SPEED OF A ROCKET
Suppose that \frac{3}{4} of the initial mass of the rocket in Example 8.15 is fuel, so the fuel is completely consumed at a constant rate in 90 s. The final mass of the rocket is m = m_0/4. If the rocket starts from rest in our coordinate system, find its speed at the end of this time.
IDENTIFY and SET UP:
We are given the initial velocity v_0 = 0, the exhaust speed v_{ex}= 2400 m/s, and the final mass m as a fraction of the initial mass m_0. We’ll use Eq. (8.40) to find the final speed v:
v=v_{0}+v_{\mathrm{ex}} \ln \frac{m_{0}}{m}=0+(2400 \mathrm{~m} / \mathrm{s})(\ln 4)=3327 \mathrm{~m} / \mathrm{s}EXECUTE:
Let’s examine what happens as the rocket gains speed. (To illustrate our point, we use more figures than are significant.) At the start of the flight, when the velocity of the rocket is zero, the ejected fuel is moving backward at 2400 m/s relative to our frame of reference. As the rocket moves forward and speeds up, the fuel’s speed relative to our system decreases; when the rocket speed reaches 2400 m/s, this relative speed is zero. [Knowing the rate of fuel consumption, you can solve Eq. (8.40) (v-v_{0}=-v_{\mathrm{ex}} \ln \frac{m}{m_{0}}=v_{\mathrm{ex}} \ln \frac{m_{0}}{m}) to show that this occurs at about t = 75.6 s.] After this time the ejected burned fuel moves forward, not backward, in our system. Relative to our frame of reference, the last bit of ejected fuel has a forward velocity
of 3327 m/s – 2400 m>s = 927 m/s.