This is a very stupid hypothetical - sorry!

Would the Space Shuttle Orbiter plus external tank have been able to leave the pad just using the SSMEs (i.e. without the solid rocket boosters ~80% of its lift-off thrust).

If so (and I'm guess it couldn't?) how fast/far could this configuration (basically the second stage) go on it's own before the ET ran dry?

With this stupid question I'm really trying to get some idea of the relationship between the power needed on the first and second stages and the contribution of the SSMEs.


2 Answers 2


The reason for the SRBs is that the stack weighs something 4.4 million lbs.

The two SRBs provide about 2.8 million lbs of thrust each. The three SSMEs provide about 600Klbs each.

If you subtract 5.6 million lbs of thrust, and the SRBs weigh 1.3 million lbs each, then the remaining stack, a little over 1.8 million lbs only has 1.18 million lbs of thrust.

So it will sit exactly where it is, burning off fuel, destroying the launch pad, until enough LOX/LH2 is burned off by the SSMEs to get the total mass of the stack down to about 1.18 million lbs, then it will start moving. But then there is not enough impulse to make it to orbit.

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    $\begingroup$ It's never going to lift off--the rockets will quickly burn through the cooling water (it's a dump of tanks, not a system that can maintain the flow) and the reflected energy is going to do very bad things to the orbiter. $\endgroup$ Commented Nov 9, 2014 at 4:24
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    $\begingroup$ @LorenPechtel That is why I said it would destroy the launch pad. No one really knows what would happen, but disaster is the most likely outcome. $\endgroup$
    – geoffc
    Commented Nov 9, 2014 at 14:44
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    $\begingroup$ @geoffc The question was about the orbiter--while I agree it would also destroy the pad that's not what he was asking. $\endgroup$ Commented Nov 9, 2014 at 20:53
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    $\begingroup$ @geoffc Wikipedia puts the orbiter at 68,585 kg empty, and the Super Lightweight External Tank at 26,500 kg empty. At 5.25 MN (535,000 kgf) thrust at ground level, with those numbers, you have about 5.6x the required liftoff thrust. I therefore suspect that it would be possible to pick a meaningful fuel load point where thrust equals mass and gets the orbiter moving without the SRBs, but I also agree that it likely would be an academic exercise. $\endgroup$
    – user
    Commented Dec 12, 2016 at 20:32
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    $\begingroup$ Something doesn't quite add up. Liftoff mass = 7.4M lbs, liftoff thrust = 5.6M+1.8M = 7.4M lbs? That would yield a net acceleration of zero. See the Wikipedia page about the Space Shuttle. They quote system mass of 4,400,000 lb and thrust of 6,780,000 lbf, so without the SRBs, thrust would be about equal to weight at around 1.8M lbs. $\endgroup$
    – Anthony X
    Commented Jan 7, 2017 at 3:34


Without SRBs, by maintaining the real stack's thrust to weight ratio at ignition, you run out of fuel at about 167 seconds with (if we maintain the flight profile of the real stack) a velocity around 1.4 km/s and an altitude of around 70 km.

Without SRBs, with an initial thrust to weight ratio of 1, you run out of fuel at about 289 seconds with (if we maintain the flight profile of the real stack) a velocity around 2.4 km/s and an altitude of around 100 km.

To attain a stable, low Earth orbit, you need around 7.4 km/s at approximately 110 km altitude.

A few data points

  • The Orbiter (specifically OV-105) had an empty weight of 68,585 kg (68.6 Mg) or 110,000 kg (110 Mg) at liftoff
  • The Super Lightweight External Tank had an empty weight of 26,500 kg (26.5 Mg) or 756,000 kg (756 Mg) when fully fueled
  • The Orbiter plus the Super Lightweight External Tank thus have a combined empty weight of 95.1 Mg.
  • The Orbiter (again OV-105) three main engines provided 1,752 kN thrust each (when running at 104% at sea level) for a total of 5,255 kN or 536 Mgf on SSMEs only
  • The two solid rocket boosters (SRBs) massed 571 Mg each at liftoff, and contributed 12,500 kN or 1,274 Mgf of thrust each

Common assumptions

Given that the SSMEs burned through 629,340 kg of LOX and 106,261 kg of LH2 in a 480 seconds normal ascent burn, the burn rate is about 1.3 Mg/second LOX and 0.22 Mg/second LH2. We also see that the LOX to LH2 mass flow ratio is about 5.92:1 as $\frac{629\,340}{106\,261} \approx \frac{5.92}{1}$.

For simplicity, I'm not allowing for any payload whatsoever. In this scenario, if you want to carry a payload, doing so will necessarily reduce the amount of fuel you can bring, because you are basically transferring liftoff mass from fuel (which you need to get off the ground) to, as far as propulsion is concerned, dead weight.

Realistic case: thrust to weight ratio at ignition same as a normal Space Shuttle including SRBs

For a normal Space Shuttle, including external tank and SRBs, the total thrust provided by all five engines (three main engines plus two SRBs) is $3 \times 1\,752 + 2 \times 12\,500 = 30\,255$ kN or about 3,085 Mgf at liftoff. At the same time, the total mass of the fully equipped spacecraft is $110 + 756 + 2 \times 571 = 2\,008$ Mg. This gives a thrust to weight ratio (TWR) of $\frac{3\,085}{2\,008} \approx 1.536$ at the moment of ignition.

