Weight and g: The radius of the earth is R. At what distance above the earth's surface will the acceleration of gravity be 4.9 m/s²? A) 0.41R B) 0.50R C) 1.0R D) 1.4R E) 2.0R

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Quizplus · Accepted Answer

The acceleration due to gravity at a distance $r$ from the center of the Earth (where $r > R$, $R$ being the Earth's radius) is given by $g' = \frac{G M}{r^2}$, where $G$ is the gravitational constant and $M$ is the mass of the Earth. At the surface of the Earth, the acceleration due to gravity is $g = \frac{G M}{R^2}$. We are told that at some height $h$ above the Earth's surface, the acceleration due to gravity is $4.9 \, \text{m/s}^2$, which is half of the standard $9.8 \, \text{m/s}^2$. This means that $g' = \frac{g}{2}$.Let $r = R + h$ be the distance from the center of the Earth to the point where $g' = 4.9 \, \text{m/s}^2$. We want to find $h$ in terms of $R$.Since $g' = \frac{G M}{r^2}$ and $g = \frac{G M}{R^2}$, setting $g' = \frac{g}{2}$ gives us $\frac{G M}{(R + h)^2} = \frac{G M}{2R^2}$.Solving for $h$ in terms of $R$, we get $R + h = \sqrt{2}R$, which implies $h = (\sqrt{2} - 1)R \approx 0.41R$.

Weight and G: the Radius of the Earth Is R