **Setting Up the Ratio**:
- At equilibrium, the force exerted by the spring is equal to the weight of the body:
\[ mg = kx \quad \text{(at Earth's surface)} \]
\[ mg' = kx' \quad \text{(at height \( h \))} \]
- Dividing these two equations gives:
\[ \frac{g'}{g} = \frac{x'}{x} \]

6. **Substituting for \( g' \)**:
- Substitute \( g' \) into the ratio:
\[ \frac{g R_e^2}{(R_e + h)^2} \div g = \frac{x'}{x} \]
- This simplifies to:
\[ \frac{R_e^2}{(R_e + h)^2} = \frac{x'}{x} \]

7. **Calculating \( x' \)**:
- Rearranging gives:
\[ x' = x \cdot \frac{R_e^2}{(R_e + h)^2} \]
- Substitute the known values:
\[ x' = 1 \, \text{cm} \cdot \frac{(6400 \, \text{km})^2}{(6400 \, \text{km} + 1600 \, \text{km})^2} \]
- Calculate \( R_e + h = 6400 + 1600 = 8000 \, \text{km} \):
\[ x' = 1 \cdot \frac{6400^2}{8000^2} \]
- This simplifies to:
\[ x' = 1 \cdot \frac{40960000}{64000000} = \frac{16}{25} \, \text{cm} \]
- Therefore:
\[ x' = 0.64 \, \text{cm} \]

8. **Final Answer**:
- The body will stretch the spring by \( 0.64 \, \text{cm} \) at a height of 1600 km above the Earth's surface.