Lesson Video: Rocket Propulsion

In this video we learn about rocket thrust and rocket speed through the Ideal Rocket Equation.

12:45

Video Transcript

In this video, we’re going to learn about rocket propulsion, that is the specific phase of rocket flight when it’s burning fuel. We’ll see how to calculate the thrust that a rocket’s engine produces as well as how to solve for a rocket’s instantaneous speed while it’s accelerating.

To start out, imagine that you’re in a rocketry competition. The rules of the competition are simple. First, every rocket must use liquid fuel. And second, the rocket that accelerates from rest to a speed of 100 meters per second over the shortest amount of time wins. You want to know what’s the minimum amount of fuel you can put in your rocket so that it reaches the speed of 100 meters per second while carrying the least weight. To answer this question, we’ll want to know something about rocket propulsion.

When it comes to rocket thrust, or rocket propulsion, this effect is due to Newton’s third law of motion. When hot, combusted gases are expelled from the bottom of a rocket’s engine, these gases produce a force. This force, which is experienced internally in the rocket’s combustion chamber, is counterbalanced by a thrust force pushing the rocket forward. By the third law of motion, these forces are equal and opposite.

Rockets create this force internally. By channeling and directing the high-speed exhaust gases, the rocket is essentially shooting mass at high speed out of its engines. And the third law takes care of pushing the rocket forward. The fuel that a rocket burns can be liquid or solid. When the fuel is ignited in the combustion chamber, it starts to burn and, as a gas, is released out the bottom of the engine.

Over time, of course, the fuel is depleted. So, the mass of the overall rocket in flight is also constantly changing while its engines are running. This makes understanding the motion of a rocket whose engines are running more difficult. We’re talking about an object whose mass is changing in time. One thing that doesn’t change in time though, so long as there’s fuel to be burned, is the thrust created by a rocket’s engine.

If we measure the speed of the gases escaping from the combustion chamber, we could call that 𝑣 exhaust. If we multiply that speed by the time rate of change of the fuel in the rocket’s engine, then that product is equal to the thrust in newtons produced on the rocket. So, the faster a rocket is able to expel the combusted gases and the faster it goes through fuel mass, the greater the thrust produced will be.

Now let’s go back to the question we started with, which had to do with the minimum amount of fuel we could put in our rocket so that it would achieve a certain speed. Starting with our equation for the thrust produced by a rocket’s engine, if we make the simplifying assumption that there’s no gravity involved acting on the rocket, then by Newton’s second law of motion, we can say that the thrust force created by the rocket is equal to the rocket’s mass times its acceleration.

If we recall that an object’s acceleration equals the time rate of change of its velocity, then we see that 𝑑𝑡 appears in the denominator on both sides of this equation. And therefore, we can cancel it out. Looking at this result, if we then divide both sides of our equation by the mass of our rocket 𝑚, that term cancels on the right-hand side of our equation. And we’re left with 𝑣 sub exhaust times 𝑑𝑚 over 𝑚 is equal to 𝑑𝑣.

We can integrate this expression with mass on the left side and velocity on the right side. We’ll integrate our mass from the initial mass value of the rocket plus its fuel to the final mass value. And we’ll integrate velocity from whatever the initial value is to its final speed. When we do this, knowing that the integral of 𝑑𝑚 over 𝑚 is the natural log of 𝑚, and when we plug in for our beginning and final points for both mass and velocity, we found a relationship known as the ideal rocket equation. Why ideal? Because we assumed no gravity.

In a situation where a rocket is in deep space, far away from any significant mass, this is a reasonable assumption. But depending on the relative magnitude of the gravitational force on a rocket to the thrust force produced by that rocket, this may not always be a valid assumption to make. But when it is valid, we found a relationship that helps us understand the speed of a rocket as its mass changes in time. This is a very helpful result for understanding rocket propulsion. Now let’s get some practice with rocket propulsion through a couple of examples.

A rocket of mass 1200 kilograms is in deep space. The rocket accelerates from a speed of 0.80 kilometers per second to a speed of 1.0 kilometers per second in a time of 5.0 seconds. Using 50 kilograms of fuel, what must be the exhaust speed to produce this acceleration?

We can name the exhaust speed we want to solve for 𝑣 sub 𝑒𝑥. And since we’re told that the rocket is in deep space, that means we can essentially neglect the force of gravity on the rocket. Therefore, the ideal rocket equation applies to describe its motion. This equation tells us that the change in a rocket’s speed is equal to the exhaust velocity of its engines times the natural log of its initial mass over its final mass, including the mass of any lost fuel.

Since we want to solve for the exhaust speed, we rearrange the ideal rocket equation to isolate this term on one side. Regarding the change in the rocket’s velocity Δ𝑣, we’re told the rocket accelerates from 0.80 to 1.0 kilometers per second. In units of meters per second, that’s a change of 200.

Now we consider the initial and final mass of our rocket, including the fuel. The overall mass of the rocket and fuel after the acceleration happened is given as 1200 kilograms. And since we used 50 kilograms of fuel during the speed-up, that must mean the initial mass was 1200 plus 50, or 1250 kilograms.

Plugging in these values for initial and final mass, we see the units of kilograms cancel out. And we’re ready to calculate the exhaust velocity, 𝑣 sub 𝑒𝑥. To two significant figures, it’s equal to 4.9 times 10 to the third meters per second, or 4.9 kilometers per second. That’s the exhaust speed needed to produce this acceleration.

