6
$\begingroup$

Suppose I want to send a rocket to Mars, or to the Moon, from earth. How much fuel (in percentage of the full load) is needed for the rocket lift-off as a function of the altitude?

I don't know if it exist an explicit formula, but for the sake of concreteness, let say that I'm interested in the following checkpoints:

  1. Exosphere: 700 to 10,000 km (440 to 6,200 miles)
  2. Thermosphere: 80 to 700 km (50 to 440 miles)
  3. Mesosphere: 50 to 80 km (31 to 50 miles)
  4. Stratosphere: 12 to 50 km (7 to 31 miles)
  5. Troposphere: 0 to 12 km (0 to 7 miles)

I guess that the most difficult one is the first one, since the air is thicker, gravity falls off with the square of the distance, and the weight of the rocket is at his maximum. An answer would be something like:

  1. 30%
  2. 40%
  3. 60%
  4. 80%
  5. 95%

If you know good references on the topic, even books, please insert them.

EDIT: To clarify, here I'm referring to a launch from the surface and to measure at each checkpoint how much fuel I've used in percentage. Such a putative formula should be parametrized by fuel mass, rocket mass, and kinematics (time spent accelerating, acceleration) and might contain some constant taking into account atmospheric factors (mainly air density and temperature I guess) and gravity (such that I could use the same formula starting from another planet but changing these constants).
It's true that I've asked this question in order to roughly figure out how (and if) an air launch can be convenient in some situations.

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ Such function would be possible in estimated form if you supply thrust of the engine and dry mass; in more precise if you add atmospheric factors. Regardless, for your purpose (Mars/Moon) - and most purposes other than suborbital flight - it's completely impractical, as you're taking almost full gravitational drag impact over the whole burn. $\endgroup$
    – SF.
    Commented Feb 6, 2017 at 11:39
  • 2
    $\begingroup$ I think maybe you misunderstand the reason for an air launch? It's not to save energy by putting the rocket higher up; it's to launch the rocket from a less dense atmosphere. $\endgroup$
    – Rikki-Tikki-Tavi
    Commented Feb 6, 2017 at 13:52
  • 4
    $\begingroup$ gravity falls off with the square of the distance - yes, but it's the distance to the center of the planet, so the gravity reduction going from the surface to LEO is tiny. $\endgroup$
    – Hobbes
    Commented Feb 6, 2017 at 14:36
  • 1
    $\begingroup$ Gravity loss is generally a much larger factor than drag. Drag loss is dependent on the rocket's size/shape as well as its speed and is proportionally less for larger rockets. Saturn V for example loses ~48 m/s to drag and about ~1743 m/s to gravity. $\endgroup$
    – Russell Borogove
    Commented Feb 6, 2017 at 15:46
  • 2
    $\begingroup$ We could use the tank volumes of the stages of an existing rocket and the flow rates to the rocket engines to calculate how much of the tank volume is used for the different heights. Of course there is some error when using this calculation, but for a stage with 10 % structural mass and 90 % of propellant mass, the error is small, at least when there is 50 % of propellants or more left in the tanks. $\endgroup$
    – Uwe
    Commented Feb 6, 2017 at 20:14

2 Answers 2

Reset to default
3
$\begingroup$

We need a fair number of approximations here...

Let's ignore atmospheric losses, since according to Russell Borogov's comment these are dwarfed by gravity losses. Let's also assume that the rocket instantly gets to orbital altitude and velocity, so that there is no influence of the trajectory on how efficient we can be (more on that later).

The difference is the potential energy being higher due to the full rocket being placed at a higher altitude. Let's calculate it with Saturn V's payload weight to LEO: 140 tonnes (to be independent of fuel mass, which we compute later). Let's also assume gravity is constant up to LEO to keep this simple.

  • 0: We set this potential energy origin to sea level: 0J.
  • 1: 10,000 km high: 13,720 TJ (TeraJoule). This is above LEO, so we can't use Saturn V's LEO payload for this though. Anyone got some high Earth orbit payload launcher data?

Note that LEO is ~2,000km: 2,74 TJ

  • 2: 700km: 960,4 GJ
  • 3: 80km: 109,76 GJ
  • 4: 50km: 68,6 GJ
  • 5: 12km: 16,4 GJ

How much fuel is that? Let's compute the total energy of the fuel from the rocket's total dV to LEO. To get there, we burn the entire first and second stages, and 2.5 minutes of the third stage (the rest being used for translunar injection). That is in total 2.58+4.13+1,47 = 8,18km/s delta-V expended (LEO orbital speed being 7.8km/s).

