## Related questions with answers

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?

Solution

Verified$\textbf{Given:}$

$M = 6.0 \times 10^{24}$ kg

$R = 6.37 \times 10^6$ m

The gravitational field strength can be calculated using the formula:

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

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

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

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

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

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