# 8.7 Introduction to rocket propulsion  (Page 2/4)

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A rocket’s acceleration depends on three major factors, consistent with the equation for acceleration of a rocket . First, the greater the exhaust velocity of the gases relative to the rocket, ${v}_{\text{e}}$ , the greater the acceleration is. The practical limit for ${v}_{\text{e}}$ is about $2\text{.}5×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ for conventional (non-nuclear) hot-gas propulsion systems. The second factor is the rate at which mass is ejected from the rocket. This is the factor $\text{Δ}m/\text{Δ}t$ in the equation. The quantity $\left(\text{Δ}m/\text{Δ}t\right){v}_{\text{e}}$ , with units of newtons, is called "thrust.” The faster the rocket burns its fuel, the greater its thrust, and the greater its acceleration. The third factor is the mass $m$ of the rocket. The smaller the mass is (all other factors being the same), the greater the acceleration. The rocket mass $m$ decreases dramatically during flight because most of the rocket is fuel to begin with, so that acceleration increases continuously, reaching a maximum just before the fuel is exhausted.

## Factors affecting a rocket’s acceleration

• The greater the exhaust velocity ${v}_{\text{e}}$ of the gases relative to the rocket, the greater the acceleration.
• The faster the rocket burns its fuel, the greater its acceleration.
• The smaller the rocket’s mass (all other factors being the same), the greater the acceleration.

## Calculating acceleration: initial acceleration of a moon launch

A Saturn V’s mass at liftoff was $2\text{.}\text{80}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{kg}$ , its fuel-burn rate was $1\text{.}\text{40}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{kg/s}$ , and the exhaust velocity was $2\text{.}\text{40}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ . Calculate its initial acceleration.

Strategy

This problem is a straightforward application of the expression for acceleration because $a$ is the unknown and all of the terms on the right side of the equation are given.

Solution

Substituting the given values into the equation for acceleration yields

$\begin{array}{lll}a& =& \frac{{v}_{\text{e}}}{m}\phantom{\rule{0.25em}{0ex}}\frac{\text{Δ}m}{\text{Δ}t}-g\\ & =& \text{}\frac{2\text{.}\text{40}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}}{2\text{.}\text{80}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{kg}}\left(1\text{.}\text{40}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{kg/s}\right)-9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\\ & =& \text{}\text{2}\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}.\end{array}$

Discussion

This value is fairly small, even for an initial acceleration. The acceleration does increase steadily as the rocket burns fuel, because $m$ decreases while ${v}_{\text{e}}$ and $\frac{\text{Δ}m}{\text{Δ}t}$ remain constant. Knowing this acceleration and the mass of the rocket, you can show that the thrust of the engines was $3\text{.}\text{36}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{N}$ .

To achieve the high speeds needed to hop continents, obtain orbit, or escape Earth’s gravity altogether, the mass of the rocket other than fuel must be as small as possible. It can be shown that, in the absence of air resistance and neglecting gravity, the final velocity of a one-stage rocket initially at rest is

$v={v}_{\text{e}}\phantom{\rule{0.25em}{0ex}}\text{ln}\phantom{\rule{0.15em}{0ex}}\frac{{m}_{0}}{{m}_{\text{r}}},$

where $\text{ln}\left({m}_{0}/{m}_{\text{r}}\right)$ is the natural logarithm of the ratio of the initial mass of the rocket $\left({m}_{0}\right)$ to what is left $\left({m}_{\text{r}}\right)$ after all of the fuel is exhausted. (Note that $v$ is actually the change in velocity, so the equation can be used for any segment of the flight. If we start from rest, the change in velocity equals the final velocity.) For example, let us calculate the mass ratio needed to escape Earth’s gravity starting from rest, given that the escape velocity from Earth is about $\text{11}\text{.}2×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ , and assuming an exhaust velocity ${v}_{\text{e}}=2\text{.}5×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ .

$\text{ln}\phantom{\rule{0.15em}{0ex}}\frac{{m}_{0}}{{m}_{\text{r}}}=\frac{v}{{v}_{\text{e}}}=\frac{\text{11}\text{.}2×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}}{2\text{.}5×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}}=4\text{.}\text{48}$

Solving for ${m}_{0}/{m}_{\text{r}}$ gives

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