If the radius of Earth somehow shrank by half without any change in Earth’s mass, what would be the value of g at the new surface? What would be the value of g above the new surface at a distance equal to the present radius?

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Answered 2 years ago

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$\textbf{Given:}$

$M = 5.97 \times 10^{24}$ kg

$R = 6.37 \times 10^6$ m

$\textbf{g} = 9.8 \dfrac{\text{N}}{\text{kg}}$

The gravitational field strength can be calculated using the formula:

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

If the Earth shrank by half its original size, $R= \dfrac{R}{2}$ , we calculate the $\textbf{g}$ at the surface using Eq (1) as follows:

\begin{align*} g_{S} &= \dfrac{GM}{\left(\dfrac{R}{2}\right)^2} \\ &= \dfrac{G \cdot (5.97 \times 10^{24})}{\left(\dfrac{6.37 \times 10^6}{2}\right)^2} \\ &= \boxed{39.28 \dfrac{\text{N}}{\text{kg}}} \end{align*}

This value is approximately four times its original $\textbf{g}$

Above the new surface, $R= R$ , we calculate $\textbf{g}$ using Eq (1) as follows:

\begin{align*} g_{R} &= \dfrac{GM}{R^2} \\ &= \dfrac{G \cdot (5.97 \times 10^{24})}{(6.37 \times 10^6)^2} \\ &= \boxed{9.82 \dfrac{\text{N}}{\text{kg}}} \end{align*}

This value is approximately the same its original $\textbf{g}$

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