Chapter 5, Problem 40P

To determine

The percentage change in the value of acceleration due to gravity at $380\text{km}$ from its value at the Earth’s surface.

Answer to Problem 40P

Solution:$11\%$

Explanation of Solution

Given:

Distance of the point from the Earth’s Surface = 380 km

Formula Used:$g=G\frac{M}{r^2}$

M is the mass of the planet, $G$ is thegravitational constant, and $r$ is the distance of the point from the planet where g is required to be found.

Calculation:

Acceleration due to gravity on earth is $9.8{\text{m/s}}^2$ .

Radius of the earth is $6.38\times {10}^6\text{m}$ .

Distance between the center of the earth and the point is:

  $\begin{array}{l}r=6.38\times {10}^6\text{m}+\left(380\text{km}\right)\left(\frac{1000\text{m}}{1\text{km}}\right)\\ r=\text{6}\text{.386}\times {10}^6\text{m}\text{+}\text{3}\text{.8}\times {\text{10}}^5\text{m}\\ r=\text{6}\text{.76}\times {\text{10}}^6\text{m}\end{array}$

  $\begin{array}{l}{g}_{Earth}=G\frac{M_{\text{Earth}}}{r_{Earth}^2}\\ =\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(6.76\times {10}^6\text{m}\right)}^2}\\ =8.73{\text{m/s}}^{\text{2}}\\ \text{percentage}\text{change}\text{in}g=\frac{\left(9.8{\text{m/s}}^2-8.73{\text{m/s}}^2\right)}{9.8{\text{m/s}}^2}\times 100\%\\ =11\%\end{array}$

Conclusion:

Therefore, the percentage change in the value of acceleration due to gravity at $380\text{km}$ from its value at the Earth’s surface is only $11\%$ .