Question

The height in terms of radius of the earth ($R$), at which the acceleration due to gravity becomes $g/9$, where $g$ is acceleration due to gravity on earth's surface, is

• A. $\sqrt{3}R$
• B. $2\sqrt{2}R$
• C. $2R$
• D. $\frac{4}{9} R$

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The acceleration due to gravity decreases as we move away from the surface of the Earth. The value of gravity $g'$ at an altitude $h$ follows the inverse-square law with respect to the distance from the center of the Earth.

Step 2: Key Formula or Approach:
Gravity at height $h$:
$g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$
where $R$ is the radius of the Earth.

Step 3: Detailed Explanation:
We are given that at height $h$, the acceleration due to gravity is $g' = \frac{g}{9}$.
Substitute this into the gravity altitude formula:
$\frac{g}{9} = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$
Cancel out $g$ from both sides:
$\frac{1}{9} = \frac{1}{\left(1 + \frac{h}{R}\right)^2}$
Take the positive square root of both sides (since altitude $h$ is positive):
$\frac{1}{3} = \frac{1}{1 + \frac{h}{R}}$
Rearrange to solve for $h$:
$1 + \frac{h}{R} = 3$
$\frac{h}{R} = 2$
$h = 2R$.

Step 4: Final Answer:
The height is $2R$.