Chapter 5, Problem 45P

To determine

The speed of the space shuttle when the satellite is being released relative to Earth’s center.

Answer to Problem 45P

Solution:

  $7463.7\text{m/s}$

Explanation of Solution

Given:

The space shuttle releases satellite at a height of 780 km from Earth’s surface.

Formula Used:

The expression for the speed $v$ of a satellite in a circular orbit $r$ around the Earth is:

  $v=\sqrt{G\frac{M_{Earth}}{r}}$

Here, $G$ is the gravitational constant and $M_{earth}$ is the mass of Earth and r is the orbital radius.

Calculation:

Convert the height from km to m:

  $\begin{array}{c}h=(780\text{km})\left(\frac{1000\text{m}}{1\text{km}}\right)\\ =7.8\times {10}^5\text{m}\end{array}$

Orbital radius:

  $\begin{array}{c}r=R_{Earth}+h\\ =(6.38\times {10}^6\text{m)+}\left(7.8\times {10}^5\text{m}\right)\\ =7.16\times {10}^6\text{m}\end{array}$

Mass of the Earth: $5.98\times {10}^{24}\text{kg}$

The speed of the shuttle is

  $\begin{array}{l}v=\sqrt{G\frac{M_{earth}}{r}}\\ =\sqrt{\left(\frac{\text{(6}{\text{.67×10}}^{\text{-11}}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{)(5}{\text{.98×10}}^{\text{24}}\text{kg)}}{\text{7}{\text{.16×10}}^{\text{6}}\text{m}}\right)}\\ =7463.\text{7}\text{m/s}\end{array}$

Conclusion:

Therefore,the speed of the shuttle is $\text{7463}\text{.7}\text{m/s}$ .