To determine the height at which the acceleration due to gravity is 1/9th of its value at the Earth's surface, we can use the formula for gravitational acceleration as a function of height.

1. Gravitational Acceleration Formula
The acceleration due to gravity (g) at a height (h) above the Earth's surface is given by:
\[ g_h = \frac{g_0}{(1 + \frac{h}{R})^2} \]
Where:
- \( g_h \) = acceleration due to gravity at height \( h \)
- \( g_0 \) = acceleration due to gravity at the Earth's surface (approximately \( 9.81 \, \text{m/s}^2 \))
- \( R \) = radius of the Earth (approximately \( 6.4 \times 10^6 \, \text{m} \))

2. Setting Up the Equation
To find the height where \( g_h = \frac{g_0}{9} \):
\[ \frac{g_0}{9} = \frac{g_0}{(1 + \frac{h}{R})^2} \]

3. Simplifying the Equation
By canceling \( g_0 \) from both sides:
\[ \frac{1}{9} = \frac{1}{(1 + \frac{h}{R})^2} \]
Taking the reciprocal:
\[ (1 + \frac{h}{R})^2 = 9 \]
Taking the square root:
\[ 1 + \frac{h}{R} = 3 \]

4. Solving for Height
Rearranging gives:
\[ \frac{h}{R} = 2 \]
Thus:
\[ h = 2R \]
Given \( R \approx 6.4 \times 10^6 \, \text{m} \):
\[ h \approx 2 \times 6.4 \times 10^6 = 12.8 \times 10^6 \, \text{m} \]
So, the height \( h \approx 12,800 \, \text{km} \) above the Earth's surface for \( g \) to equal \( \frac{g_0}{9} \).