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The gravitational field strength at Earth’s surface is 9.8 N/kg. What is the field strength at the center of Earth? At a distance one Earth radius beyond the surface?

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Answered 2 years ago
Answered 2 years ago
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Given:\textbf{Given:}

M=6.0×1024M = 6.0 \times 10^{24} kg

R=6.37×106R = 6.37 \times 10^6 m

The gravitational field strength can be calculated using the formula:

g=GMR2\begin{gather} \textbf{g} = \dfrac{GM}{R^2} \end{gather}

At the center of the Earth, R=0R= 0, we calculate the gravitational field strength using Eq (1) as follows:

gC=GMR2=G(6.0×1024)(0)2=0\begin{align*} g_C &= \dfrac{GM}{R^2} \\ &= \dfrac{G \cdot (6.0 \times 10^{24})}{(0)^2} \\ &= \boxed{0} \end{align*}

At the distance one Earth radius beyond the surface, R=2RR = 2R, we calculate the gravitational field strength using Eq (1) as follows:

g2R=GM(2R)2=G(6.0×1024)(26.37×106)2=2.47 N\begin{align*} g_{2R} &= \dfrac{GM}{(2R)^2} \\ &= \dfrac{G \cdot (6.0 \times 10^{24})}{(2 \cdot 6.37 \times 10^6)^2} \\ &= \boxed{2.47 \text{ N}} \end{align*}

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