Well congratulations! If you get through (and understand) this and the last three posts on Newton’s laws, consider yourself a rocket scientist third class.
In my high school days, my friend Byron and I formed a two-member rocket club. We would pool our money and get and build model rockets. We also got solid propellant cartridges that were inserted into the rocket. We would put the rocket on a launch pad (a wooden block) and ignite the propellant, either with a fuse or electrically with a small coil of wire that was heated with an electric current. The rocket would ascend very fast, then a small charge at the end of the propellant cartridge would push out the nose cone and the folded parachute inside. Then the fun part was to try and retrieve the rocket, especially on windy days!
So today, I am going the find the equation of motion and the velocity equation of these rockets using Newton’s second law, F = ma.
We are going to make several assumptions. I will relax some of these later, but seeing that you are only a third class rocket scientist, let’s start out making things as simple as possible.
- The force exerted by the cartridge is constant from ignition to the time it cuts off.
- The mass of the rocket is constant throughout its flight.
- The rocket is going straight up and down. (We will not be using a parachute.)
- The flight time is short so we will not take into account the rotation of the earth.
- The force due to gravity is constant throughout the flight.
- Air drag is ignored.
A first class rocket scientist would not be making these assumptions and would include the effects of these into the equations. Assumption 1 is not valid for real rockets and the varying force would have to be accounted for.
Assumption 2 would make the equations very inaccurate in reality. Depending on the type of propellant, the rocket fuel can be from 83% to 96% of the total mass of the rocket. As the fuel is burned, the mass of the rocket is getting less and less, and as you can see from F = ma, this means that the acceleration must increase if m×a is to continue equalling the constant force.
Assumption 3 is made so that the velocity, acceleration, and position are measured along the same line. For big rockets, measurements of these three things are measured with something called vectors which provide a direction as well as a value.
Assumption 4 is made because the earth does rotate and this rotation adds to the speed, not in the vertical direction but in a sideways direction.
Assumption 5 is OK for our model rocket because it will not go that high, but the force of gravity does lessen with altitude and this must be taken into account for larger rockets.
Assumption 6 allows us to keep the equations simple (as do all of them).
Now let’s look at an actual model rocket. These numbers are realistic numbers that are possible for model rockets.
Suppose we launch a model rocket that weighs 400 g. The engine provides 9 N of force for 3.3 s. Let’s answer the following questions:
- How high does the rocket go?
- What is the maximum speed of the rocket?
- What is the total flight time?
Before I set this up, let’s agree on what’s positive and what’s negative. Let’s agree that up is positive and down is negative. So the force due to the rocket engine is positive because it is pushing the rocket up and the force due to gravity is negative because it is pulling the rocket down.
Now there are two phases of the rocket’s flight: the first phase is when the engine is burning and the second phase begins when the engine stops. Let’s look at F = ma during the first phase.
Phase 1 – engine is burning: There are two forces acting on the rocket, the one due to the engine and the other due to gravity. In part 2 of this series of posts, I explained how to find the force due to gravity. The gravity force points down so it is negative. The other force, of course, is due to the rocket engine and it is positive. So putting this into F = ma:
9 N – (0.4 kg)(9.8 m/s²) = 0.4 kg × a
I use 0.4 kg instead of 400 g because we are using SI units (explained in last post). So simplifying this equation and removing the units gives:
5.08 = 0.4a . This is the starting equation for the first phase.
Phase 2 – engine stops: Now let’s look at the second phase. After the engine cuts off, the only force acting on the rocket is gravity. So the only acceleration the rocket is experiencing is only due to gravity. So F = ma becomes:
– (0.4 kg)(9.8 m/s²) = 0.4 kg × a
But you can see from this that a = -9.8 m/s² to make this a true equation. This simple equation is the starting equation for the second phase.
This post is now longer than I thought it would be, so I will continue with this in my next post.