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.
Search Nedrilad ::




Custom Search