3
$\begingroup$

I want to launch the heaviest booster + spaceship possible in about 1.5 of earths gravity, using the technology we currently have. To make things easier, imagine there was the same atmosphere as on earth.

Anyways:

  • Is there a weight limit to rockets, where fuel can't be burnt
    quickly enough to accelerate itself over a long enough period of
    time?
  • Would we be close (considering limited resources and weak material) to such a limit with 1.5 of earths gravity?

My english isn't that great, so I'm not sure if 'limit' is the exact word I mean. Thank you very much for your time.

$\endgroup$
  • $\begingroup$ Have you seen Sea Dragon? $\endgroup$ – Zeiss Ikon Oct 15 '19 at 12:30
  • $\begingroup$ @Zeiss Ikon - Now I have, thanks. There seem to be many advantages... $\endgroup$ – justthisonequestion Oct 15 '19 at 12:39
  • $\begingroup$ Your question is somewhat underspecified (eg. what sort of atmosphere? more like mercury, or more like venus?) so getting a better answer than a list of giant rocket designs is going to be impossible. And a list of giant rocket designs falls a bit outside the remit of this site and risks question closure, so maybe tighten up the question a bit? $\endgroup$ – Starfish Prime Oct 15 '19 at 12:44
  • $\begingroup$ @Starfish Prime - Thank you, sure. Come again in a minute or two. $\endgroup$ – justthisonequestion Oct 15 '19 at 12:48
  • 1
    $\begingroup$ Second question: have you seen Kerbal Space Program? $\endgroup$ – Zeiss Ikon Oct 15 '19 at 13:07
4
$\begingroup$

This is quite a difficult question to answer fully, because rocket science is hard.

There are three things you need to do.

  1. Boost your rocket high enough that it leaves the atmosphere so you can enter a suitable orbit (I'll ignore boosting straight into an escape trajectory for now).
  2. Boost your rocket out quickly enough that you don't spend more time (and hence fuel) fighting against the force of gravity than you really need to. This is called gravity drag.
  3. Boost your rocket out of the thickest part of the atmosphere as quickly as possible so as to avoid waisting fuel pushing through the atmosphere (atmospheric drag), but not so fast that the dynamic pressure your rocket experiences smashes it to bits (Max q).

Part (1) might actually be easier on your super-earth... the scale height of the atmosphere is inversely proportional to the strength of the planetary gravity, so for the same surface temperature the scale height will be smaller as gravity is smooshing the atmosphere into a thinner layer. Surface air pressure will be higher, but the thickest part of the atmosphere will be thinner, making (3) hard to work out.

For a 200km orbit on earth, you need an orbital velocity of about 7.8km/s. Actual rockets need a delta-V budget about 1.5km/s higher than this as a result of those gravity drag and atmospheric drag losses.

Assuming your planet has the same average density to Earth, it'll have a 50% greater radius and hence have about 3.4x Earth's mass. Orbital velocity at 200km is, surprise! about 1.5x the velocity at the same altitude on earth.

Lets be optimistic and assume that the additional atmospheric and gravity drag losses are also 1.5x their terran equivalent. You'll therefore need a delta-V budget of about 14km/s, which is about 4.7km/s bigger than that required on Earth. This is a punishing amount extra. From the rocket equation, $\Delta_v = v_e\log{(m_0/m_f)}$, where $v_e$ is the rocket exhaust velocity, $m_0$ is the fully fuelled and ready to go rocket mass and $m_f$ is the final empty mass of the rocket (or dry mass). For the same rocket technology, you either need to double the amount of fuel in your rocket (without increasing the empty mass of the rocket!), or halve the weight of your rocket and its payload (and still carry the same amount of fuel!). You'll also need 50% more thrust to maintain the same sort of trajectory... anything less than that and your gravity losses mount up rapidly as you'll be spending much longer trying to thrust up using your weedy rockets.

If you could handwave the issue of thrust away, and reduce your problem to one of delta-V, then you could just throw in an extra stage by taking a really, really big rocket design and replacing its intended payload with another rocket.

The Sea Dragon (as suggested by Zeiss Ikon in the OP comments) can lift a substantial 550 tonnes to LEO. The Delta-IV Common Booster Core has a fuelled mass of about 232 tonnes and an empty mass of about 28 tonnes. Given the performance of its rocket engine (full specs here) it could be strapped to a 65 tonne payload and have a delta-V of about 4.7km/s. If you were able to scale that design up linearly, which seems plausible, you'd be able to push about 120 tonnes into orbit using that super-CBC as a third stage using to all the 550kg payload available with the Sea Dragon. Your rocket's launch weight is about 18000 tonnes, giving it a mass ratio of about 1:150.

Handwaving the issue of thrust away isn't really something you'll find you're able to do in the real world. Rocket thrust is defined as $F = \dot{m} v_e$, where $\dot{m}$ is the mass of fuel you're throwing through the engine per second, and $v_e$ is the exhaust velocity. You can trivially increase thrust by adding extra rocket engines... but now you're burning through your fuel much more quickly so you have to carry more fuel, and your dry mass has gone up which reduces your stage's delta-V (because the rocket equation ruins everything) and so on and so forth. Complex staging mechanisms are the solution in Kerbal Space Program, but they're tricky to implement in the real world... SpaceX recently cancelled their fuel-crossfeed project for example, at least in part because the engineering is Quite Hard.

