# Newton’s Laws, Part 6

Now let’s use a bigger rocket and remove two of the assumptions we made with the model rocket. If you have not read the previous posts in this series, I suggest that you do before starting cold on this one.

As I said before, the assumption that the rocket’s mass stays the same throughout its flight is not realistic for larger rockets. The fuel is the majority of the mass of a rocket and that is spent as the rocket ascends. The assumption that gravity is constant throughout a rocket flight is also not realistic as gravity does get weaker the higher the rocket goes.

Let’s assume a hypothetical rocket that has a fuelled mass of 16,000 kg. The unfuelled rocket (the rocket structure and payload) weigh 550 kg. We will continue to assume a constant thrust (again not realistic) of 2,800,000 N that lasts 270 seconds. As the fuel burns, let’s assume a constant rate of fuel being ejected. The mass of the rocket, as time t increases is:

m(t) = 16000 – 57.2222 t

this makes sense since at t = 0, the mass of the rocket is 16000 kg and after 270 seconds, the mass is 550 kg. Note that this mass equation is only valid during the thrust phase of the rocket. As before, we have to deal with two separate phases of the rocket’s flight because the equations to be solved are different in each phase.

The change in gravity requires a little explanation. The acceleration due to gravity is due to the attraction between two bodies (you and the earth). The acceleration due to gravity follows what is called the inverse square law. This means that the acceleration due to gravity decreases by the square of the distance between them.

Now the distance between you and the center of the earth is about 6400 km. We use the center of large objects from which to measure distance. As our rocket travels up, this distance is getting greater so gravity is becoming weaker. If x(t) is the distance from the launchpad after t seconds, then g(t), the acceleration due to gravity at t seconds is:

${g}{(}{t}{)}\hspace{0.33em}{=}\hspace{0.33em}{9}{.}{8}{\left({\frac{6400}{{6400}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}$

So during the burn phase, there are two forces acting in the rocket: gravity and the rocket engine. We will follow the same convention as before, anything pointing down is negative, and up is positive. So for phase 1, Newton’s second law, F = ma is

$\begin{array}{l} {{F}{=}\hspace{0.33em}{2}{,}{800}{,}{000}\hspace{0.33em}{-}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{9}{.}{8}{\left({\frac{6400}{{6400}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}}\\ {{=}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{a}} \end{array}$

As you can see, removing those two assumptions really complicates the equation. Things get a little simpler for phase 2, after the engines burn out because the mass is now constant at 550 kg and this can be divided out from both sides of the equation:

$\begin{array}{c} {{F}{=}{-}{550}\hspace{0.33em}\times\hspace{0.33em}{9}{.}{8}{\left({\frac{6400}{{6400}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{550}\hspace{0.33em}\times\hspace{0.33em}{a}}\\ {\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{-}{9}{.}{8}{\left({\frac{6400}{{6400}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{a}} \end{array}$

These equations from both phases are rather nasty ones. To get technical, these are non-linear second order differential equations. You can see why rocket scientists make the big bucks.

So this post is long enough so I will say more about these equations next time.

Posted on Categories Physics, Pre-VCE