Assuming earth to be spherical, at what height above the north pole, value of g is same as that on earth's surface at equator?

Explanation

To solve the problem of finding the height above the North Pole where the value of g is the same as that on Earth's surface at the equator, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Acceleration due to Gravity at the Equator**:
The acceleration due to gravity at the equator (g_eq) can be expressed as:
\[ g_{eq} = g - \omega^2 r \]
where:
- \( g \) is the acceleration due to gravity at the surface of the Earth,
- \( \omega \) is the angular velocity of the Earth,
- \( r \) is the radius of the Earth.

2. **Acceleration due to Gravity at Height h**:
The acceleration due to gravity at a height \( h \) above the surface of the Earth can be expressed as:
\[ g_h = g \left(1 - \frac{2h}{r}\right) \]

3. **Set the Two Expressions Equal**:
We need to find the height \( h \) where:
\[ g_h = g_{eq} \]
Thus, we set the equations equal:
\[ g \left(1 - \frac{2h}{r}\right) = g - \omega^2 r \]

4. **Simplify the Equation**:
Rearranging the equation gives:
\[ g - \frac{2hg}{r} = g - \omega^2 r \]
Cancelling \( g \) from both sides:
\[ -\frac{2hg}{r} = -\omega^2 r \]
This simplifies to:
\[ \frac{2hg}{r} = \omega^2 r \]

5. **Solve for h**:
Rearranging for \( h \):
\[ h = \frac{\omega^2 r^2}{2g} \]

6. **Substitute Values**:
We need to substitute the known values:
- The radius of the Earth \( r \approx 6400 \times 10^3 \) m,
- The angular velocity \( \omega = \frac{2\pi}{T} \), where \( T = 24 \times 3600 \) seconds (the time period of rotation of the Earth).

First, calculate \( \omega \):
\[ T = 24 \times 3600 = 86400 \text{ seconds} \]
\[ \omega = \frac{2\pi}{86400} \text{ rad/s} \]

Now, calculate \( \omega^2 \):
\[ \omega^2 = \left(\frac{2\pi}{86400}\right)^2 \]

7. **Final Calculation for h**:
Substitute \( \omega^2 \) and \( r \) into the equation for \( h \):
\[ h = \frac{\left(\frac{2\pi}{86400}\right)^2 \cdot (6400 \times 10^3)^2}{2g} \]
Using \( g \approx 9.81 \, \text{m/s}^2 \):
\[ h = \frac{(4\pi^2) \cdot (6400 \times 10^3)^2}{2 \cdot 9.81 \cdot (86400)^2} \]

8. **Calculate the Numerical Value**:
After performing the calculations, we find:
\[ h \approx 10 \text{ km} \]

### Conclusion:
The height above the North Pole where the value of g is the same as that on Earth's surface at the equator is approximately **10 kilometers**.