Chapter 13, Problem 64A

Many people mistakenly believe that the astronauts that orbit the Earth are 'above gravity.' Earth’s mass is $6\times {10}^{24}\text{kg}$, and its radius is $6.38\times {10}^6\text{m}\left(6380\text{km}\right)$. Use the inverse-square law to show that in space-shuttle territory, 200 kilometers above Earth’s surface, the force of gravity on a shuttle is about 94% that at Earth’s surface.

To determine

To show:The force of gravity on a shuttle is about $94\%$ that at Earth’s surface.

Explanation of Solution

Given:

The mass of Earth is $M=6\times {10}^{24}\text{ kg}$ .

The radius is $R=6.38\times {10}^6\text{ m}$ .

The height is $h=200\text{ km}$ .

Formula used:

The expression for acceleration due to gravity is

  $g=\frac{GM}{r^2}$

Here, $G$ is the universal gravitational constant, $M$ is the mass, and $r$ is the radial distance.

The radial distance is,

  $r=R+h$

Here, $R$ is the radius and $h$ is the height.

Calculation:

Consider the universal gravitational constant as $G=6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}$ and the acceleration due to gravity on Earth as $g=9.8\text{N}\cdot \text{m}/{\text{s}}^{\text{2}}$ .

The acceleration due to gravity on shuttle is,

  $\begin{array}{l}g=\frac{GM}{r^2}\\ g_{\text{shuttle}}=\frac{GM}{{\left(R+h\right)}^2}\\ g_{\text{shuttle}}=\frac{\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(6\times {10}^{24}\text{ kg}\right)}{{\left(\left(6.38\times {10}^6\text{ m}\right)+\left(200\text{ km}\times \frac{1000\text{ m}}{1\text{ km}}\right)\right)}^2}\\ g_{\text{shuttle}}=9.24\text{N}\cdot \text{m}/{\text{s}}^{\text{2}}\end{array}$

So, the percentage of acceleration due to gravity on shuttle territory in Earth’s surface is,

  $\begin{array}{l}\text{Percentage}=\frac{g_{\text{shuttle}}}{g}\times 100\%\\ \text{Percentage}=\frac{9.24\text{N}\cdot \text{m}/{\text{s}}^{\text{2}}}{9.8\text{N}\cdot \text{m}/{\text{s}}^{\text{2}}}\times 100\%\\ \text{Percentage}=94\%\end{array}$

Conclusion:

Thus, the force of gravity on a shuttle is about $94\%$ that at Earth’s surface.