Question

1. at what altitude above earths surface is the acceleration due to gravity equal to g/4?

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 is $g_{h}=\frac{GM}{(R + h)^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, $R$ is the radius of the Earth, and $h$ is the altitude.

Step2: Set up the ratio

We are given that $g_{h}=\frac{g}{4}$. Substituting the expressions for $g$ and $g_{h}$, we get $\frac{GM}{(R + h)^{2}}=\frac{1}{4}\times\frac{GM}{R^{2}}$.

Step3: Simplify the equation

Cancel out $GM$ from both sides of the equation. We have $\frac{1}{(R + h)^{2}}=\frac{1}{4R^{2}}$. Cross - multiply to get $4R^{2}=(R + h)^{2}$.

Step4: Take the square root of both sides

Taking the square root of both sides, we get $2R=R + h$ (we consider the positive square root since distance cannot be negative).

Step5: Solve for $h$

Subtract $R$ from both sides of the equation: $h = R$.

Answer:

The altitude above the Earth's surface is equal to the radius of the Earth.