Question

The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)

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

Hint:

The acceleration due to gravity decreases with the square of the distance from the center of the Earth. Use \(g_h = g \left( \frac{R}{R + h} \right)^2\) for calculations involving heights above the Earth's surface.

Answer 1

The Correct Option is B

Step 1: Formula for acceleration due to gravity at a height \(h\) from Earth's surface.
The acceleration due to gravity at a height \(h\) is given by:\[g_h = g \left( \frac{R}{R + h} \right)^2,\]where: \(g_h\): Acceleration due to gravity at height \(h\), \(g\): Acceleration due to gravity on the surface of the Earth, \(R\): Radius of the Earth.Step 2: Setting \(g_h = \frac{g}{4}\).
Substitute \(g_h = \frac{g}{4}\) into the equation:\[\frac{g}{4} = g \left( \frac{R}{R + h} \right)^2.\]Step 3: Simplifying the equation.
Cancel \(g\) from both sides:\[\frac{1}{4} = \left( \frac{R}{R + h} \right)^2.\]Take the square root of both sides:\[\frac{1}{2} = \frac{R}{R + h}.\]Cross-multiply:\[R + h = 2R.\]Step 4: Solve for \(h\).
\[h = 2R - R = R.\]Step 5: Conclusion.
The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is:\[\boxed{R}.\]