9.7 Rocket propulsion  (Page 3/8)

Thrust on a spacecraft

1. What is the thrust (the force applied to the rocket by the ejected fuel) on the spacecraft?
2. What is the spacecraft’s acceleration as a function of time?
3. What are the spacecraft’s accelerations at t = 0, 15, 30, and 35 s?

Strategy

1. The force on the spacecraft is equal to the rate of change of the momentum of the fuel.
2. Knowing the force from part (a), we can use Newton’s second law to calculate the consequent acceleration. The key here is that, although the force applied to the spacecraft is constant (the fuel is being ejected at a constant rate), the mass of the spacecraft isn’t; thus, the acceleration caused by the force won’t be constant. We expect to get a function a ( t ), therefore.
3. We’ll use the function we obtain in part (b), and just substitute the numbers given. Important: We expect that the acceleration will get larger as time goes on, since the mass being accelerated is continuously decreasing (fuel is being ejected from the rocket).

Solution

1. The momentum of the ejected fuel gas is
   
   $p=m_{g}v.$
   
   The ejection velocity $v=2.5\times {10}^2\text{m/s}$ is constant, and therefore the force is
   
   $F=\frac{dp}{dt}=v\frac{d{m}_g}{dt}=\text{−}v\frac{dm}{dt}.$
   
   Now, $\frac{d{m}_g}{dt}$ is the rate of change of the mass of the fuel; the problem states that this is $2.0\times {10}^2\text{kg/s}$ . Substituting, we get
   
   $\begin{array}{cc}\hfill F& =v\frac{d{m}_g}{dt}\hfill \\ & =\left(2.5\times {10}^2\frac{\text{m}}{\text{s}}\right)\left(2.0\times {10}^2\frac{\text{kg}}{\text{s}}\right)\hfill \\ & =5\times {10}^4\text{N.}\hfill \end{array}$
2. Above, we defined m to be the combined mass of the empty rocket plus however much unburned fuel it contained: $m=m_R+m_g$ . From Newton’s second law,
   
   $a=\frac{F}{m}=\frac{F}{m_R+m_g}.$
   
   The force is constant and the empty rocket mass $m_R$ is constant, but the fuel mass $m_g$ is decreasing at a uniform rate; specifically:
   
   $m_g=m_g(t)=m_{g_0}-\left(\frac{d{m}_g}{dt}\right)t.$
   
   This gives us
   
   $a(t)=\frac{F}{m_{g_{\text{i}}}-\left(\frac{d{m}_g}{dt}\right)t}=\frac{F}{M-\left(\frac{d{m}_g}{dt}\right)t}.$
   
   Notice that, as expected, the acceleration is a function of time. Substituting the given numbers:
   
   $a(t)=\frac{5\times {10}^4\text{N}}{2.0\times {10}^4\text{kg}-\left(2.0\times {10}^2\frac{\text{kg}}{\text{s}}\right)t}.$
3. At $t=0\text{s}$ :
   
   $a(0\text{s})=\frac{5\times {10}^4\text{N}}{2.0\times {10}^4\text{kg}-\left(2.0\times {10}^2\frac{\text{kg}}{\text{s}}\right)(0\text{s})}=2.5\frac{\text{m}}{{\text{s}}^2}.$
   
   At $t=15\text{s},a(15\text{s})={2.9\text{m/s}}^2$ .
   At $t=30\text{s},a(30\text{s})=3.6{\text{m/s}}^2$ .
   Acceleration is increasing, as we expected.

Significance