# How much fuel would one need to launch a 1kg object from 100,000 feet?

Back in the old days of the cold war the US and Russia were trying to develop a Nuclear powered aircraft. Problems with radiation shielding and "safety" held them back.

Let's say that they got this technology to work and developed a large flying fortress type aircraft capable of flying at 600mph at 100,000 feet. Now if you wanted to launch a space payload off that platform (weighing 1kg) how would you adjust the launch equations to calculate how much cryogenic fuel you will need? Also, taking into consideration that the atmosphere up there is very cold and there is very little air pressure. This would mean that the fuel would stay colder longer and need lighter tanks. Also since the pressure is reduced the nozzle velocity of the propellent would be higher at those altitudes.

Edit: Similar to the Pegasus, in the fact that it is launched from 40,000 feet from a carrier aircraft. Inclination: 51.6, Apogee: 421Km, Perigee: 431 (basically the ISS' orbit), launched eastward.

Expanding further: I want to create a model in c# that will calculate and plot the amount of fuel required to launch 1kg into orbit from several altitudes and speeds. Over time I want to add in factors like drag etc, to determine what the performance of the carrier craft should be.

• You mean something like the Pegasus? Jul 19, 2013 at 7:16
• @RodyOldenhuis, yes like that, what are it's launch calculations for the payload taking into account the 100K feet altitude (as opposed to the 40K feet).
– user39
Jul 19, 2013 at 8:07
• Your question is non-determinate. We need to know what type of orbit (inclination, apogee, perigee) and what type of propulsion system (Isp, propellent, oxidizer) at a minimum. In addition to that, can we stage? What latitude is the launch site at? Can we launch due east? Your reference to Virgen Galactic confuses me because SS1 (and SS2) are suborbital craft.
– Erik
Jul 20, 2013 at 5:58

Since, as asked, the question isn't answerable, I'll discuss the elements which affect how one finds the answer.

The exact amount of propellant is going to vary depending upon what propellant, what the efficiency of the stage is, what the current speed is while crossing 100,000', the structural mass needed for that fuel, the desired course (Orbital type and/or escape), and the current weather.

Note that at 100K feet, the only real effect of the weather is going to be mild changes in density; this is probably one of the smallest effects.

The specific fuels will make tremendous differences; which fuel determines how much the casing mass is going to be as well as the fuel mass to accelerate the casing and the payload. H2/O2 is high efficiency, but loses due to increased casing mass. Aluminium-Rubber is significantly lower energy, but also much more dense, and thus has a much lower casing mass. (Casing mass is everything that isn't fuel nor payload.)

The altitude alone reduces the density of air and the local gravity.

Local gravity effect at 100,000': approximate earth equitorial diameter 41850758', we get a radius of 20925379'; adding 100000 we get 21025379'. Applying the inverse square law, 1/((21025379/20925379)^2), we get about 0.9905G. This means that the payload will need less fuel, as well.

Atmospheric density at 100,000' the air pressure averages 10millibars; "standard" is 1 bar. This will vary slightly by temperature, but not nearly as much as near-surface.

To get one extreme limit, let's consider the Centaur upper stage, one of the most fuel-efficient chemical rockets there is and let's hypothetically say we can scale it down while preserving the proportion of fuel/oxidiser to dry mass (about 10:1). We will also (for the sake of establishing a lower bound) ignore air resistance and gravity losses.

It's specific impulse is about 450s, so the rocket equation tells us that to reach orbital velocity of 7800 m/s we start at the equator, travelling East at 250 m/s relative to the atmosphere for a total initial velocity of 750 m/s. The rocket equation tells us that $$\Delta v=v_{\text{e}}\ln {\frac {m_{0}}{m_{f}}}$$

$v_e$ is $450g$ ie about 4500 m/s, so we can solve and get $m_0/m_f \approx 4.7$ Now $m_f = 1 + 0.1 * (m_0-m_f)$ (since we need a 1kg payload plus 1/10 of the fuel for tankage and engines. So we solve and get $m_f \approx 1.5kg$ and so $m_0 \approx 7.5 kg$.

So, under these extremely optimistic assumptions, you would need about 6kg of propellant (plus 0.5kg of tanks and rocket) to launch your 1kg payload.

Elon Musk said they considered it, and found that the fuel saving from launching from high altitude would be about 5%, offset by needing a wing to do an aerial turning maneuver.

At altitudes over 20kilometers, the air is already so thin that drag is negligible compared to gravity and inertia.

It's important to realize that orbit is not about being high, but being fast. You need 8k m/s velocity to get into orbit. Starting off with a speed of 250 m/s is not going to help much with this. The main benefit is the thinner air.

To write a program to analyze this, you just take small time deltas (say .001 seconds), and in a for loop, repeatedly update:

1. Fuel burned (remove weight)
2. Change in velocity (thrust from fuel / new weight)
3. Slow down from drag (drag equation--you can find coefficients for various spacecraft online as well as atmospheric density at different altitudes)
4. Change in height (new velocity * time delta)

If your rocket also has stages, you will need to remove the appropriate weight when certain fuel milestones are reached to account for stages falling off. Keep repeating that cycle until you run out of fuel (and plummet back to your earthly grave), or reach 8000 m/s (hurray, orbit). Throw the whole thing inside another for loop, and you can have it inc

• Your last sentence seems to trail off in mid-word. Apr 25, 2018 at 17:52