I have 2 questions about the cost of liquid rocket fuel to launch (anything) to LEO orbit. Let's assume we use fuels used in commercial spaceflight today (probably LOX+LH2 or LOX+RP-1 etc.).

  1. What is the theoretical fuel cost to launch 1 kg of payload to orbit on an ideal rocket (rocket with 0 kg dry mass)?

  2. What is the fuel cost of todays (most economical) launch systems (probably SpaceX Falcon 9, Russian Proton etc.) per kilogram?

I am not asking about the total cost to orbit (rocket + fuel) but rather the fuel cost only. There are some relevant questions, which are not duplicates of mine though (they neither ask nor answer what I want):


2 Answers 2


The Falcon 9 burns somewhere around \$200k-300k in propellant (stated to be \$200k in 2015, but the vehicle's grown in size since then). For non-expendable launches, it puts about 16000 kg into orbit, so that's about \$20/kg.

Starship burns cheaper methane fuel, and propellant cost is estimated at about \$500k/launch when purchased in volume. Total payload early on is probably going to be closer to 100 t than 150 t, so that'd be \$5/kg.


What is the theoretical fuel cost to launch 1 kg of payload to orbit on an ideal rocket (rocket with 0 kg dry mass)?

We can use the rocket equation to get a rough idea of the fuel required.

$$\delta V = v_e ln \frac{m_0}{m_f}$$

  • $\delta V$ required to reach LEO is 9.4 km/s
  • $v_e$ is the exhaust velocity of the rocket, 3 km/s is pretty good for a chemical rocket
  • $m_0$ is the initial total mass, including fuel.
  • $m_f$ is the final mass, 1 kg.

We need to solve for $m_0$.

$$m_0 = m_f e^{\frac{\delta V}{v_e}}$$

Plugging in the numbers...

$$m_0 = 1 \text{ kg} \times e^{\frac{9.4 km/s}{3 km/s}}$$

$$m_0 = e^{3.13} \text{kg}$$

$$m_0 = 23 \text{ kg}$$

An initial mass of 23 kg means 22 kg of fuel to get a 1 kg of payload on a zero mass rocket to LEO.

According to this answer a Falcon 9 uses 2:1 LOX to RP-1 so that's about 14 kg LOX and 7 kg RP-1. And they say LOX is about \$0.20/kg while RP-1 is \$1.20/kg.

  • 14 kg of LOX at $0.20/kg is \$2.80.

  • 7 kg of RP-1 at $1.20/kg is \$8.40.

About \$11. Though so little probably won't get you SpaceX's bulk discount.

However chemical rockets are used for lift off because they have the necessary oomph to lift the many tons of rocket, fuel, and payload against the force of gravity. With just 1 kg you might be able to get away with a more efficient, but less powerful, method of propulsion.

1 kg in Earth gravity exerts only 10 N of force. Our most efficient engines are ion thrusters. There's a whole raft of reasons it's a bad idea to use these inside an atmosphere, but let's say they do work. The problem remains that existing ion thrusters have thrusts measured in micro Newtons. Some Magnetoplasmadynamic thrusters (MPDT) on the drawing board can, theoretically, provide the necessary thrust.

Let's assume we have a zero mass MPDT with enough thrust to lift 1 kg. How much fuel would it need? These have exhaust velocities up to 60 km/s.

$$m_0 = 1 \text{ kg} \times e^{\frac{9.4 km/s}{60 km/s}}$$

$$m_0 = e^{0.157} \text{kg}$$

$$m_0 = 1.17 \text{ kg}$$

1.17 kg initial mass means 0.17 kg of fuel to lift 1kg of mass into orbit. Our hypothetical zero-mass MPDT would need about 12 N of trust to lift its payload an fuel. That is inside what we believe to be achievable with an MPDT (though zero mass and operating inside an atmosphere is not).

However, this is 0.17 kg of a noble gas. Traditional ion thrusters use Xenon propellant. At roughly \$850/kg we're looking at about \$150. However, MPDTs could use much cheaper propellant such as helium, hydrogen, or lithium.

Unlike chemical rockets, ion thrusters use electricity to accelerate ions. They need a power source. Typically these are solar panels, but an MPDT requires far more power such as a small nuclear reactor or power beamed via ground-based lasers. We would also have to assume the power source is zero mass.

Let's put this to the limit. What if the exhaust velocity was the speed of light, a photon rocket! Let's be clear, this is like trying to move your car with a flashlight. There is no way it has enough thrust to launch 1 kg, this is just an exercise.

$$m_0 = 1 \text{ kg} \times e^{\frac{9.4 km/s}{300,000 km/s}}$$

$$m_0 = e^{0.0000313} \text{kg}$$

$$m_0 = 1.00003 \text{ kg}$$

A photon rocket needs 0.03 grams of fuel to lift 1kg of payload to LEO. That is the hypothetical best we could do assuming we can build a zero mass rocket.

  • 2
    $\begingroup$ You need a zero-mass power source providing hundreds of kilowatts to go along with that zero-mass MPDT. Also, MPDTs are designed to operate in vacuum, ionizing and accelerating propellant at much lower pressures. These high specific impulse engines aren't generally going to work as well or at all within an atmosphere. $\endgroup$ Commented Jan 15, 2021 at 22:14
  • 1
    $\begingroup$ @ChristopherJamesHuff Yes, it's a numbers exercise, but the electricity required by an ion thruster (does that count as "fuel"?) is worth mentioning. $\endgroup$
    – Schwern
    Commented Jan 15, 2021 at 22:39
  • $\begingroup$ Yes, from what I remember the most ambitious single stage to orbit proposals or calculations give numbers very similar to this; 95 to 96% rocket+fuel, 5 to 4% payload. update: Found it! Don Petit's The Tyranny of the Rocket Equation Also see Is there an absolute limit to single stage ∆V when tank mass is a fixed faction of propellant mass? $\endgroup$
    – uhoh
    Commented Jan 15, 2021 at 23:37
  • $\begingroup$ "There is no way it has enough thrust to launch 1 kg, this is just an exercise." Theoretically, a photon rocket could launch 1kg at 1g. You would just need to multiply the force the rocket is generating by the speed of light to determine the amount of energy the laser would need to consume, IIRC, so it would take an extremely powerful laser consuming an ungodly amount of power to do so. $\endgroup$
    – nick012000
    Commented Jan 16, 2021 at 3:53
  • 1
    $\begingroup$ How did you get the 500 and 100 Dollar numbers? That would be >> 10 million for a Falcon 9 launch. $\endgroup$
    – asdfex
    Commented Jan 16, 2021 at 9:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.