Now let’s work through an example involving calculating rocket thrust.

A spacecraft is moving uniformly in deep space, where gravitational forces are negligible. At the instant 𝑡 sub zero, the spacecraft’s engine is activated for a 30-second time interval, during which exhaust is ejected from the spacecraft at a rate of 2.0 times 10 to the two kilograms per second with the exhaust moving at a speed of 2.5 times 10 to the two meters per second. The spacecraft’s mass before ejecting any exhaust is 2.0 times 10 to the fourth kilograms. What magnitude force is applied to the spacecraft by the engine while it is activated? What is the magnitude of the spacecraft’s acceleration at the instant 𝑡 sub zero? What is the magnitude of the spacecraft’s acceleration at the instant 𝑡 equals 15 seconds? What is the magnitude of the spacecraft’s acceleration at the instant 𝑡 equals 30 seconds?

In this four-part exercise, we can label the magnitude force applied by the spacecraft during activation, 𝐹 sub 𝑒. We can call the acceleration of the spacecraft at time 𝑡 equals 𝑡 sub zero 𝑎 of 𝑡 sub zero. And we can make similar labels for when 𝑡 equals 15 and 𝑡 equals 30 seconds.

In the problem statement, we’re told the time rate of change in kilograms per second of fuel being burned by the spaceship. We’re also told the speed of the exhaust gases coming out of the engine and the initial mass of the spaceship. Because of the activation of the engine creating exhaust gases and a fuel burn rate, we know that there’s a thrust force created by the spaceship. This thrust force is equal to the product of the exhaust gas speed times the time rate of change of mass.

Since we’re given both of these values, we can plug in and solve for 𝐹 sub 𝑒. When we do and enter this product on our calculator, we find, to two significant figures, it’s equal to 5.0 times 10 to the fourth newtons. That’s the thrust force that the spacecraft produces.

Next, we wanna solve for the acceleration of the spacecraft at various moments during its engine activation. We recall Newton’s second law of motion that says that the net force acting on an object equals its mass times its acceleration. When we write the forces acting on our spaceship, we’re told that the force of gravity is negligible. So, it’s the thrust force, 𝐹 sub 𝑒, that equals the spacecraft’s mass times its acceleration. We’ll have to be careful with this equation though because we know the mass of our spacecraft is not constant. It’s changing as fuel is burned up. And therefore, its acceleration is not constant either.

When 𝑡 is equal to 𝑡 sub zero, the acceleration of our spacecraft is equal to the thrust force divided by the initial mass, 𝑚 sub 𝑖, of the craft altogether. We solved for 𝐹 sub 𝑒 in part one. And we’re given the mass of the spacecraft initially. We’ve called it 𝑚 sub 𝑖. This fraction equals 2.5 meters per second squared. That’s the spacecraft’s acceleration at the outset of engine activation.

Next, we want to solve for the acceleration of the spacecraft after its engines have been firing for 15 seconds. This is equal to the thrust force, 𝐹 sub 𝑒, divided by the mass of the spacecraft at this point in time. We’ve called it 𝑚 sub 15, where 𝑚 sub 15 is equal to the initial mass of the spacecraft minus 15 seconds times the burn rate 𝑑𝑚 𝑑𝑡.

When we plug in for 𝑚 sub 𝑖 and for 𝑑𝑚 𝑑𝑡, we find that the mass of the spacecraft after 15 seconds of fuel burning is 17000 kilograms, down from its original 20000. When we plug in for 𝐹 sub 𝑒 and 𝑚 sub 15, calculating this fraction to two significant figures, we find a result of 2.9 meters per second squared. That’s the spacecraft’s acceleration 15 seconds after turning on its engines.

Finally, we wanna solve for the acceleration of the spacecraft 30 seconds after engine activation. This is equal to the thrust force, 𝐹 sub 𝑒, divided by the mass of the spacecraft at that point in time. 𝑚 sub 30 is equal to the initial mass of the spacecraft minus 30 seconds times the fuel burn rate 𝑑𝑚 𝑑𝑡. This is equal to 14000 kilograms. Plugging in for 𝐹 sub 𝑒 and 𝑚 sub 30, we find an acceleration of 3.6 meters per second squared. That’s the spacecraft’s acceleration after its engines have been running for 30 seconds.

Let’s summarize what we’ve learned about rocket propulsion. We’ve seen that rocket propulsion is based on Newton’s third law of motion. That for every action, say a force from object 𝑎 on object 𝑏, there’s an equal and opposite reaction, the force of object 𝑏 on object 𝑎. We’ve also seen that the thrust force generated by a rocket, we’ve called it 𝑡, is equal to the exhaust speed of the gases coming out of the rocket’s engine multiplied by the fuel mass burn rate. Written as an equation, 𝑡 is equal to 𝑣 sub 𝑒𝑥 times 𝑑𝑚 𝑑𝑡.

And finally, we’ve seen that the ideal rocket equation, which ignores the forces due to gravity, says that the change in a rocket’s speed, Δ𝑣, is equal to its exhaust gas speed times the natural log of the initial mass of the rocket, rocket plus fuel, divided by its final mass. Understanding rocket propulsion is greatly helped by keeping in mind the two equations that account for thrust as well as rocket’s speed as the mass of the rocket changes.