Game Development Reference

In-Depth Information

where
u
is the relative velocity between the expelled mass and the object (the rocket in

this case).

For a rocket traveling straight up, neglecting air resistance and the pressure at the ex‐

haust nozzle, the only force acting on the rocket is due to gravity. But the rocket is

expelling mass (burning fuel). How it expels this mass is not important here, since the

forces involved are internal to the rocket; we need to consider only the external forces.

Let the fuel burn rate be
-m'
. The equation of motion (in the vertical direction) for the

rocket is as follows:

ΣF = m dv/dt + dm/dt u

−mg = m dv/dt − m' u

If you rearrange this so that it looks like there's only an
ma
term on the right of this

equation, you get:

m' u − mg = m dv/dt = ma

Here you can see that the thrust that propels the rocket into the air is equal to
m'u
. Since

the fuel burn rate is constant, the mass of the rocket at any instant in time is equal to:

m = m
o
− m' t

where
m
o
is the initial mass, and the burn rate,
m'
, is in the form mass per unit time.