A Classical Approach of Newtonian Mechanics

6 CONSERVATION OF MOMENTUM 6.4 Rocket science

!

∫

M

f

p

assumed to be in outer space. It follows that the total momentum of the system is
a constant of the motion. Hence, we can equate the momenta evaluated at times

t and t + dt:

(M + m) v = (M + m + dm) (v + dv) + (−dm) (v − u). (6.24)

Neglecting second order quantities (i.e., dm dv), the above expression yields

0 = (M + m) dv + u dm. (6.25)

Rearranging, we obtain

dv dm

= −

u M + m

. (6.26)

Let us integrate the above equation between an initial time at which the rocket

is fully fueled—i.e., m = mp, where mp is the maximum mass of propellant that
the rocket can carry—but stationary, and a final time at which the mass of the

fuel is m and the velocity of the rocket is v. Hence,

v dv m

= − dm^.^ (6.27)^

It follows that

0 u^

"

v

#v=v

mp M^ +^ m^

(^) m=m
which yields
u (^) v=0 (^) = − [ln(M + m)]
m=m ,^ (6.28)^
v = u ln
M + mp

. (6.29)
M + m
The final velocity of the rocket (i.e., the velocity attained by the time the rocket

has exhausted its fuel, so that m = 0 ) is

v = u ln

1 +

mp

!

. (6.30)

Note that, unless the initial mass of the fuel exceeds the fixed mass of the rocket

by many orders of magnitude (which is highly unlikely), the final velocity vf of

the rocket is similar to the velocity u with which fuel is ejected from the rear

of the rocket in its instantaneous rest frame. This follows because ln x ∼ O(1),

unless x becomes extremely large.

∫