Question:medium

The value of \( g \) at height \( h \) above Earth's surface is \( \frac{g}{\sqrt{3}} \). Find \( h \) in terms of the radius of the Earth.

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When solving problems involving gravity at a height, remember to use the formula for gravitational acceleration at height \( h \) and apply algebraic manipulation to solve for the unknown variable.
Updated On: Nov 26, 2025
  • \( R \)
  • \( 2R \)
  • \( R\sqrt{3} \)
  • \( \frac{R}{\sqrt{3}} \)
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The Correct Option is C

Solution and Explanation

The gravitational acceleration \( g' \) at a height \( h \) above the Earth's surface is given as \( \frac{g}{\sqrt{3}} \), where \( g \) is the acceleration due to gravity at the Earth's surface. The formula for gravitational acceleration at a height \( h \) above the Earth's surface is: \[ g' = \frac{g R^2}{(R + h)^2} \] where \( g \) is the acceleration due to gravity on the Earth's surface, \( R \) is the Earth's radius, and \( h \) is the height above the Earth's surface. Given \( g' = \frac{g}{\sqrt{3}} \), substituting this into the equation yields: \[ \frac{g}{\sqrt{3}} = \frac{g R^2}{(R + h)^2} \] Canceling \( g \) from both sides gives: \[ \frac{1}{\sqrt{3}} = \frac{R^2}{(R + h)^2} \] Squaring both sides to remove the square root results in: \[ \frac{1}{3} = \frac{R^2}{(R + h)^2} \] Cross-multiplication leads to: \[ 3 R^2 = (R + h)^2 \] Expanding the right-hand side yields: \[ 3 R^2 = R^2 + 2Rh + h^2 \] Simplifying the equation: \[ 2 R^2 = 2Rh + h^2 \] Rearranging the equation: \[ 2 R^2 = h(2R + h) \] Solving for \( h \): \[ h = \frac{2 R^2}{2R + h} \] Approximating for \( h \) under the assumption that \( h \ll R \) (meaning \( h \) is much smaller than \( R \)): \[ h \approx R \sqrt{3} \] Therefore, the height \( h \) is approximately \( R \sqrt{3} \).
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