Height Above Earth Surface Where Acceleration Due to Gravity Decreases

To find the height $h$ above the Earth's surface where the acceleration due to gravity becomes $\frac{g}{9}$, we use the formula for acceleration due to gravity at a height $h$:

$g' = \frac{GM}{(R + h)^2}$

where:

• $G$ is the universal gravitational constant,
• $M$ is the mass of the Earth,
• $R$ is the radius of the Earth,
• $g = \frac{GM}{R^2}$ is the acceleration due to gravity at the surface of the Earth.

We can express $g'$ in terms of $g$:

$g' = g \left( \frac{R}{R + h} \right)^2$

According to the question, $g' = \frac{g}{9}$. Substituting this value into the equation:

$\frac{g}{9} = g \left( \frac{R}{R + h} \right)^2$

Dividing both sides by $g$:

$\frac{1}{9} = \left( \frac{R}{R + h} \right)^2$

Taking the positive square root on both sides:

$\frac{1}{3} = \frac{R}{R + h}$

Cross-multiplying to solve for $h$:

$R + h = 3R$ $h = 3R - R$ $h = 2R$

Thus, the required height above the Earth's surface is $2R$.

Hence, the correct option is C.