I am confused with energy requirement equations...

Am I anywhere near with my following assumptions:

$\Delta v$ from surface to LEO is 9000 m/s typical exhaust velocity $v_e$ = 4500 m/s

and mass of payload $m_l = 40\mathrm{t}$ and $m_p$ = mass of propellant

$=> \Delta v = v_e \, \ln \frac{m_p+m_l}{m_l}$

$=> m_p = m_l (\mathrm{e}^{\frac{\Delta v}{v_e}}-1)$

$=> m_p = 40 000 (\mathrm{e}^\frac{9000}{4500}-1) = 256 \mathrm{t}$ of propellant

The energy with 100% efficiency would then be:

$E = 0.5 \, m_p \, v_e^2$

or (gets me the same result?)

$E = 0.5 \, m_l (\mathrm{e}^\frac{\Delta v}{v_e}-1) \, v_e^2$

$=> E = 2588 \mathrm{GJ}$

Am I in a right order of magnitude here...?

  • $\begingroup$ You would get the amount of energy with which you accelerated your exhaust gasses. Is that what you want to know? $\endgroup$ Oct 13, 2014 at 22:21
  • $\begingroup$ note that "to" LEO is somewhat confusing. LEO is not a place, but a particular speed. $\endgroup$
    – Fattie
    Oct 14, 2014 at 10:10

2 Answers 2


The 'sending' part of your question is what makes this a tangle. Because sending is a process, it isn't possible to speak in terms of perfect efficiency. You can't isolate the fuel it took to put the payload into orbit from the fuel it took to put the rocket carrying the payload into orbit.

It may be that what you seek is a theoretical minimum energy required to put 40 tons in orbit. In that case maybe it means something to say what the kinetic energy of 40 tons of material moving at orbital velocity is, because if there was a way to get it there with absolutely no energy losses, that is all the energy you would need. Orbital speed at 200km above the Earth, which is where low earth orbit is considered to begin, is 6.9 km/s.

0.5 x 40,000kg x 6900 (m/s) $^2$ = 952.2 GJ

Maybe if there was a space elevator to get above the atmosphere, and you had a way to make the mass of the cargo pod hauling the payload negligible compared to the payload itself, and you were able to minimize losses due to friction, and then the payload could be accelerated sideways to orbital speed with maybe a high-efficiency magnetically levitating mass driver, which had a cargo pod whose mass was also negligible, you could start to approach that amount. Exactly how close we could ever get to that ideal figure is highly debatable.

Thinking about engineering problems in terms of the theoretical ideal has real application. It is good to know exactly how much room there is for improvement sometimes. For instance, it has often been noted that building a rocket and then only using it once is awfully inefficient. But it doesn't really serve a purpose unless you can identify a place where the process can be improved, and design a method to make it so.

Failing that, the amount of energy it currently takes to get 40 tons of material to orbit, stated in terms of the energy in the fuel it takes to launch it, requires that we choose a rocket to do the job and then identify what fuel it uses, and how much. There is no rocket currently in service that can carry a payload of 40 tons. So let's imagine a Falcon 9 that is 3 times as big, since we are only going for a rough estimate anyhow and I was able to find enough data to go on. That data didn't include the oxidizer to fuel ratio the Merlin engine uses, but we are already pretty abstract here, let's use the 2.56 ratio referenced on Wikipedia for LOX/RP-1.

In that case, it would burn about 400,000 kg of RP-1 to get to LEO, which at a specific energy of roughly 46 MJ/kg is 18,400 GJ, or 19 times the theoretical minimum mentioned above.

  • $\begingroup$ With your calculation about Falcon 9, is it necessary to consider the oxidizer to fuel ratio (2.56)? (en.wikipedia.org/wiki/RP-1) Or is the calculation as straightforward as you did? $\endgroup$
    – Nina K
    Oct 18, 2014 at 20:34
  • $\begingroup$ You are right about the oxidizer - my bad. I'll go fix it. I should mention i took a shot at answering this question because it is at the 'introductory' level i'm at. If you want a sense of what factors are normally considered when calculating fuel, look at 'How much fuel would one need to launch a 1kg object from 100,000 feet?'. $\endgroup$
    – kim holder
    Oct 19, 2014 at 20:16
  • $\begingroup$ For another example of exactly how much more complex this issue is that what is being very roughly estimated here, look at the points raised just when i asked if oxidizer to fuel ratio is based on mass or volume. $\endgroup$
    – kim holder
    Oct 19, 2014 at 23:19

Not sure, if I should "answer" or "comment", but let's put some more questions...

