Written By Keshav Mohta (Grade 11)

Launching objects into outer space is expensive. Very expensive. But why so?

Launching a rocket means giving it energy. Raising an object into the atmosphere requires energy that we term as gravitational potential energy (GPE). Imagine you are lifting a heavy 10 kg dumbbell high in the air and then keep it stationary. You used energy in your body to do this lifting. This energy got transferred to the dumbbell as GPE. To launch a 1 kg object into low earth orbit (LEO), we can calculate how much GPE is required:

Where:

*m*is the mass of the object*g*is the acceleration due to Earth’s gravity*h*is the height the object has been raised (here we assume 400 km to be the altitude)

Below is a calculation from the amount of GPE that needs to be provided to lift the 1 kg object.

But, to keep an object in orbit, it also must have velocity. Kinetic energy (KE) shows the energy of moving objects. Throwing a tennis ball laterally requires energy. Your arms provide this energy for the tennis ball to move, resulting in the moving tennis ball now possessing KE. Similarly, we must supply KE to the 1 kg object as well, to allow it to start orbiting the Earth.

Where:

*m*is the mass of the object*v*is the velocity with which the object is moving

To orbit the Earth at a height of 400 km, an object requires a velocity of approximately 8 km/s, that is 8000 m/s. Going into the mathematical detail for this calculation is beyond the scope of this article. We can calculate the amount of kinetic energy needed to keep this object orbiting.

And thus, the total energy needed to put a 1 kg object in space is:

And so, it should cost only ₹85.5 to produce the energy needed to place a 1 kg object into orbit around the Earth. In reality, the cost to put 1 kg in orbit is approximately $20,200 which works out to ₹17,17,000. So, what explains this disproportionate increase in rocket launching costs?!

It is very simply FUEL! We haven’t accounted for the fuel that is required to put the 1 kg object in orbit. We also must carry that fuel up, which increases the total mass of the rocket. Now, we need more fuel to carry the already needed fuel + the 1 kg object. And very soon, you see an exponential growth in the quantity of fuel needed to put a 1 kg object into orbit. This is what makes rocket launches so expensive.

To calculate the true fuel requirements for launching into space, we can use the Ideal Rocket Equation (often referred to as just the Rocket Equation).

We can split the initial launch mass of the rocket into three components: the mass of propellant (m_{p}), mass of payload (m_{pay}) and mass of the rocket’s structure (m_{s}).

Initially when the rocket is at the launchpad its mass is: m_{i} = m_{p} + m_{pay} + m_{s}.

After the rocket is launched into the sky and is orbiting, its mass is: m_{f} = m_{pay} + m_{s}. This is because all the propellant has been used to provide gravitational potential energy and kinetic energy to the rocket.

Therefore, we can state the following: m_{f} = m_{i} – m_{p}.

When a rocket launches, it uses the basic principle of Newton’s 3^{rd} Law. It pushes propellant out at a high velocity downwards, leading to an equal and opposite upward force on the rocket.

Using the formula for force from Newton’s 2^{nd} law, that is force equals a change in momentum divided by a change in time, we can calculate the thrust produced by the rocket engine. Momentum (p) is simply mass times velocity. Changing requires a change in mass or a change in velocity. In this case mass is expelled from the rocket engine at a certain velocity. This gives us the following formula for the thrust on the rocket:

Where:

- dm/dt is the rate at which mass (of propellant) is expelled from the rocket
- v
_{ex}is the speed with which propellant is expelled from the rocket

It is important to note the negative sign with v_{ex} because the exhaust velocity is in the opposite direction to the force acting on the rocket.

Additionally, Newton’s 2^{nd} law also states that *F = ma*, where *m* is the instantaneous mass of the rocket at any given time and *a* is the acceleration.

Conceptually, we can vary either mass with time, or velocity with time to create force on an object. Both formulae should give us the same result. We can now do some algebra to simplify this.

Where:

- Δv is the change in velocity the rocket experiences
- v
_{ex}is the velocity with which propellant is expelled out of the rocket - m
_{i}is the initial launch mass of the rocket - m
_{f}is the final mass of the rocket once it gets to orbit and has used all propellant

This is the fundamental rocket equation. Through further algebra, we can try to find the ratio of propellant to initial mass.

The ratio m_{p}:m_{i} is also called the propellant fraction. It tells us what proportion of the total mass of the rocket must be propellant to create a certain velocity change in the rocket.

The most important variable in this equation is v_{ex} and it is the primary driver of the mass of propellant you need. On average, rockets have v_{ex} between 2000 to 4500 m/s.

Different v_{ex} can affect how much propellant is required to put a payload into orbit. The typical Δv required to reach low earth orbit (LEO) is approximately 9400 m/s. For a rocket with v_{ex} = 2000 m/s, we can calculate the propellant fraction.

This means that 99% of the rocket’s launch mass must be propellant, and only 1% can be used for the structure and payload. This is of course very impractical. But what happens if we double the exhaust velocity.

This is far more doable. Only 90% of the rocket’s launch mass must be propellant and 10% can be used for the structure and payload. The key here is that the rocket equation is exponential. Changing the exhaust velocity by a factor of 2, changed the non-propellant fraction by a factor of 10.

In conclusion, the Rocket Equation shows us how by slightly improving the engine efficiency, thus increasing exhaust velocity, we can make rocket launches far cheaper. It is the key to cheaper space travel.

Featured Image Courtesy – Revoluntised