Rocket Equations

This summer I decided that I would spend some time with my 13-year-old daughters building model rockets. The process of doing this would teach them how to measure and build as well as engineer something that actually did something cool such as fly.  Building rockets is a good way to study a wide range of physics, engineering, math and science. I built rockets when I was a kid and it was such great fun I thought I would do it again.

I brushed up on my rocket skills and read the official NAR handbook. In the book they described rocket equations to estimate the height and velocity that a rocket will go theoretically ignoring aerodynamic drag. As I started looking at the equations I was slightly confused as to how they were derived. I looked around on the Internet and did not see a clear  representation of the math in one place that my kids would understand. My daughters have had basic algebra but most of the derivations on the net leave a lot of the steps out. I decided to derive these equations by brushing off my college physics book and going through it in a very verbose way. I’m writing about this because along with my daughters I thought it would help others who are teaching younger students on the basics of the equations of motion. 


Between reading the NAR handbook and this you will have a good understanding of the forces on a rocket.  The equations of motion are well documented in general. I’m going to glue together the basics of finding out how fast a model rocket will go and how high theoretically it will go. We will calculate the velocity of the rocket after its initial burn and then after the coasting of the rocket in the air. Remember this only works ignoring aerodynamic drag which is a very big piece of how a rocket flies.  Later I may address the more complex math but for now we’ll start with the easy stuff.

By definition these equations work with constant acceleration. We use Newton’s laws of motion.  All of Newton’s laws work on a rocket and affect how it flies. Remember this is ideal.  For much better real world models use RockSim or OpenRocket.  

Given the graph below with a line in a coordinate plane where Y is equal to velocity and X is equal to time. In this basic graph of velocity versus time I have defined acceleration as the change in velocity over the change in time. If we add up the area under the line this would be the area of a rectangle plus the area of the triangle. This area is called the displacement. For our equations this will be the altitude of a rocket.

Again looking at the graph of the line of acceleration as:

Remember that acceleration is the slope of the graph.  Rearranging gives us:

This final equation looks a lot like the basic equation for a straight line that everyone is used to.

We want to find the area under the line which is our displacement. In order to do this accurately we will use calculus and integrate velocity with respect to time.

Plugging in equation 1 to the integration and solving gives us:

By integrating we add up all the area under the line, that’s all integration really is. “s” is the disposition.  Physicist use s for some reason (note s = d in our graph).  This equation, we will call equation 2, has acceleration in it. In order for us to use it for our rocket equation we need to get acceleration out by substituting our definition of acceleration. We’re going to do some substitution and simplify.

The next equation we will take equation 3 and eliminate time.  The algebra on this is a little messy but I have all the steps and I’m using the binomial theorem.  Substitute t in equation 3 and simplify.

After all that work equation 3 is our altitude at burnout and equation 4 is our coasting altitude.

In order to get the altitude at burn out we will need the velocity at the time our motor runs out of burn. In order to get this we will use Newton’s Second Law F=ma:

T = thrust or force to put the rocket in the air 

t = motor burn time

g = acceleration due to gravity. 9.8m/s^2

wavg = average weight of rocket

vm =   maximum velocity during motor burn

In order to use this law we need to include gravity with mass to figure the force on our rocket and that is the gravity for acceleration. Given our equation above for velocity we will substitute.

Finally we can calculate some ideal values for our rocket. Given the picture of our rocket below lets figure out our altitude at burnout and our maximum altitude theoretically.

Using my kids Alpha rocket with an A8-3 motor in it weighs 38.8 grams at lift off. The weight of the propellant is 3.12 grams giving an average weight of 37.2 grams during the thrust of the flight. The motor thrusts for .32 seconds and has an impulse of 2.50 Newton seconds. In our equations you will want to convert grams to kilograms as the standard unit of measurement.

The maximum velocity of our rocket ideally will be 64.03 m/s or 143 mph.  Now let’s figure our altitude after the rocket motor burns.

Using equation 3:

Our rocket altitude at motor burnout is 10.24 meters or 32 feet above the ground theoretically.

Using equation 4 we can find out the total altitude theoretically that our rocket will go.

Our rocket theoretically will go 219.22 m or 719 feet in the air ignoring aerodynamic drag with constant acceleration.

A few notes on this last equation to solve for the coasting altitude. Many of the units I’ve used here are called vectors. They have a magnitude and direction. Gravity has a  downward direction. This is why the 9.8 is in the negative direction. It works out because negative divided by a negative is a positive and altitude needs to be positive. I didn’t mention vectors throughout because it may have been more confusing. Be aware however that a rocket doesn’t go just straight up it has an X, Y and Z coordinate.  We know when we shoot off a rocket it tends to point into the wind which would decrease its overall altitude.