91Ó°ÊÓ

Given that a small rocket burns 0.05 kg of fuel per second, ejecting it as a gas with a velocity relative to the rocket of magnitude 1600 m/s.

Mass of burning of gas per sec $\frac{d\mathrm{m}}{d\mathrm{t}}=-0.05\mathrm{kg}/\mathrm{s}$

Magnitude of velocity, $\mathrm{v}=1600\mathrm{m}/\mathrm{s}$

In rocket propulsion, the mass of a rocket changes as the fuel is used up and ejected from the rocket. Analysis of the motion must contain the momentum carried away by the spent fuel as well as the momentum of the rocket itself.

Let $\frac{\mathrm{dm}}{\mathrm{dt}}$is mass lost as the rocket accelerates, v is the exhaust speed.

Hence thrust can be found as below.

$\mathrm{F}=-\mathrm{v}\frac{\mathrm{dm}}{\mathrm{dt}}$ ….. (1)

Substitute 1600 m/s for v and -0.05 kg/s for $\frac{\mathrm{dm}}{\mathrm{dt}}$values into equation (1).

$\begin{array}{rcl}\mathrm{F}& =& -\left(1600\mathrm{m}/\mathrm{s}\right)\left(-0.05\mathrm{kg}/\mathrm{s}\right)\\ & =& 80\mathrm{N}\end{array}$

Since, the motion of the rocket depends on the conservation of momentum, the rocket will operate in outer space and it will operate better because of the absence of air resistance.

The rocket could be steered by ejecting the gas in a direction with a component perpendicular to the rocket’s velocity and ejecting it in a direction parallel (as opposed to anti-parallel) to the rocket’s velocity.