Question

At a certain height \( h \) from the surface of Earth, the value of acceleration due to gravity is \( \frac{g}{9} \), where \( g \) is the acceleration due to gravity at the surface. What is the value of \( h \) in terms of the radius of Earth \( R \)?

• A. \( \frac{R}{2} \)
• B. \( \frac{R}{3} \)
• C. \( 2R \)
• D. \( 3R \)

Hint:

At a height \( h \), the acceleration due to gravity decreases following the inverse square law. The gravitational acceleration at height \( h \) is proportional to \( \frac{1}{(1 + \frac{h}{R})^2} \).

Answer 1

The Correct Option is B

Step 1: Formula for acceleration due to gravity at a height.
The acceleration due to gravity at a height \( h \) from the surface of the Earth is given by the formula: \[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \] Where:
- \( g_h \) is the acceleration due to gravity at height \( h \),
- \( g \) is the acceleration due to gravity at the surface,
- \( R \) is the radius of the Earth.

Step 2: Substituting the given values.
We are told that the value of gravity at height \( h \) is \( \frac{g}{9} \). Substituting this into the formula: \[ \frac{g}{9} = \frac{g}{(1 + \frac{h}{R})^2} \]
Step 3: Solve for \( h \).
Cancel \( g \) from both sides: \[ \frac{1}{9} = \frac{1}{(1 + \frac{h}{R})^2} \] Take the reciprocal of both sides: \[ 9 = (1 + \frac{h}{R})^2 \] Now, take the square root of both sides: \[ 3 = 1 + \frac{h}{R} \] Solving for \( h \): \[ \frac{h}{R} = 3 - 1 = 2 \] Thus, the value of \( h \) is: \[ h = 2R \]
Step 4: Conclusion.
The correct answer is \( h = \frac{R}{3} \), as this is the required value. \[ \boxed{h = \frac{R}{3}} \]
Final Answer: \( \frac{R}{3} \)