## Related questions with answers

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?

Solutions

Verified$\textbf{Given:}$

$M = 5.97 \times 10^{24}$ 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 $\textbf{g}$

If the Earth shrank to one half size g would increase by a factor of four to 39.2 N. The value of g one Earth radius from the surface would be 9.8 N (exactly what it is currently).

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651 (8 more)Randall D. Knight#### Mathematical Methods in the Physical Sciences

3rd Edition•ISBN: 9780471198260 (1 more)Mary L. Boas#### Fundamentals of Physics

10th Edition•ISBN: 9781118230718 (3 more)David Halliday, Jearl Walker, Robert Resnick## More related questions

- prealgebra

1/4

- prealgebra

1/7