SOLUTION The rocket of a spacecraft is fired in the direction of motion. The deceleration, the velocity change, and the thrust are to be determined.
Assumptions 1 The flow of combustion gases is steady and one-dimensional during the firing period, but the flight of the spacecraft is unsteady. 2 There are no external forces acting on the spacecraft, and the effect of pressure force at the nozzle outlet is negligible. 3 The mass of discharged fuel is negligible relative to the mass of the spacecraft, and thus, the spacecraft may be treated as a solid body with a constant mass. 4 The nozzle is well designed such that the effect of the momentum-flux correction factor is negligible, and thus, 𝛽 ≅ 1.
Analysis (a) For convenience, we choose an inertial reference frame that moves with the spacecraft at the same initial velocity. Then the velocities of the fluid stream relative to an inertial reference frame become simply the velocities relative to the spacecraft. We take the direction of motion of the spacecraft as the positive direction along the x-axis. There are no external forces acting on the spacecraft, and its mass is essentially constant. Therefore, the spacecraft can be treated as a solid body with constant mass, and the momentum equation in this case is, from Eq. 6–29,
\vec{F}_{\text {thrust }}=m_{\text {body }} \vec{a}=\sum_{\text {in }} \beta \dot{m} \vec{V}-\sum_{\text {out }} \beta \dot{m} \vec{V} (6-29)
\vec{F}_{\text {thrust }}=m_{\text {spacecraft }} \vec{a}_{\text {spacecraft }}=\sum_{\text {in }} \beta \dot{m} \vec{V}-\sum_{\text {out }} \beta \dot{m} \vec{V}
where the fluid stream velocities relative to the inertial reference frame in this case are identical to the velocities relative to the spacecraft. Noting that the motion is on a straight line and the discharged gases move in the positive x-direction, we write the momentum equation using magnitudes as
m_{\text {spacecraft }} a_{\text {spacecraft }}=m_{\text {spacecraft }} \frac{d V_{\text {spacecraft }}}{d t}=-\dot{m}_{\text {gas }} V_{\text {gas }}
Noting that gases leave in the positive x-direction and substituting, the acceleration of the spacecraft during the first 5 seconds is determined to be
a_{\text {spacecraft }}=\frac{d V_{\text {spacecraft }}}{d t}=-\frac{\dot{m}_{\text {gas }}}{m_{\text {spacecraft }}} V_{\text {gas }}=-\frac{80 kg / s }{12,000 kg }(+3000 m / s )=- 2 0 m / s ^{2}
The negative value confirms that the spacecraft is decelerating in the positive x direction at a rate of 20 m/s².
(b) Knowing the deceleration, which is constant, the velocity change of the spacecraft during the first 5 seconds is determined from the definition of acceleration to be
\begin{aligned}d V_{\text {spacecraft }} &=a_{\text {spacecraft }} d t \rightarrow \Delta V_{\text {spacecraft }}=a_{\text {spacecraft }} \Delta t=\left(-20 m / s ^{2}\right)(5 s ) \\&=-100 m / s\end{aligned}
(c) The thrusting force exerted on the space aircraft is, from Eq. 6–29,
F_{\text {thrust }}=0-\dot{m}_{\text {gas }} V_{\text {gas }}=0-(80 kg / s )(+3000 m / s )\left(\frac{1 kN }{1000 kg \cdot m / s ^{2}}\right)=-240 kN
The negative sign indicates that the trusting force due to firing of the rocket acts on the aircraft in the negative x-direction.
Discussion Note that if this fired rocket were attached somewhere on a test stand, it would exert a force of 240 kN (equivalent to the weight of about 24 tons of mass) to its support in the opposite direction of the discharged gases.
