# Rocket Equation and Energy

#### PorCesar Cisneros

Jul 27, 2022

Rockets are limited in size. You can only put a certain amount of propellants on board.

So there’s a real question of how much performance you can get out of a given amount of propellant you can load on board.

This is where we get a bit into the math.

Why is it so difficult to get to space?

It takes a lot of energy to get there.

How much energy do you really need to put 1 kilogram into orbit?

Well, if you remember your high school physics classes, there are two forms that energy can take. There is potential energy you need to lift 1 kilogram off the surface of the Earth.

Let’s say we’re going about 400 kilometers and let’s not pay attention to the fact that Earth’s gravity is an inverse square law so it gets slightly weaker as you go, but let’s say it’s 9.8 meters by second squared, which is what is here on the surface of the Earth.

So if we take 1 kilogram, the potential energy you need to lift it a certain distance is mass * gravitational acceleration * height – (mgh)

And we calculate the numbers here.

1 kilogram

9.8 meters per second

400,000 meters. We’re using the MKS system and we’re getting about 3.9 megajoules of energy.

Now the other energy we need is kinetic energy because if we just put something 250 miles up in the air and drop it, it’s going to fall back to Earth.

We will prepare another blog on orbital mechanics later.

But in general, to stay in orbit, you need to go at a reasonably high speed, only about 8 kilometers per second, to be in orbit a few kilometers above the surface of the Earth.

And we know from your high school physics classes that kinetic energy is half mass times speed squared.

So, we take the mass of 1 kilogram and square the speed of 8 kilometers per second.

So, we can see in low earth orbit that kinetic energy dominates over potential energy, now, how about we add them up?

We get 36 MegaJoules.

Now we are going to convert energy because 1 Joule per second is equal to one watt.

So, one Joule equals 1 Watt/second, and 36 megajoules equals 3.6 * 10 to the seventh Watt/second.

And we can convert that to Kilowatt-hours, which are the units that we use to measure electrical energy consumption.

Usually, in Mexico, we pay a little more than \$1.70 MexicanPeso  kWh

So 10-kilowatt hours cost around 20 pesos.

That’s the energy it takes to put 1 kilogram into orbit.

And I hope you find it surprising.

Because, if we could put that energy right into that 1 kilogram, it’s very cheap.

It would revolutionize the space business if we could do that.

But of course, we have to use rockets to get to space, and we have to carry the propellant you need to get out in the last stage. So you need propellant to carry the propellant you’re going to use, and then you need more propellant to carry that, and that’s where it gets chaotic.

So this is a surprising result for most people.

It shows what is sometimes referred to as the curse of the rocket equation

because this is what we have to deal with.

We cannot put 1 kilogram into orbit for the cost of \$20 pesos, it costs many millions.

References:

Ecuación del cohete de Tsiolkovski

Energía potencial

Energía cinética

Tarifas CFE 2022 | Precios kWh | Tipos de tarifas domésticas

#### Por Cesar Cisneros

##### Un comentario en «Rocket Equation and Energy»
1. Rubén Rosete dice:

Me hiciste recordar mis clases de Física de hace más de 25 años, gracias