Question

At a height \( h \) above the Earth's surface, the acceleration due to gravity becomes \( \frac{g}{\sqrt{3}} \). What is the value of \( h \) in terms of the Earth's radius \( R \)?

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

Hint:

When working with gravitational problems involving height above the Earth's surface, remember to use the formula for gravity variation with height: \( g_h = \frac{g}{(1 + \frac{h}{R})^2} \).

Answer 1

The Correct Option is A

The acceleration due to gravity at a height \( h \) above the Earth's surface is defined by the equation: \[ g_h = \frac{g}{\left( 1 + \frac{h}{R} \right)^2} \]. Here, \( g \) represents the acceleration due to gravity at the Earth's surface, \( g_h \) denotes the acceleration due to gravity at height \( h \), and \( R \) is the Earth's radius. Given that the acceleration due to gravity at height \( h \) equals \( \frac{g}{\sqrt{3}} \), we can write: \[ \frac{g}{\sqrt{3}} = \frac{g}{\left( 1 + \frac{h}{R} \right)^2} \]. By canceling \( g \) from both sides: \[ \frac{1}{\sqrt{3}} = \frac{1}{\left( 1 + \frac{h}{R} \right)^2} \]. Taking the square root of both sides yields: \[ \frac{1}{\sqrt{3}} = \frac{1}{1 + \frac{h}{R}} \]. Solving for \( h \): \[ 1 + \frac{h}{R} = \sqrt{3} \] \[ \frac{h}{R} = \sqrt{3} - 1 \] \[ h = R(\sqrt{3} - 1) \] Consequently, the height \( h \) is \( R \). The correct selection is option (a).