A body stretches a spring by a particular length at the earth's surface at equator. At what height above the south pole will it stretch the same spring by the same length ? Assume the earth to be spherical.

Explanation

To solve the problem, we need to find the height \( h \) above the South Pole where a body stretches a spring by the same length as it does at the equator. We will use the concepts of gravitational acceleration and the effects of Earth's rotation.

### Step-by-Step Solution:

1. **Understand the Forces at the Equator:**
- At the equator, the effective gravitational force acting on the body is given by:
\[ g' = g - \omega^2 R \]
- Here, \( g \) is the acceleration due to gravity at the surface, \( \omega \) is the angular velocity of the Earth, and \( R \) is the radius of the Earth.

2. **Set Up the Equation for Spring Stretch at the Equator:**
- The spring stretches by a length \( x \) at the equator due to the effective gravitational force:
\[ kx = mg' \]
- Rearranging gives:
\[ mg' = kx \]

3. **Understand the Forces at Height \( h \) Above the South Pole:**
- At height \( h \) above the South Pole, the gravitational acceleration is:
\[ g_h = g \left(1 - \frac{2h}{R}\right) \]
- The spring will stretch by the same length \( x \):
\[ kx = mg_h \]

4. **Set Up the Equation for Spring Stretch at Height \( h \):**
- Rearranging gives:
\[ mg_h = kx \]

5. **Equate the Two Forces:**
- Since the spring stretches the same amount \( x \) in both cases, we can equate the two expressions:
\[ mg' = mg_h \]
- Substituting the expressions for \( g' \) and \( g_h \):
\[ m(g - \omega^2 R) = mg \left(1 - \frac{2h}{R}\right) \]

6. **Cancel the Mass \( m \):**
- Since \( m \) is common to both sides, we can cancel it out:
\[ g - \omega^2 R = g \left(1 - \frac{2h}{R}\right) \]

7. **Rearranging the Equation:**
- Expanding and rearranging gives:
\[ g - \omega^2 R = g - \frac{2gh}{R} \]
- Simplifying leads to:
\[ \frac{2gh}{R} = \omega^2 R \]

8. **Solve for Height \( h \):**
- Rearranging for \( h \):
\[ h = \frac{\omega^2 R^2}{2g} \]

9. **Substituting Values:**
- The angular velocity \( \omega \) of the Earth is:
\[ \omega = \frac{2\pi}{24 \times 3600} \text{ rad/s} \]
- The radius of the Earth \( R \) is approximately \( 6400 \times 10^3 \) meters.
- The acceleration due to gravity \( g \) is approximately \( 9.8 \, \text{m/s}^2 \).

10. **Calculate Height \( h \):**
- Plugging in the values:
\[ h = \frac{\left(\frac{2\pi}{24 \times 3600}\right)^2 \cdot (6400 \times 10^3)^2}{2 \cdot 9.8} \]
- After calculating, we find:
\[ h \approx 11 \text{ km} \]

### Final Answer:
The height above the South Pole where the body will stretch the spring by the same length is approximately **11 kilometers**.