How much kinetic energy is that delta-V ? That is 4,69 TJ. Note that this is 171% of LEO potential energy, since orbiting is not just about the altitude, but also about the speed, to avoid falling down (see also: Vis-viva equation)

Now, we can substract the potential energy gained from launching at a higher altitude from that kinetic in order to approximate the required delta-V to get to orbit from a given altitude:

Energy/deltaV still required:

  • 0: 4,69 TJ, ~8,18km/s i.e. 0 m/s dV saved
  • 2: 3,73 TJ, ~7,30km/s i.e. 880 m/s dV saved
  • 3: 4,58 TJ, ~8,09km/s i.e. 90 m/s dV saved
  • 4: 4,62 TJ, ~8,12km/s i.e. 60 m/s dV saved
  • 5: 4,67 TJ, ~8,17km/s i.e. 10 m/s dV saved

Then, using the rocket equation, we can compute how much fuel is needed, and give the result as a ratio of the original fuel weight.

To simplify, we'll assume the first stage is entirely expended (makes calculations harder since its specific impulse is lower than the two next stages) and substract its 2.58km/s deltaV.

  • 0: 5.6km/s to LEO in second + third stages, 2160+456+39=~2660T of fuel in all 3 stages, 456+39=~500T expended fuel from 2nd+3rdstage
  • 2: 4,72km/s @421s ISP -> 440T wet mass -> 300T fuel -> 200T saved
  • 3: 5,51km/s @421s ISP -> 523T wet mass -> 392T fuel -> 108T saved
  • 4: 5,54km/s @421s ISP -> 536T wet mass -> 396T fuel -> 104T saved
  • 5: 5,55km/s @421s ISP -> 537T wet mass -> 397T fuel -> 103T saved

As a ratio to the total 2660T fuel mass:

  • 0: 0km: 100,0% fuel mass
  • 2: 700km: 93%
  • 3: 80km: 96%
  • 4: 50km: 96%
  • 5: 12km: 96%

Notice that 3-5 should actually be closer to 99%, I think I got the fuel mass wrong on the last stage since I forgot the mass of the lunar modules.

Let's conclude with saying that the fuel savings as a ratio only get significant when you launch from an altitude that gets very close to LEO -- since your energy is expended both to get to orbital speed and orbital altitude.

Notes:

Improve this answer
$\endgroup$
5
  • $\begingroup$ It is an interesting answer. Usually energy-based do not work so well. Thrust is all about momentum. I have a hunch this won't lead to useful results. $\endgroup$
    – uhoh
    Commented Feb 15, 2017 at 13:01
  • $\begingroup$ We should be able to get good approximations, but not exact results of course. If anything, these numbers would be a higher bound on how much fuel we can expect to save. $\endgroup$
    – Florian Castellane
    Commented Feb 15, 2017 at 13:03
  • $\begingroup$ Honestly, unless you can do this with relative delta-v, it will be hard to defend. It doesn't really help to say that you believe it is a good approximation. I've never seen this done successfully using energy. $\endgroup$
    – uhoh
    Commented Feb 15, 2017 at 13:43
  • 1
    $\begingroup$ The savings are not so high, and I am also using numbers from Saturn V instead of using an energy-based approach for everything, that is why I would expect reasonable results. $\endgroup$
    – Florian Castellane
    Commented Feb 15, 2017 at 14:03
  • 2
    $\begingroup$ Thank you for your answer, but I can't grant the bounty here. As I expected trying to use energy does not really work for launch to orbit problems. Look at a launch on YouTube where they show the altitude - the first 20% of the propellant gives nearly zero potential energy, and if you think about it, trying to add up kinetic energy doesn't work because the 90% of the mass disappears during the launch. I appreciate the effort but this answer is really no help here. $\endgroup$
    – uhoh
    Commented Feb 16, 2017 at 14:52
1
$\begingroup$

It should be possible to extract the fuel by using this formula for the $\Delta V$ of a rocket's lift off:

$$\Delta V = g I_{sp} \ln(m_i/m_f) - \int_{t_i}^{t_f}dt (g \sin(\gamma) +D/m)$$

where the contributions are respectively Tsiolkovsky rocket equation, gravity loss and drag loss. See Reference

The idea is to look at the position $(x(t),y(t),z(t))$ of a known rocket launch. From there it should be possible to extract the angle $\gamma$ as a function of time and the $\Delta V$ at any given time interval. Moreover we can estimate the drag force $D = \tfrac12\, \rho\, V^2\, C_D\, A$ at any time (knowing the drag coefficent C_D and the shape of the rocket. As pointed out, $\rho = \rho(y)$ ).

Then we can numerically solve for the remaining wet mass $m_f$ mass at some given $y(t)$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Drag force depends on altitude (thinner atmosphere as you gain height), so you'll need to expand D into a formula that produces D from some known terms and has height as a variable. $\endgroup$
    – Hobbes
    Commented Dec 31, 2017 at 6:51
  • $\begingroup$ Correct, we need density of the atmosphere as a function of the altitude. It's fair to say that the whole thing can only be done numerically. $\endgroup$
    – Rexcirus
    Commented Dec 31, 2017 at 12:14

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.