Chapter 16, Problem 79A

To determine

The mass of gas released.

Answer to Problem 79A

The mass of gas released is $171\text{ kg}$ .

Explanation of Solution

Given:

The mass of a space probe is $m_{\text{p}}=7.600\times {10}^3\text{ kg}$ .

The speed of a space probe is $v_{\text{p}}=125\text{m}/\text{s}$ .

The course correction $\theta =30.0°$ .

The speed gas expelled by the rocket is $v_{\text{g}}=3.200\text{km}/\text{s}$ .

Formula used:

The expression for the course correction is,

$\tan \theta =\frac{m_{\text{g}}{v}_{\text{g}}}{m_{\text{p}}{v}_{\text{p}}}$

Here, $m_{\text{g}}$ is the mass of the gas, $v_{\text{g}}$ is the speed of the gas expelled, $m_{\text{p}}$ is the mass of the space probe, and $v_{\text{p}}$ is the speed of the space probe.

Calculation:

The mass of the gas released is,

$\begin{array}{l}\tan \theta =\frac{m_{\text{g}}{v}_{\text{g}}}{m_{\text{p}}{v}_{\text{p}}}\\ m_{\text{g}}=\frac{m_{\text{p}}{v}_{\text{p}}\tan \theta }{v_{\text{g}}}\\ m_{\text{g}}=\frac{\left(7.600\times {10}^3\text{ kg}\right)\left(125\text{m}/\text{s}\right)\tan 30.0°}{\left(3.200\text{km}/\text{s}\times \frac{1,000\text{ m}}{1\text{ km}}\right)}\\ m_{\text{g}}=171\text{ kg}\end{array}$

Conclusion:

Thus, the mass of gas released is $171\text{ kg}$ .