That just leaves you increasing your exhaust velocity. You can't really do that with a chemical rocket, because the liquid hydrogen/liquid oxygen combination is about as good as it gets. If you want an engine that shoots stuff out faster and provides the sort of massive thrust you'll need to efficiently escape your super earth, you're almost certainly going to have to take the Nuclear Option.

Now, nuclear rockets have been built in the past and even operated from static test cells, though none have ever actually lifted off. That puts them at the edge of plausibility for "technology we currently have", but these things aren't vapourware or handwavium. According to Project Rho, the Dumbo Nuclear Thermal Rocket looks like it has the sort of outrageous thrust-to-weight ratio and specific impulse that you need. With a bunch of those, the Sea Dragon/CBC couple become a 3-stage nuclear rocket, littering the world below it with spent fuel rods as it headed for the stars, but it would be able to get there and carry a decent amount of stuff with it.

So there you go. Pure chemical rocket? Tenuous. Nuclear rocket? Probably fine.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Just a nit-picky ilttle detail. From what I read in a fact article in Analog magazine forty-some years ago, Dumbo didn't use fuel rods; the fuel was enclosed between corrugations in a rolled up sheet (think cardboard with one flat surface peeled off). The reactant was the moderator, so it was subcritical until you turned on the pumps for thrust. Shutdown was left as an exercise, as I recall, though it may have run cool enough not to melt down when reactant flow stopped. $\endgroup$ – Zeiss Ikon Oct 15 '19 at 16:55
  • $\begingroup$ @ZeissIkon eh, no-one asked for this to be clean or re-useable. Quite frankly, the people below should be grateful I didn't suggest Orion. $\endgroup$ – Starfish Prime Oct 15 '19 at 19:44
  • $\begingroup$ thanks for all the links $\endgroup$ – justthisonequestion Oct 16 '19 at 6:26
  • $\begingroup$ I like the idea of scratching at the limit of chemical rockets, so I might tune the planets size down a little. I'll probably casually place a high mountain somewhere below the planned escape trajectory, to skip a part of the thickest atmosphere. Is there another obvious factor I'm missing, that might help similarly to the mountain? $\endgroup$ – justthisonequestion Oct 16 '19 at 9:57
  • $\begingroup$ @justthisonequestion There is the reason we always launch east unless there's a good reason not to: the planet's rotation supplies a significant fraction of orbital velocity. On Earth, it's close to 500 m/s out of 7800 needed for orbit. If your planet (a little bigger than Earth) had a 10 hour day, that could be above 1200 m/s out of your 10,000, and make it worth considerable cost, effort, even conflict, to launch from the equator. Even if it's impossible to make a polar orbit... $\endgroup$ – Zeiss Ikon Oct 16 '19 at 11:10
3
$\begingroup$

There are three significant limitations that enter into this question.

First, thrust to weight ratio. This determines how much fuel you can actually lift off the launch pad. If the rocket weighs more than the thrust the engines can deliver, it'll just sit there until it burns off enough propellant to be able to rise. The usual rule of thumb is that thrust to weight should be between 1.5 and 2 at launch -- higher wastes less propellant in gravity losses, but lower is accepted because it gives a little longer for acceleration to build speed.

Second is the size or reliability of individual rocket engines. Today, it's no big deal for SpaceX to launch a Falcon Heavy with 27 engines firing at launch -- sixty years ago, it wasn't a given that a single engine would ignite when you wanted it to (and there were no second tries -- in those days, restartable engines were a dream, not a reality). Bigger engines don't require you to ignite as many at once; lots of smaller ones make the design more tolerant of an individual failure (but increase the likelihood of that individual failure, as there are more "almost reliable" items).

Third is how much gravity you have to overcome, and this comes in two forms: first, surface gravity (you've selected 1.5 G) -- this raises the thrust required to lift off the pad, which increases propellant consumption rate, and so on. It's been written that near this level of gravity (given a rocky planet otherwise similar to Earth) is the point where it will never be possible to escape the world with chemical rockets, because they can't lift enough fuel off the ground to build up enough velocity to get away.

Which brings up the second form of gravity problem: escape velocity. A tiny, compact planet (say, one with Earth's mass but higher density) that has 1.5 G at surface is much easier to launch from than one that's Earth's density but that little big bigger with the same surface gravity. Orbit is always the same fraction of escape, but if escape is a large enough value, you won't get there with a chemical rocket that can lift its own fuel and upper stages.

In the end, (each stage of) your rocket can reach a final velocity that's a log function of the mass fraction. For reference, the Falcon 9 has a mass ratio of around 27 -- that is, the airframe plus propellants mass about 27 times what the empty ("dry") airframe masses. That dry mass includes the wet mass of the second stage and any payload, including its propellant(s). The higher your planet's escape velocity, the larger that ratio has to get, or (what I haven't talked about, because we're near the limit on this for chemical rockets with current Space Shuttle Main Engines and SpaceX Raptor) you have to increase the exhaust velocity (commonly expressed as Specific Impulse) of your engines to get more push out of a given mass of propellant.

