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

Explanation:

Step1: Recall gravity - height formula

The acceleration due to gravity at the surface of the Earth $g=\frac{GM}{R^{2}}$, and at a height $h$ above the 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.

Step2: Set up the ratio

We are given that $g_h=\frac{1}{4}g$. So, $\frac{GM}{(R + h)^{2}}=\frac{1}{4}\times\frac{GM}{R^{2}}$. Canceling out $GM$ from both sides, we get $\frac{1}{(R + h)^{2}}=\frac{1}{4R^{2}}$.

Step3: Cross - multiply and solve for $h$

Cross - multiplying gives $4R^{2}=(R + h)^{2}$. Taking the square root of both sides, $2R=R + h$ (we consider the positive root since distance is non - negative). Subtracting $R$ from both sides, we find $h = R$.

Answer:

A. Equal to the radius of Earth