Question

Ion-propulsion rockets have been proposed for use in space. They employ atomic ionization techniques and nuclear energy sources to produce extremely high exhaust velocities, perhaps as great as$8.00\times {10}^{6}m/s$. These techniques allow a much more favorable payload-to-fuel ratio. To illustrate this fact: (a) Calculate the increase in velocity of a$20000kg$space probe that expels only$40.0kg$of its mass at the given exhaust velocity. (b) These engines are usually designed to produce a very small thrust for a very long time—the type of engine that might be useful on a trip to the outer planets, for example. Calculate the acceleration of such an engine if it expels$4.50\times {10}^{6}kg/s$ at the given velocity, assuming the acceleration due to gravity is negligible.

Short answer

(a)The increase in velocity of the space probe is$16\mathrm{k}\mathrm{m}/\mathrm{s}$.

(b) The acceleration is $36\times 10^7\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Definition of Acceleration

The acceleration can be referred to as the rate at which velocity changes.

Given Data

The mass of space probe is${\mathrm{m}}_1=20000\mathrm{k}\mathrm{g}$.

The mass of expulsion is${\mathrm{m}}_2=40.0\mathrm{k}\mathrm{g}$.

The exhaust velocity of rocket is ${\mathrm{v}}_2=8.0\times 10^6\mathrm{m}/\mathrm{s}$.

Calculation of change in velocity

Using the conservation of momentum along vertical direction we get,

${\mathrm{m}}_1{\mathrm{v}}_1={\mathrm{m}}_2{\mathrm{v}}_2$

So,

$\begin{array}{c}{\mathrm{v}}_1=\frac{40\times 8\times 10^6}{20000}\\ {\mathrm{v}}_1=16\mathrm{k}\mathrm{m}/\mathrm{s}\end{array}$

Calculation of acceleration

The impulse is,

$\mathrm{F}\mathrm{d}\mathrm{t}=\mathrm{v}\mathrm{d}\mathrm{m}$

The acceleration is,

$\begin{array}{c}\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\\ =\frac{\mathrm{v}\frac{\mathrm{d}\mathrm{m}}{\mathrm{d}\mathrm{t}}}{\mathrm{m}}\end{array}$

Substituting the obtained values,

$\begin{array}{c}\mathrm{a}=\frac{16\times 10^3\times 4.5\times 10^{-6}}{20000}\\ =36\times 10^7\mathrm{m}/{\mathrm{s}}^2\end{array}$