Chapter 7: Q.50 (page 180)

A rocket burns fuel at a rate of $5.0 k g / s$ , expelling the exhaust gases at a speed of $4.0 k m / s$ relative to the rocket. We would like to find the thrust of the rocket engine.
a. Model the fuel burning as a steady ejection of small pellets, each with the small mass $∆ m$ . Suppose it takes a short time $∆ t$ to accelerate a pellet (at constant acceleration) to the exhaust speed $v_{e x}$ . Further, suppose the rocket is clamped down so that it can’t recoil. Find an expression for the magnitude of the force that one pellet exerts on the rocket during the short time while the pellet is being expelled.
b. If the rocket is moving, $v_{e x}$ is no longer the pellet’s speed through space but it is still the pellet’s speed relative to the rocket. By considering the limiting case of $∆ m$ and $∆ t$ approaching zero, in which case the rocket is now burning fuel continuously, calculate the rocket thrust for the values given above.

Short Answer

Expert verified

(a) Thrust $= 20000 N$ ; Force expression for one pellet, $F = \frac{∆ m}{∆ t} ∆ v$

(b) Thrust in outer space with burning fuel continuously, $56000 N$

Step by step solution

Given information

fuel burn rate $\frac{∆ m}{∆ t} = 5 k g / s$

exhaust gas speed $v_{e x} = 4.0 k m / s = 4000 m / s$

mass of the pellet $∆ m$

short time of the pellet

Explanation (part a)

Thrust of the rocket $= \frac{∆ m}{∆ t} v_{e x} = 5 \times 4000 = 20000 N$

At initial condition, pellet is at rest with respect to rocket and its velocity changes by $∆ v$ within a short period of time $∆ t$ . which is shown in free body diagram.

The equation for the magnitude of force exerted on the pellet is as follows:

$F = ∆ m a = ∆ m \frac{∆ v}{∆ t}$

Explanation (Part b)

If the rocket is moving, $v_{e x}$ is no longer the pellet’s speed through space but it is still the pellet’s speed that becomelocalid="1649862760382" $∆ v = e s c a p e s p e e d = 11.2 k m / s$ relative to the rocket. therefore the thrust equation is given by,

$F = \frac{∆ m}{∆ t} ∆ v$

By considering the limiting case of $∆ m$ and $∆ t$ approaching zero, in which case the rocket is now burning fuel continuously, calculate the rocket thrust for the values given above.

$F = 5 \times 11200 = 56000 N$