If I have a two stage launch and I still want to stick up with the $9000 \text {km/s} \Delta V$, should I then first calculate the fractions of propellant I need for the two stages?

$$M_f = 1 - (M_1/m_0) = 1 - e^{-\Delta V/V_e}$$

And If the first stage would be until $5000 \text {m/s}$ and second until $9000 \text {m/s}$ with $V_e = 4500 \text {m/s}$, I would get:

$$M_{f1} = 1 - e^{-5000/4500} = 67.1% \text {propellant}$$ $$M_{f2} = 1 - e^{-4000/4500} = 58.9% \text {propellant}$$

With the second one, I need to use the deltaV from $5000$ to $9000 \text {m/s}$ ($=4000 \text {m/s}$), right?

With assumption of $8%$ from the mass being construction, I end up with

First stage: - construction $8%$ - propellant $67.1%$ - Second stage $24.9%$

Second stage: - construction $8%$ ($2.0%$ from initial mass) - Propellant $58.9%$ ($14.7%$ from initial mass) - Final payload $33.1%$ $(8.3%$ initial mass)

So in TOTAL the shares are: - $8.25%$ the final payload - $81.76%$ propellant - $9.99%$ construction

As the payload needs to be $40 \text t$, I end up with:

First stage: - Construction $38.8 \text t$ - Propellant $325.2 \text t$ - Second stage $120.8 \text t$ TOTAL: $484.8 \text t$

  • Second stage:
  • Construction $9.7 \text t$
  • Propellant $71.1 \text t$
  • Final payload $40 \text t$ TOTAL $120.8 \text t$

So far so good (even though I am skipping many details as this is not really my field of science at all, I am trying to keep the focus of only getting the order of magnitude of the energy use...)

Now I am in trouble though... Which energy equation should I use:

$$E = 0.5 \times (m_0-m_1) \times Ve^2$$ ($m_0-m_1$ would be the mass of the propellant)


$$E = 0.5 \times m_1 \times (e^{\Delta V/V_e}-1) \times V_e^2$$

With the first equation I get $3293 \text {GJ}$ and $0.720 \text {GJ}$, so $4013 \text {GJ}$ in TOTAL.

With the second I got confused with m1. In first stage the $m_1$ is the mass of the second stage ($120.8 \text t$)? But with the second stage it must be the mass of the payload and the construction, right...? Or just the payload ($40 \text t$)? Or is this matter of definition, how loose boundaries do I want to set to these calculations? (the payload is ONLY the useful payload ($= 40 \text t$) or also the whole construction of the second stage (= useful payload + engines, empty tanks etc)

Thanks a lot, if there's still more people willing to help in this interdisciplinary research of mine...

  • $\begingroup$ My feeling is that stating energy this way has no real meaning. It may be useful to know what sort of comparison you are trying to make. I'll make some general comments. To estimate propellant mass fraction for a general case, start with the payload and work backwards. The estimate of 8% is a possible number for the % of a stage that is engines and structure - that number can get lower. Unless you are launching a 40t metal slug on that and you don't care what happens to it, you also need to put in a mass for a vessel. The 4500 m/s is possible for LH2 fuel in a vacuum. At sea level it is lower. $\endgroup$
    – kim holder
    Oct 19, 2014 at 20:39
  • $\begingroup$ Using the kinetic energy formula for calculating the energy involved has no meaning here. The energy content of a given quantity of fuel can be calculated by figuring out how much fuel you are going to need and then stating it in terms of its energy density. If all you are looking for is an 'order of magnitude' accuracy, that will get you well within that. But i don't see how that can be useful. $\endgroup$
    – kim holder
    Oct 19, 2014 at 20:47
  • $\begingroup$ This is what happens, when you step out of your own field of science and you try to get get an answer even though the whole questions might be strange... LOL I am studying energy and environmental sciences and am expected to deliver a comparison of sending water from Earth to space and delivering it from an asteroid.... So I chose 40t, because that is the expected amount ARRM would deliver (in theory). This has been quite an task (it's my training thesis topic) without knowing anything in the beginning about rocketry, planetary science, astronomy etc. $\endgroup$
    – Nina K
    Oct 20, 2014 at 7:03
  • $\begingroup$ Water in space is a huge topic. With searches you'd find lots of questions here that deal with it, and you could certainly ask more. If you want to get a general background to tackle this, check the resources listed here. That was me a month ago. $\endgroup$
    – kim holder
    Oct 20, 2014 at 14:36
  • $\begingroup$ Delivery from Earth is probably best modeled based on Falcon 9s. They are currently the cheapest launch out there, and if they manage to make the rocket reusable, that is a very big deal. Modeling the case for water mined from asteroids is a much more complicated issue, making estimates based on delivery from Earth is a piece of cake by comparison. It has definitely been modeled many times before, however. If you want help finding papers on that, people here will know of lots of places. $\endgroup$
    – kim holder
    Oct 20, 2014 at 14:47

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.