91Ó°ÊÓ

Given that a 70 kg astronaut floating in space in a 110 kg MMU (manned maneuvering unit) experiences an acceleration of 0.029 m/s² when he fires one of the MMU’s thrusters.

Mass of astronauts,$\mathrm{m}=70\mathrm{kg}$

Mass of unit, $\mathrm{M}=110\mathrm{kg}$

Acceleration, $\mathrm{a}=0.029\mathrm{m}/{\mathrm{s}}^2$

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 $\frac{d\mathrm{m}}{d\mathrm{t}}$is mass lost as the rocket accelerates and v is the exhaust speed.

Hence, thrust can be found as,

$\mathrm{F}=-\mathrm{v}\frac{d\mathrm{m}}{d\mathrm{t}}$ ….. (1)

Calculate the total weight as below.

$\begin{array}{rcl}\mathrm{W}& =& 110+70\\ & =& 180\mathrm{kg}\end{array}$

The speed, $\mathrm{v}=490\mathrm{m}/\mathrm{s}$

Determine the lost mass as given below.

$\begin{array}{rcl}\frac{d\mathrm{m}}{d\mathrm{t}}& =& -\frac{\mathrm{m}}{\mathrm{v}}\mathrm{a}\\ & =& -\left(\frac{180}{490}\right)\left(0.029\right)\\ & =& -0.0106\mathrm{kg}/\mathrm{s}\end{array}$

Loss of mass in 5 seconds is,

$\begin{array}{rcl}\mathrm{m}& =& 0.0106\mathrm{kg}/\mathrm{s}×5\mathrm{s}\\ & =& 0.053\mathrm{kg}\end{array}$

Thrust of the thruster is define by substituting known values into equation (1).

$\begin{array}{rcl}\mathrm{F}& =& -\mathrm{v}\frac{d\mathrm{m}}{d\mathrm{t}}\\ & =& -\left(490\right)\left(-0.0106\right)\\ & =& 5.19\mathrm{N}\end{array}$

Hence, the thrust of the thruster is $5.19\mathrm{N}$.