Chapter 14, Problem 50A

Use the equation $T=2\pi \sqrt{\frac{r^3}{GM}}$ to show that the period of the space shuttle 200 km (200,000 m) above Earth’s surface is about 90 minutes. The radius of Earth is 6370 km (6,370,000 m).

To determine

To show: Period of space shuttle is $T=90\text{minutes}$ .

Explanation of Solution

Given:

Distance of space shuttle from Earth’s surface is $D=200\text{km}$ .

Formula used:

The given equation is, $T=2\pi \sqrt{\frac{r^3}{GM}}$

Calculation:

As radius of Earth is around 6370 km.

So, the distance of space shuttle from Earth’s center is,

  $\begin{array}{l}r=6.37\times {10}^6\text{m}+0.2\times {10}^6\text{m}\\ =6.\text{57}\times {\text{10}}^6\text{m}\end{array}$

Now,

Gravitational constant, $G=6.67\times {10}^{-11}{\text{m}}^3{\text{kg}}^1{\text{s}}^{-2}$ and

Mass of Earth, $M=5.97\times {10}^{24}\text{kg}$

The time period is,

  $\begin{array}{c}T=2\pi \sqrt{\frac{{\left(6.\text{57}\times {\text{10}}^6\right)}^3}{\left(6.67\times {10}^{-11}\right)\times 5.97\times {10}^{24}}}\\ =5302.47\text{s}\\ =\frac{5302.47}{60}\text{min}\\ T=88.37\text{min}\\ \approx 90\text{min}\end{array}$

Conclusion:

Hence, it is shown that the period of space shuttle is approximately $T=90\text{minutes}$ .