9.7 Rocket propulsion  (Page 4/8)

Rocket in a gravitational field

Let’s now analyze the velocity change of the rocket during the launch phase, from the surface of Earth. To keep the math manageable, we’ll restrict our attention to distances for which the acceleration caused by gravity can be treated as a constant g .

The analysis is similar, except that now there is an external force of $\overrightarrow{F}=\text{−}mg\widehat{j}$ acting on our system. This force applies an impulse $d\overrightarrow{J}=\overrightarrow{F}dt=\text{−}mgdt\widehat{j}$ , which is equal to the change of momentum. This gives us

and so

where we have again neglected the term $d{m}_{g}dv$ and dropped the vector notation. Next we replace $d{m}_g$ with $\text{−}dm$ :

Dividing through by m gives

and integrating, we have

Unsurprisingly, the rocket’s velocity is affected by the (constant) acceleration of gravity.

Remember that $\text{Δ}t$ is the burn time of the fuel. Now, in the absence of gravity, [link] implies that it makes no difference how much time it takes to burn the entire mass of fuel; the change of velocity does not depend on $\text{Δ}t$ . However, in the presence of gravity, it matters a lot. The − g $\text{Δ}t$ term in [link] tells us that the longer the burn time is, the smaller the rocket’s change of velocity will be. This is the reason that the launch of a rocket is so spectacular at the first moment of liftoff: It’s essential to burn the fuel as quickly as possible, to get as large a $\text{Δ}v$ as possible.

Summary

• A rocket is an example of conservation of momentum where the mass of the system is not constant, since the rocket ejects fuel to provide thrust.
• The rocket equation gives us the change of velocity that the rocket obtains from burning a mass of fuel that decreases the total rocket mass.

Key equations

Conceptual questions

Problems