Because absent the SRBs we have 536 Mgf of thrust, we can allow for $\frac{536}{1.536} \approx 349$ Mg of gross liftoff weight while maintaining the thrust to weight ratio. Subtracting the empty weight of the Orbiter and the SLET leaves about 254 Mg that we can use for fuel.

Keeping the 5.92:1 LOX/LH2 ratio, we can load up about 36.7 Mg of LH2 and 217 Mg of LOX.

By burning 1.3 Mg/second LOX and 0.22 Mg/second LH2, this fuel load lasts $\frac{217}{1.3} \approx \frac{36.7}{0.22} \approx 167$ seconds. Consequently, we run out of gas in about 167 seconds.

Absolute best theoretical case: thrust to weight ratio = 1 at ignition

Since the orbiter plus external tank with no payload whatsoever weighs 95 Mg, and you have 536 Mgf of thrust, this leaves you with about 440 Mg that you can use for fuel and still get the spacecraft to move at all.

Keeping the LOX/LH2 mass flow ratio the same, in 440 Mg we can fill the ET with some 376 Mg LOX and 63.6 Mg LH2 ($376 \approx 63.6 \times 5.92$). Assuming that the mass flow rate is unchanged with this lower initial mass, this fuel load will be depleted after 289 seconds, give or take a fraction of a second or so.

Note that this configuration will require very different launch facilities to keep the spacecraft from being destroyed by reflected energy before it has time to climb sufficiently that this is not a major problem. But since the Space Shuttle absent the SRBs is already a quite different spacecraft from what we had, I think it's safe to say that we can also adjust the launch facilities to accomodate the different spacecraft design.

Where does that leave us?

I couldn't find any nice official NASA graph for altitude versus time for the Space Shuttle (if anyone knows of one, please comment!), but velocity versus time was easier.

This xkcd forums post by davidstarlingm references the NASA STS-30 press kit, which among a few other notable points in the trajectory puts "Negative Return" at 238 seconds with a velocity of 6,915 ft/s which is a hair under 2,108 m/s and an altitude of about 319,000 feet or 97 km. While I doubt the real curve is this nice (taking into account for example the engine dethrottling late in the ascent to maintain a maximum 3 G acceleration should result in a more S-shaped curve), looking at the plot of the velocity versus time data and the function describing the data points available:

y = 0.0002x^3 - 0.0753x^2 + 38.683x - 362.56

puts the Space Shuttle velocity around the 290 second mark at somewhere around 8,000 ft/s or 2,400 m/s. That's the theoretical best case, where we started out with a TWR = 1. As a quick approximation, we actually got to space today in terms of altitude; the Kármán line is at 100 km altitude, and we at least passed 97 km altitude some time before our engines shut down due to fuel exhaustion.

In the more realistic case maintaining the original stack's TWR = 1.536 at T-0, at 167 seconds the spacecraft's velocity is around 5,000 ft/s or 1,500 m/s. At the same time you are about halfway between the real spacecraft's SRB staging at 125 seconds (153,405 ft altitude in the case of STS-30) and negative return at 238 seconds (STS-30: 319,008 ft altitude), so let's split the difference and call it 236,200 ft or 72 km. In this case, we are nowhere near space.

Compare these to the real STS-30 trajectory, which was attained with main engines cutoff (MECO) at 511 seconds (so actually a somewhat longer burn) and 24,286 ft/s (7.4 km/s) velocity at about 362,000 ft or 110 km altitude.

No matter how you slice it, from 1.5 km/s or 2.4 km/s to 7.4 km/s is quite a long way to go, and especially in the case of TWR = 1.536, we also still have a fair distance to climb.

As a consequence, at these points the Space Shuttle Orbiter is an unguided, unpowered ballistic projectile on its way back toward the ground.

  • $\begingroup$ You can get a few ascent traj data points from the Ascent ADI Cue Card included in the Ascent Checklist - found here nasa.gov/centers/johnson/pdf/567068main_ASC_135_F_1.pdf There is a tiny table there of Time, Theta, Altitude, and Altitude rate. PDF page 193. $\endgroup$ Commented Jan 6, 2017 at 23:46
  • $\begingroup$ @OrganicMarble Thank you! I will have a look at it later and maybe incorporate into the answer as appropriate. $\endgroup$
    – user
    Commented Jan 7, 2017 at 6:01
  • $\begingroup$ No. You're assuming the same flight profile--but your rocket starts out with an acceleration of zero at the moment of ignition. Besides, such a slow liftoff would result in it's destruction from reflected energy. $\endgroup$ Commented Jan 7, 2017 at 20:51
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    $\begingroup$ @Sean Fair point regarding fuel starvation/exhaustion. $\endgroup$
    – user
    Commented May 31, 2018 at 20:21
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    $\begingroup$ @Sean "How much additional velocity and\or altitude would we be able to get by firing the OMS engines after running the ET dry?" Probably not much. IIRC the OMS offered a few hundred ft/s worth of delta-v. Even if that recollection of mine is off by an order of magnitude, it's still too little to matter. $\endgroup$
    – user
    Commented May 31, 2018 at 20:22

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