The Wikipedia had the spectrum of SpaceX Raptor's specific impulse to be 330s at sea level and 380 in vacuum. Using an updated effective exhaust velocity at sea level to = 3280 m/s and in the vacuum to be = 3750 m/s.

The thing was, with the specification from SpaceX's Website or Wikipedia, the amount of fuel in Super Heavy and Starship could not sustain enough energy to escape the earth's gravitational potential.

Without accounting air resistance, which reduces the energy, but account for the reduced air pressure, since it had positive contribution to the launch with respect to the increased effective exhaust velocity, the final kinetic energy of Starship + gravitational potential of Starship =(1.15918e+12 -6.97824e+12)J=-5.81906e12J

This has nothing to do with the amount of the raptors used during the launch. The simulation used 36 Raptors for Super Heavy as indicated from the website(72/2), and another 7 for Starship(since 6 could not sustain a circulation speed in the final attitude).

In fact, even with Rocketdyne RS-25's effective exhaust velocity through out the launch (4400 m/s), and without the air resistance, the Starship lacked an amount of energy (9.57629e+12 -1.07148e+13)J=-1.13851e12J

What went wrong with the spectrum of the Starship and Super Heavy? What kind of specific impulse did they actually want?

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    $\begingroup$ Do you mean "specification" for "spectrum" here? $\endgroup$ – Russell Borogove Mar 4 at 22:51
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    $\begingroup$ I'm not completely certain what you're asking here, but Starship was never expected to be a useful SSTO. Starship + Super Heavy Booster is a two-stage to orbit launcher, like many others. $\endgroup$ – Russell Borogove Mar 4 at 22:52
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    $\begingroup$ Several ideas. (1) Your simulation is wrong. A naive application of the ideal rocket equation suggests a delta V of 12.7 km/s using the numbers on the linked wikipedia page, more than enough to get a good sized payload to LEO. (2) The numbers on the wikipedia page may well be wrong. This would not be the first time. (3) You misread the 13.5 seconds. That is the delay between main engine cutoff and second stage ignition for a very specific engine. The main engine burnt for over four minutes. (4) You used the parameters for the wrong vehicle. Starship ≠ Delta II. $\endgroup$ – David Hammen Mar 4 at 23:35
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    $\begingroup$ The word "spectrum" does not exist in the link you provided. I'm baffled by what you mean and your comments make it even more confusing. $\endgroup$ – Organic Marble Mar 4 at 23:37
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    $\begingroup$ @uhoh Without details on the OP's calculations, I do not think this question is rescuable. $\endgroup$ – David Hammen Mar 5 at 0:14

Calculations of launches based on kinetic energy generally fail since you need a specific frame of reference in which to do them, and the rockets are continually changing speed. You can't apply kinetic energy to these problems very easily.

However there is a straightforward way to do this calculation using momentum.

The Wikipedia article linked in the question SpaceX Starship gives the following:

First Stage
Gross mass         3,580,000 kg
Propellant mass    3,400,000 kg 
exhaust velocity   3.2 km/s

Second Stage
Gross mass         1,320,000 kg
Propellant mass    1,200,000 kg
exhaust velocity   3.7 km/s

For the first stage flight, the initial and final masses $m_o$ and $m_f$ are 4,900,000 and 1,500,000 kg. Applying the Tsiolkovsky rocket equation

$$\Delta v = v_E \ln\frac{m_o}{m_f}$$

gives 3.2 km/s times 1.18 or 3.79 km/s.

For the second stage flight the initial and final masses $m_o$ and $m_f$ for a zero mass payload scenario are now 1,320,000 and 120,000 kg. Applying Tsiolkovsky again gives an additional 8.87 km/s

The total delta-v for a zero payload launch is then 12.84 km/s.

Ignoring atmospheric drag and other issues, Earth's minimum escape velocity (required to "escape earth gravitational potential") is

$$v_{esc} = \sqrt{\frac{2GM}{r}}$$

where Earth's standard gravitational parameter $GM$ is 3.986e+05 km^3/s^2 and it's equatorial radius $r$ is 6378 km. That gives 11.17 km/s as a lower theoretical minimum, but realistic launches typically will add 1 to 1.5 km/s to cover gravity loss and losses due to lowering power near max-Q minus the benefits of launching prograde and using Earth's ~0.4 km/s rotational speed at low to mid latitudes.

That means escape would need 12.2 to 12.8 km/s which is coincidentally what the ship can do.

This means that an empty 2nd stage cold possibly just barely escape Earth and end up in an orbit around the Sun in the ecliptic at circa 1 AU, and from time to time drift past the Earth.

But wait, there's more!

As we can read about in discussion of Starship's predecessor design in Can five refillings of the BFR second stage be useful to get to the Moon? To Mars? All five in Earth orbit? a launch of a second stage will be to low Earth orbit only. The plan is to use a additional launches to put second stages full of fuel (i.e. "tankers") in Earth orbits so that the first stage can fill up again and burn again, perhaps several times, in order to escape Earth's gravitational potential with a full payload.

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    $\begingroup$ @ShoutOutAndCalculate yes it seems so; "The system was not designed to escape the earth gravity field" seems basically correct. It's designed to bring a large payload to Earth orbit, and then for those payloads destined for deep space to then undergo one or more refuelings before leaving Earth completely. $\endgroup$ – uhoh Mar 6 at 0:01
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    $\begingroup$ @ShoutOutAndCalculate as far as I can determine from your incomplete description of your calculations, you are assuming a straight vertical ascent. Gravity losses for such a trajectory are far worse than for any realistic trajectory to orbit or escape. $\endgroup$ – Christopher James Huff Mar 6 at 13:35
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    $\begingroup$ @ShoutOutAndCalculate there is no limit to the number of questions you can ask (I've asked over 2,000 in this site alone). If you post that in the form of a new question it provides much more room for responses in the form of new answers. $\endgroup$ – uhoh Mar 7 at 3:36
  • $\begingroup$ @ChristopherJamesHuff (Calculation updates): You were correct, using "go up to low air density and start to reach for horizontal velocity to counter act gravity drag", with adjustment start at 30km and to horizontal in 30s(not changing during the 13.5 s second stage ignition), the empty payload were able to reach a speed of 11265 m/s at attitude 93036 m with kinetic energy +potential energy =(7.62508e+12 -7.39963e+12)J escaped gravity field.(barely and unaccounted for air resistance, a refuel could to go for other missions.) Thank you for the hints. $\endgroup$ – ShoutOutAndCalculate Mar 7 at 10:23

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