The free motion of a rocket-propelled body

Imagine a rocket-propelled body moving in a region where aerodynamic drag and lift and gravitational force may be neglected, i. e. in space remote from any planets, etc. At time t let the mass of the body plus unburnt fuel be M, and the speed of the body relative to some axes be V. Let the fuel be consumed at a rate of m, the resultant gas being ejected at a speed of v relative to the body. Further, let the total rearwards momentum of the rocket exhaust, produced from the instant of firing to time t, be /relative to the axes. Then, at time t, the total forward momentum is

H{ = MV-I (9.64)

At time (/ + 6t) the mass of the body plus unbumt fuel is (M – m6t) and its speed is (V + 6V), while a mass of fuel m6t has been ejected rearwards with a mean speed, relative to the axes, of (v — V — j6V). The total forward momentum is then

tf2 = {M – m8t){V + 8V) – rh6t{v – V-^6V)-I

Now, by the conservation of momentum of a closed system:

#, =#2

i. e.

MV – I =MV + M6V – mV6t – mStSV – mv6t + mV6t + ^mStSV-I

which reduces to

M6V — – mStSV — mvSt = 0 2

Dividing by 6t and taking the limit as 6t —> 0, this becomes

Note that this equation can be derived directly from Newton’s second law, force = mass x acceleration, but it is not always immediately clear how to apply this law to bodies of variable mass. The fundamental appeal to momentum made above removes any doubts as to the legitimacy of such an application. Equation (9.65) may now be rearranged as

dV _ m