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\(20000\;kg\)space probe that expels only\(40.0\;kg\)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.

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

The mass of space probe is\({{\rm{m}}_{\rm{1}}}{\rm{ = 20000}}\;{\rm{kg}}\).

The mass of expulsion is\({{\rm{m}}_{\rm{2}}}{\rm{ = 40}}{\rm{.0}}\;{\rm{kg}}\).

The exhaust velocity of rocket is \({{\rm{v}}_{\rm{2}}}{\rm{ = 8}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{6}}}\;{\rm{m/s}}\).

Using the conservation of momentum along vertical direction we get,

\({{\rm{m}}_{\rm{1}}}{{\rm{v}}_{\rm{1}}}{\rm{ = }}{{\rm{m}}_{\rm{2}}}{{\rm{v}}_{\rm{2}}}\)

So,

\(\begin{array}{c}{{\rm{v}}_{\rm{1}}}{\rm{ = }}\frac{{{\rm{40 \times 8 \times 1}}{{\rm{0}}^{\rm{6}}}}}{{{\rm{20000}}}}\\{{\rm{v}}_{\rm{1}}}{\rm{ = 16}}\;{\rm{km/s}}\end{array}\)

The impulse is,

\({\rm{Fdt = vdm}}\)

The acceleration is,

\(\begin{array}{c}{\rm{a = }}\frac{{\rm{F}}}{{\rm{m}}}\\{\rm{ = }}\frac{{{\rm{v}}\frac{{{\rm{dm}}}}{{{\rm{dt}}}}}}{{\rm{m}}}\end{array}\)

Substituting the obtained values,

\(\begin{array}{c}{\rm{a = }}\frac{{{\rm{16 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ \times 4}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}}}{{{\rm{20000}}}}\\{\rm{ = 36 \times 1}}{{\rm{0}}^{\rm{7}}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)