I highly recommend Kerbal Space Program and the optional Real Solar System and Realism Overhaul mods, to compare how easy it is to launch to orbit from Kerbin (1G at surface, but 1/10 the diameter of Earth) vs. launching from Earth. There are also ways to change the surface gravity and resize planets, so you can actually model launching from your own fictional world, and see what it takes to get into orbit (and don't forget, once you're in orbit, you're halfway to anywhere in terms of energy).

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Whilst I like your answer, I think you probably want to sprinkle a few references over it in order to meet the recently-arrived hard-science tag. $\endgroup$ – Starfish Prime Oct 15 '19 at 15:16
  • $\begingroup$ @StarfishPrime Honestly, I barely remember how to calculate this stuff. This is what computers and software are for. In KSP, I let Kerbal Engineer or MechJeb handle this kind of calculation. The references to support this (based on how/when/where I learned it) would be bigger than the answer, if I could remember all of them. And not to be too blunt, few people would understand this in "hard science tag" math. I'll try to add a little of it, though. $\endgroup$ – Zeiss Ikon Oct 15 '19 at 15:40
  • $\begingroup$ Eh, just some pointers to some relevant material would do ;-) and you can't just tease with stuff like "if escape is a large enough value, you won't get there with a chemical rocket that can lift its own fuel and upper stages." without hazarding a guess as to what that value is. $\endgroup$ – Starfish Prime Oct 15 '19 at 15:43
  • $\begingroup$ @StarfishPrime I did mention that a world with Earth's density and 1.5 G would be close to that limit. From my reading, the limit is barely above 1.5 G for a rocky planet with roughly Earth's composition, but there are a bunch of variables. If it's got a 10 hour day, you might just make an equatorial orbit with an equatorial launch, but polar orbits are forever barred, for instance. $\endgroup$ – Zeiss Ikon Oct 15 '19 at 15:50
  • $\begingroup$ thanks. very interesting read. I'll try the KSP then... $\endgroup$ – justthisonequestion Oct 16 '19 at 6:25
0
$\begingroup$

Escaping from the Earth is extremely hard. In order to reach orbit ~90% (depending on a range of parameters) of a rockets mass must be propellants leaving just 10% for the rocket engines, rocket structure, avionics (and payload). So a planet with 1.5x the gravitational pull would probably not be practical to escape from as an exponential relationship exists in the rocket equation. Doing some checks:

Escape velocity = SQRT(2GM/r) Rocket equation: Mo/Mf = 1 – e^ (-Change in Velocity/Isp x g) G = Gravitational constant = 6.67x10^-11 m^3/kg/s^2 M = Mass of the planet = 5.972 x10^24 kg (Earth) x1.5 = 8.96 x 10^24 kg (planet x) r = radius of planet x = 6371000 m (same as Earth) Isp = rocket specific Impulse for example SpaceX Raptor Engine 330s nuclear ~800-1000s g = gravitational acceleration = 9.8 (Earth) x1.5 = 14.7m/s^2 planetx Mo/Mf = ratio of rocket mass empty to mass full

Doing the maths on the escape velocity of planet : sqr(2 x 6.67x10^-11 x 8.96 x10^24/6371000) =13.7 km/s. So the change in velocity required to escape is 13.7 km/s

Doing the maths on the rocket equation to get the mass fraction: Mo/Mf = 1 – 2.718 ^ (-13700/350 x 14.7) mass fraction = ~1 so the entirety of the rocket more or less would need to be propellant. No mass left for rocket engines, propellant tanks or payload. And hardly better using nuclear Isp @1000s

Staging would help a bit but not enough. If a rocket engine could be developed with a specific impulse of 10,000 s then it might be doable but as things stand chemical rocket engines are ~ 300-500 s and nuclear engines perhaps 800 - 1000s so we’re out by a factor of 10 even with nuclear rockets.

So it’s not possible with existing technology and would require some very exotic set up to make it work.

https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

https://en.wikipedia.org/wiki/Nuclear_thermal_rocket

https://en.wikipedia.org/wiki/Escape_velocity

|improve this answer|||||
$\endgroup$
  • $\begingroup$ The ratio is actually a lot worse than 90% fuel to 10% rocket -- Space shuttle was 15x as much fuel as vehicle, Falcon 9 is higher than that. $\endgroup$ – Zeiss Ikon Oct 15 '19 at 17:50
  • $\begingroup$ very interesting. thank you $\endgroup$ – justthisonequestion Oct 16 '19 at 6:28
  • $\begingroup$ Total empty mass of Falcon first and second stages = 21.5mt Total propellant mass of Falcon first and second stages = 514.8mt Mass ration >95% with zero payload spacecraft.ssl.umd.edu/academics/483F19/483F19L03.performance/… Remember that both Falcon 9 and the Space Shuttle were designed to be partial reusable so include “wasteful” elements such as wings and grid fins. $\endgroup$ – Slarty Oct 16 '19 at 19:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.