Question 8.14: Launch of a rocket A rocket floats next to an interplanetary...
Launch of a rocket
A rocket floats next to an interplanetary space station far from any planet. It ignites its engine. In the first second of its flight, it ejects \frac{1}{60} of its mass with a relative velocity magnitude of 2400 m/s. Find its acceleration.
SET UP AND SOLVE We are given that ∆m = -m/60 and ∆t = 1.0 s. We use the same coordinate system as in Figure 8.28, with the +x axis to the right.
The acceleration is given by Equation 8.26:
a=\frac{\Delta \upsilon }{\Delta t}=-\frac{\upsilon _{ex}}{m}\frac{\Delta m}{\Delta t}
=-\frac{2400 m/s}{m}\left(\frac{-m /60}{1.0 s} \right)=40 m/s^2.
REFLECT At the start of the flight, when the velocity of the rocket is zero, the ejected exhaust is moving to the left, relative to our coordinate system, at 2400 m/s. At the end of the first second, the rocket is moving at 40 m/s. The speed of the exhaust relative to our system is 2360 m/s. We could now compute the acceleration at t = 1.0 s (using the decreased rocket mass), find the velocity at t = 2.0 s, and so on, stepping through the calculation one second at a time until all the fuel is used up. As the mass decreases, the acceleration in successive 1 s intervals increases.
Detailed calculation shows that, after about 22 s, the rocket’s velocity in our coordinate system exceeds 2400 m/s. The exhaust ejected after this time therefore moves forward, not backward, in our system. The final velocity acquired by the rocket can be greater in magnitude (and is often much greater) than the relative speed vex.
Practice Problem: Find the rocket’s velocities 2.0 s and 3.0 s after launch. Answers: 81 m/s, 122 m/s.