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Question

How would the gravitational field at Earth’s surface be affected if Earth shrank in size without any change in mass? What would be its relative strength at the new surface if Earth shrank to half size? To one-tenth size?

Solution

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

M=5.97×1024M = 5.97 \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}

If the Earth shrank by half its original size, R=R2R= \dfrac{R}{2}, we calculate the gravitational field strength using Eq (1) as follows:

ghalf=GM(R2)2=G(5.97×1024)(6.37×1062)2=39.28Nkg\begin{align*} g_{half} &= \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 g\textbf{g}

If the Earth shrank by one-tenth its original size, R=R10R= \dfrac{R}{10}, we calculate the gravitational field strength using Eq (1) as follows:

gonetenth=GM(R10)2=G(5.97×1024)(6.37×10610)2=982Nkg\begin{align*} g_{one-tenth} &= \dfrac{GM}{\left(\dfrac{R}{10}\right)^2} \\ &= \dfrac{G \cdot (5.97 \times 10^{24})}{\left(\dfrac{6.37 \times 10^6}{10}\right)^2} \\ &= \boxed{982 \dfrac{\text{N}}{\text{kg}}} \end{align*}

This value is approximately one hundred times its original g\textbf{g}

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