at what height above the surface of the earth does the acceleration due to gravity become one - fourth of…

Explanation:

Step1: Recall gravitational - acceleration formula

The acceleration due to gravity at the surface of the Earth is given by $g=\frac{GM}{R^{2}}$, and at a height $h$ above the Earth's surface, $g_{h}=\frac{GM}{(R + h)^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, $R$ is the radius of the Earth, and $h$ is the height above the Earth's surface.

Step2: Set up the ratio

We want to find $h$ when $g_{h}=\frac{1}{4}g$. Substituting the formulas for $g$ and $g_{h}$ into the ratio $\frac{g_{h}}{g}=\frac{1}{4}$, we get $\frac{\frac{GM}{(R + h)^{2}}}{\frac{GM}{R^{2}}}=\frac{1}{4}$.

Step3: Simplify the ratio

The $GM$ terms cancel out, and we have $\frac{R^{2}}{(R + h)^{2}}=\frac{1}{4}$. Taking the square - root of both sides, $\frac{R}{R + h}=\frac{1}{2}$ (we consider the positive root since distances are non - negative).

Step4: Solve for $h$

Cross - multiply to get $2R=R + h$. Then, subtract $R$ from both sides to find $h = R$.

Answer:

A. Equal to the radius of Earth